2021/116) If the product of two numbers is 2,400 and their LCM is 96, then what is their HCF?

The question is ill formed because

Product = 2400

LCM = 96

So HCF = $\frac{product}{LCM} = \frac{2400}{96}=25$

But LCM has to be a multiple of HCF . ths is so because LCM is a multiple of each of the numbers and each number is multiple of HCF

As 25 does not divide 96. so the problem is incorrect, or this case is not possible 

2021/115) What are the maximum and minimum values of $3x+4y$ on the circle $x^2+y^2=1$

as $x^2 + y^ 2 = 1$ we can choose $x = \sin\, t$, $y = \cos\, t$

$3x + 4 y= 3 \sin\, t + 4 \cos\, t$

to convert $3 x + 4y = 3 \sin\,t + 4 \cos\, t$ to the form $A \sin (x+ t)$

$A \sin (x+t) = A \sin\,t \cos\, x + A \cos\, t \sin\, x$

we can choose $3 = 5 \cos\, x$ and $4 = 5 \sin x$  (as $3^2 + 4^2 = 25 = 5^2$

= $5 \cos\, x \sin\, t + 5 \cos\, t \sin\, x = 5 \sin (x-t)$

it is maximum when $\sin (x-t) = 1$ and maximum value = 5

minumum when $\sin (x-t) = -1$ and minimum value = 5

2021/114 solve in integers $3x^2 + 5y^2 = 345$

working in mod 3 we have $5y^2 = 0 \pmod 3$ or $y =0 \pmod 3$

so y = 3a for some a

similarly x = 5 b for som b

so ge get $75 b^2 + 45 y^2 = 345$ 

deviding by 15 we get $5b^2 + 3a^2 = 23$

we need to check for $5b^2 < 23$ or $b <=2$

putting b = 1 we get  $3a^2 = 18$ or $a^2 = 6$ not an integer

b = 2 gives $3a^3 =3$ or a = 1

so we have a= 1 , b= 2 giving x = 10 and y = 3 

2021/113) prove the following identity: $\binom{n}{k}=\binom{n-2}{k}+2\binom{n-2}{k-1}+\binom{n-2}{k-2}$

We can solve the same in 2 ways. Combinotorics way or alegraic ways

We present here to solve in combinonorics way

From n objects we can choose k objects in   $\binom{n}{k}$ ways

let us group the n objects into (n-2,1,1) ways

for choosing k objects this can be done in 3 ways

k  objects from n-2 objects that is from 1st set 0 from 2nd set and 0 from3rd set in $\binom{n-2}{k}$ ways

k-1  objects from n-2 objects that is from 1st set 1 from 2nd set or  1 from 3rd set in $2 * \binom{n-2}{k-1}$ ways

k-2  objects from n-2 objects that is from 1st set 1 from 2nd set and  1 from 3rd set in $\binom{n-2}{k-2}$ ways

as all above 3 are mutually exclusive so no of ways  =$\binom{n-2}{k} + 2 * \binom{n-2}{k-1} + \binom{n-2}{k-2 }$

in  2 ways we have computed the number of choosing k objects from n obects so they must be same or

$\binom{n}{k} = \binom{n-2}{k} + 2 * \binom{n-2}{k-1} + \binom{n-2}{k-2 }$

2021/112) Let G= $\{a+bi$ in complex: $a^2 +b^2=1\}$. Is G a group under multiplication?

For it to be group folllowing must be tue.

1) it should be closed

that is if  x = a + ib and y = c + id and $(a^2+b^2) = 1$ and $c^2+d^2=1$ 

and xy = m + ni then $m^2+n^2 =1$

we have $xy = m + ni = (a+ib)(c+id) = (ac - bd) + (bc + ad)i$

we have m = ac - bd and n = bc + ad

$m^2 + n^2 = (ac-bd)^2 + (bc + ad)^2 = a^2c^2 - 2abcbd + b^2d^2 + b^2 c^2 + 2abcd + a^2d^2$

 $= a^2c^2 + b^2d^2 + b^2c^2 + a^2d^2 = (a^2+b^2)(c^2 + d^2) = 1$ 

So it is closed 

2) It should have an identity

1 or 1+0i is identity element as $(a+bi)(1+0i) = a+ bi$

3) it should have an inverse  

because $a^2+b^2=1$ so it it not zero and hence it has inverse and we need to show that if

m+in is inverse then $m^2+ n^2 =1$ that is the inverse is in this group

$m + in = \frac{1}{a+ib} = \frac{a-ib}{(a+ib)(a=ib)} = \frac{a-ib}{a^2+b^2} = a - ib$

'so m = a , n = - b and $m^2 + n^2 = a^2 + (-b)^2 = a^2 + b^2 =1$

so it has an inverse

4) assosiativity law holds as unde rcomplex number multiplication assosiativity holds 

  

2021/111) For which primes p, 7p+4 is a perfect square?

 7p + 4 is a perfect square say $m^2$

so $7p = m^2 -4 = (m+2)(m-2)$

now there are 2 cases

1. p is 2 which gives 7p + 4 = 18 which is not a perfect square

so

2. p is odd

so 7p is odd and  $m^2 -4$ is odd

so m+2 and m-2 are co-primes as they differ by 4

so m+2 = 7, m-2 = p gives p = 3 which is prime and m = 5

or m+2 = p and m-2 = 7 giving m = 9 and p =11 which is a prime

so p = 3 or 11 

2021/110) express $\cos\, 5t$ in term of power of $\cos\, t$

We have

$\cos\, 5t = (\cos\, 5 t + \cos\, t) - \cos\, t$

$= 2 \cos\, 3t \cos\, 2t - \cos\, t$ using $\cos\, A + \cos\, B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$

$= 2 * (4 \cos ^3 t - 3 \cos\, t)(2\cos ^2 t - 1) - \cos\, t$ using formula for $cos 3t$ and $cos 2t$

$= 16 \cos^5 t - 20 \cos^3 t + 6 \cos t -\cos t$

$= 16 \cos^5 t - 20 \cos^3 t + 5 \cos t$

2021/109) Find the sum of the series $\sum_{n=1}^{\infty}\frac{1}{n^2}$

We have tailor  expansion of $\sin\, x$  as  

$P(x) = \sin\,x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots  (1) $

This is a polynomial of degree infinite with zeroes at  0 and npi so

this is  $Ax(1- \frac{x}{\pi})(1+ \frac{x}{pi})(1-\frac{x}{2pi})(1-\frac{x}{2\pi})\cdots$

or $P(x) =Ax(1- \frac{x^2}{\pi^2})(1- \frac{x^2}{2^pi}^2)(1-\frac{x^2}{3^2pi^2})\cdots$

comparing above with (1) we get A = 1

So $P(x) =x(1- \frac{x^2}{\pi^2})(1- \frac{x^2}{2^pi}^2)(1-\frac{x^2}{3^2pi^2})\cdots$

The coefficient of $x^3$ is    $- \sum_{n=1}^{\infty}\frac{1}{n^2\pi^2}$

from (1)  coefficient of $x^3$ is $-\frac{1}{6}$

as both are same so  $- \sum_{n=1}^{\infty}\frac{1}{n^2\pi^2} = -\frac{1}{6}$

multiplying both sdes by $- \pi^2$ we get

 $\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}$ 

2021/108) The polynomial: $P(x) = 1 + a_1x +a_2x^2+...+a_{n-1}x^{n-1}+x^n$ with non-negative integer coefficients has $n$ real roots. Prove, that $P(2) \ge 3^{n}$

Because all coefficients are positive so all n roots are -ve and hence

$P(x) = \prod_{k=1}^{n} (x+ a_k)$ where all $a_k$ are positive

Further $\prod_{k=1}^{n} (a_k) = 1$

So $P(2) = \prod_{k=1}^{n}(2+a_k)\cdots(1)$

Now taking AM GM between 1,1 $a_k$ we get $(2+a_k) >= 3\sqrt[3]{(a_k)}\cdots(2)$

So from (1) and (2)

$P(2) >= 3^n \sqrt[3]{\prod_{k=1}^{n} (a_k)}  = 3^n$ and hence $P(2) >= 3^n$

2021/107) Show that $\tan^{-1}(k) = \sum_{n=0}^{k-1}\tan^{-1} \left ( \frac{1}{n^2+n+1} \right )$ - and deduce that $\sum_{n=0}^{\infty}\tan^{-1} \left ( \frac{1}{n^2+n+1} \right ) = \frac{\pi}{2}$

We have $n^2+n+1= 1+n(n+1) = \frac{1+n(n+1)}{(n+1) - n}$

Or $\frac{1}{n^2+n+1} = \frac{(n+1)-n}{1+(n+1)n}$

Using $\tan^{-1}\frac{a-b}{1+ab} = \tan^{-1} a - \tan^{-1}{b}$

We get  $\tan^{-1} \frac{1}{n^2+n+1} = \tan ^{-1}(n+1)-  \tan ^{-1}n$

Adding from 0 to k-1 we get as telescopic sum

Hence $\sum_{n=0}^{k-1} \tan^{-1} \frac{1}{n^2+n+1} =  \tan ^{-1}k-  \tan ^{-1}0 = \tan ^{-1}k$

Taking limit as $k = \infty$

$\sum_{n=0}^{\infty} \tan^{-1} \frac{1}{n^2+n+1}  = \tan ^{-1}\infty= \frac{\pi}{2}$

2021/106) Evaluate closed form of $1^2+2^2+3^2+\cdots+n^2$

We have

$(k+1)^3 = k^ 3 + 3k ^2 + 3k + 1$

Or $(k+1)^3 - k^ 3 = 3k ^2 + 3k + 1$

Adding from 1 to n we get

$\sum_{k=1}^n((k+1)^3 - k^ 3) = 3\sum_{k=1}^n k^2 + 3\sum_{k=1}^nk + \sum_{k=1}^n1$

The LHS is a telespcopic sum = $(n+1)^3-1$

We know $\sum_{k=1}^nk = \frac{n(n+1)}{2}$

so we get $(n+1)^3 - 1 = 3\sum_{k=1}^n k^2 + 3\frac{n(n+1)}{2} + n$

or  $3\sum_{k=1}^n k^2 = (n+1)^3 - 1 -  3\frac{n(n+1)}{2} - n$

$=n^3 + 3n^2 + 3n + 1 - 3\frac{n^2+n}{2} - n$

$= \frac{1}{2}(2n^3 + 6n^2 + 6n - 3n^2 -3n -n )$

$= \frac{1}{2}(2n^3 + 3n^2 + 2n )$

$=\frac{1}{2}n(2n^2+ 3n + 2)$

$= \frac{1}{2}n(n+2)(2n+1)$

so $\sum_{k=1}^n k^2 = \frac{1}{6}(n(n+2)(2n+1)$

2021/105) Prove that if k = mn and k is a perfect square and m and n are co-primes them m and n are perfect squares

 Now let p be a prime factor of k.

So p is a prime factor of m or n but not both because GCD(m, n) = 1

Now because k is a square p shall occur even number of times say 2m

All the 2m occurences must be factor of m (as we have mention p is factor of m) 

So any prime factor of k whcich is a factor of m shall occur even number of times in m and which is not a factor of m shall occur even number of times in n making both m and n perfect squares.

2021/104) Prove that for positive integer n we have $n^2 | (n+1)^n-1$

We have

$(n+1)^n - 1$

$= \sum_{k=0}^{n}  {n \choose k} n^{n-k} -1 $

$= \sum_{k=0}^{n-2}  {n \choose k} n^{n-k} + {n \choose n-1} n^{n-(n-1)} + {n \choose n} n^{n-n} -1$

$= \sum_{k=0}^{n-2}  {n \choose k} n^{n-k} +n * n + 1 - 1$

$= \sum_{k=0}^{n-2}  {n \choose k} n^{n-k} +n^2$

now each term in sum is having $n^2$ as a term and hence the expression is divisible by $n^2$

2021/103) Show that there are infinite numbers of the form $10^n+3$ that are composite

Because the number is not divsible by 2,3,5 for any n so let us check if is divisible by 7 for some n. 

Now as 7 is co-prime to 10 so as per fermats little theorem 

$10^6 \equiv 1 \pmod 7\cdots(1)$ 

By checking from1 to 6 we see that 

$3^4 = 81 \equiv  4 \pmod 7$

or  $10^4 = 81 \equiv  4 \pmod 7$ as $ 10\equiv 3 \pmod 7$

using (1) we get 

$10^{6k+ 4} = \equiv  4 \pmod 7$

or  $10^{6k+ 4} + 3 = \equiv  0 \pmod 7$

hence divisible by 7 and hence composite

so there are infinite numbers as k goes from 1 onwards are composite for n = 6k + 4

2021/102) What is the largest natural number n below 50 such that LCM (n, n + 1, . . . , 50) = LCM (1, 2, . . . , 50), where LCM stands for least common multiple.?

 it is 27. because if we leave out 27 then none of the numbers above 27 is divisible by 27 so we shall not have a factor 27 of the LCM but for numbers 1 to 50 LCM shall have a factor 27.

27 is not by magic. it has to a prime number or a power of a prime. if it is composite other than power of a prime then is has got smaller factor and this is taken care of because smaller numbers are taken care of. so we should look for a number greater than 1/2 of the number because if a is taken care of the 2a is taken care of. in LCM. so we look through numbers 26,27 etc and find 27.

2021/101) Show that any number > 6 can be written as sum of 2 co-prime numbers

Let us consider 3 cases. Let the number be n 

1) n is odd then 2 numbers  are (2, n-1)

2) n is even and of the form 4n. Then the 2 numbers (2n-1, 2n+1)

3) n is even and of the form 4n +2. Then the 2 numbers (2n -1 , 2n + 3)  

2021/100) prove: $\sum_{k=0}^{n} \binom{n}{k}^2 =\dfrac{(2n)!}{(n!)^2}$

Out of 2n objects n objects can be chosen in

$\dfrac{(2n)!}{(n!)^2}$ ways

Now let us make 2n objects into 2 groups of n objects each.

For picking n objects from the set we need k objects from 1st set and n-k from 2nd set and n varies from 0 to n so number of ways

$\sum_{k=0}^{n} \binom{n}{k} \binom{n}{n-k}$

The 2 above are same as it shows the number of ways in 2 different ways

So

$\dfrac{(2n)!}{(n!)^2}= \sum_{k=0}^{n} \binom{n}{k} \binom{n}{n-k}$

Now as $\binom{n}{k} = \binom{n}{n-k}$ so we get the result

2021/099) Find the minimum value of $\sqrt{​x^2+​4​x+​13}+​\sqrt{​x^2-​8​x+​41}$

 We have 

$\sqrt{​x^2+​4​x+​13}+​\sqrt{​x^2-​8​x+​41}$

= $\sqrt{​(x+​2)^2+​9}+​\sqrt{​(x-4)^2 + 25}$

The 1st term that is $\sqrt{​(x+​2)^2+​9}$ is distance from (x,0) to (-2,-3) and second term $\sqrt{​(x-4)^2+25}$ is distance from (x,0) to (4,-5).

clearly sum of the distanace to (x,0) is lowest when (x,0), (-2,3) and (4,5) are in a straight line that is (x,0) is on the line from (-2,3) to (4,5)  now as (x,0) lies in between the minimum is that distance from (-2.-3) to (4,5) or $\sqrt{(4+2)^2 + (5+3)^2} = \sqrt{100} = 10$ 

2021/098) Find integers x,y z, such that $x + y +z = xyz$

 Because of symmetry if (x,y,z) is a solution then any permutation of (x,y,z) is also a solution

without loss of generality let us assume that $ x<=y<=z$ 

So $x+y+z <= 3z$ 

puttying in the given equation we get

$3z >= xyz$

or $3>=xy$

This gives the following set in (x,y)

(1,1) giving 2 + z = z which does not have a solution

(1,2) giving 3 + z = 2z or z = 3

(1,3) giving 4 + z = 3z giving z = 2 which is a contradiction as it should not be less than 3

so solution set (1,2,3) or a permutation of the same.

2021/097) Solve the following equation: $[x]^2=[2x]-1$ where [x] is the floor value of the x real No

Let x = n + r where n is the integer part and r is the fractional part

we have

$\lfloor x \rfloor ^2 = 2(x+r) - 1$

so $n^2 = 2n -1$ when $ r < \frac{1}{2}$ or $n^2 = 2n$  and $ \frac{1}{2} \le   r <  1$

$n^2 = 2n -1$ when $ r < \frac{1}{2}$

gives $n^2 - 2n + 1 = 0$ or $(n-1)^2 =0 $ or n= 1 giving $ 1 \le x < 1.5$

$n^2 = 2n$  and $\frac{1}{2} \le   r <  1$

gives n = 0 or 2  giving $ .5 \le x < 1$ or  giving $ 2.5 \le x < 3$

combining them we have  $ .5 \le x < 1. 5 $ or  $ 2.5 \le x < 3$

 

2021/096) Given an arithmetic series a with a common difference d $a_1 + a_{2n+1}= z$ and $\sum_{k=1}^{n} a_{2k-1}^2 = x\cdots(1)$ $\sum_{k=1}^{n} a_{2k}^2 = y\cdots(2)$ Show $d = \frac{y-x}{2nz}$

We are given

$\sum_{k=1}^{n} a_{2k-1}^2 = x\cdots(1)$

$\sum_{k=1}^{n} a_{2k}^2 = y\cdots(2)$

subtract (1) from (2) to get

$\sum_{k=1}^{n} (a_{2k}^2- a_{2k-1}^2)  = y-x$

Or $\sum_{k=1}^{n} (a_{2k}- a_{2k-1})(a_{2k} + a_{2k-1})   = y-x$

But $(a_{2k}- a_{2k-1}= d$ common difference so we get

$\sum_{k=1}^{n} d(a_{2k} + a_{2k-1})   = y-x$

or $d \sum_{k=1}^{n} (a_{2k} + a_{2k-1})   = y-x$

Or $d \sum_{k=1}^{2n} (a_{k})   = y-x\cdots($

as $a_k = a_1 + (k-1) d$ for any k so we have

Now $a_k + a_{2n+1-k} = a_1 + (k-1)d + a_1 + (2n+1-k-1)d = 2a_1 + (2n-1) d = = a_1 + a_1 + (2n-1) d = a_1 + a_{2n}$

so $a_n + a_{n+1}d = a_1 + a_{2n} = z$

so $a_k + a_{2n+1-k} = z$

so 

 $d \sum_{k=1}^{2n} (a_{k}) $

  $= d \sum_{k=1}^{n} (a_{k} + a_{2n+1-k})$

  $= d \sum_{k=1}^{n} z$

  = 2dnz

 

  So $2dnz = y-x$

    

2021/095) Let $a,\,b,\,c,\,d,\,e,\,f$ be real numbers such that the polynomial

 $P(x)=x^8-4x^7+7x^6+ax^5+bx^4+cx^3+dx^2+ex+f$ factorizes into eight linear factors $x-x_i$ with $x_i>0$ for $i=1,\,2,\,\cdots,\,8$.

Determine all possible values of $f$.

Solution 

Using Vieta's formula  we have

$\sum_{i=1}^8 x_i = 4\dots(1)$

$\sum_{i=1}^7 \sum_{j=i+1}^8   x_ix_j = 7\cdots(2)$

$\prod_{i=1}^8 x_i = f\cdots(3)$

We have

$\sum_{i=1}^8 x_i^2= (\sum_{i=1}^8 x_i)^2 - 2 \sum_{i=1}^7 \sum_{j=i+1}^8   x_ix_j $

$= 4^2 - 2 * 7 = 2$

or $\sum_{i=1}^8 x_i^2 = 2$

Subtracting (1) from above

$\sum_{i=1}^8 (x_i^2 - x_i)  = -2$

adding $\frac{1}{4}$ to each term on LHS that is 2 and adding 2 on RHS we get

$\sum_{i=1}^8 (x_i^2 - x_i + \frac{1}{4})  = 0$

or $\sum_{i=1}^8 (x_i - \frac{1}{2})^2  = 0$

so $x_i = \frac{1}{2}$ for each i.

this satisfies the criteria that $x_i$ is positive

putting this in (3) we get $f = \frac{1}{\sqrt[8]2}$

2021/094) Let a and b be positive real numbers such that $a+b=1$. Prove that $a^a+b^b <=1$

We are given 

$1= a+ b = a^{a+b} + b^{a+b}$

So $1- (a^ab^b + a^b b^a)$

$=  a^{a+b} + b^{a+b} - (a^ab^b + a^b b^a)$

$= a^a(a^b-b^b) + b^a(b^b-a^b) = (a^a - b^a)(a^b - b^b)$

For a > b both the terms are non -ve so we have and if b > a then both terms are -ve and hence above is positive

$1- (a^ab^b + a^b b^a) >=0$ and hence the result

2021/093) Given $2^x = 3^y = 6^{-z}$ evaluate $\frac{1}{x} +\frac{1}{y} + \frac{1}{z}$

we are given $2^x = 3^y = 6^{-z}$

so $2 = 6^{\frac{-z}{x}}$

and $3 = 6^\frac{-z}{y}$

so $2 * 3 = 6^{(-\frac{z}{x} - \frac{z}{y})}$

or $6^1 = 6^{(-\frac{-z}{x} - \frac{z}{y})}$

or $1 = -\frac{z}{x} - \frac{z}{y}$

or  $\frac{z}{x} +\frac{z}{y} + 1 = 0 $

or $\frac{1}{x} +\frac{1}{y} + \frac{1}{z}= 0$

2021/092) The sum of two numbers is 15. What is the minimum sum of the resultant cubes of the two numbers?

Let the numbers be x and y

We have x+ y = 15

$(x+y)^3 = x^3 + y^3 + 3xy(x+y)$

or $15^3 = x^3 + y^3 + 3xy * 15$

or $x^3 + y^3 = 15^3 - 45xy$

this is mininum when xy is maximum

$x + y = 15$

we have $4xy = (x+y)^2 - (x-y)^2 = 15 - (-x-y)^2$

so xy is maximumum when x = y

or $x^3 + y^3$ is minumum when $x = y = 7.5$ and value is $2 * 7.5^3 = 843.75$

2021/091) if ab = cd prove that $a^2+b^2 + c^2 + d^2 $ is composite

 We have

ab = cd

or $\frac{a}{c}= \frac{d}{b} = \frac{m}{n}$ where m and n are in lowest terms or gcd(m,n) = 1

so an = cm and dn = bm

now $n^2(a^2+b^2+ c^2+d^2) $

$= (na)^2 + (nb)^2 + (nc)^2 + (nd)^2$

$= (cm)^2 + (bn)^2 +(nc)^2 + (nd)^2= (b^2+c^2)(m^2 + n^2)$

or $a^2+b^2+c^2+d^2 = \frac{b^2+c^2}{n^2} (m^2+n^2)$

as n is less than $b^2+c^2$ so it is product of 2 numbers > 1 so composite 

2021/090) If $p+q+r=0$ then what is $\frac{(p+q)^2}{3pq}+\frac{(q+r)^2}{3qr} +\frac{(r+p)^2}{3rp}$

We have p+q + r = 0

So $p^3 + q^3 + r^3 = 3pqr\cdots(1)$

And  p+q = -r

or $(p+q)^2 = r^2\cdots(2)$

Similary $(q+r)^2 = p^2\cdots(3)$

$(r+p)^2 = q^2\cdots(4)$

Hence $\frac{(p+q)^2}{3pq}+\frac{(q+r)^2}{3qr} +\frac{(r+p)^2}{3rp}$

$= \frac{r^2}{3pq}+\frac{p^2}{3qr} +\frac{q^2}{3rp}$ (from (2), (3), (4)

$= \frac{r^3+q^3+ p^3}{3pqr} = \frac{3pqr}{3pqr} = 1$  (using (1))

2021/089) Show that $2^n$ is not a factor of $3^n+1$ for n >1

We shall prove the same for 2 cases 

1) n is even

2) n is odd 

let us 1st prove for n even

case 1 :For n even say 2k $(>=2)$

$3^n + 1 =  3^{2k} + 1 = 9^k + 1 \equiv 2 \pmod 4$

so $3^n + 1$ is not divisible by 4 so cannot be divisible by $2^n$

    

case 2: for n odd  say 2k + 1 $( >=3)$

$3^n + 1 = 3^{2k+ 1} + 3 = 9^k.3 + 1 \equiv 4 \pmod 8$

so  so $3^n + 1$ is not divisible by 8 so cannot be divisible by $2^n$

2021/088) For real x,y z if x +y + z = 3 prove that $x^2+y^2 + z^2 >= 3$

We are given 

$x+y+z = 3 \cdots(1)$

now $(x-1)^2\ge 0$

or $x^2-2x + 1 \ge 0$

or $x^2 \ge 2x -1\cdots(2)$

similarly $y^2 \ge 2y-1\cdots(3)$

and $z^2\ge 2z -1\cdots(4)$

addding (2), (3), (4) we get $x^2+y^2 + z^2\ge 2(x+y+z) - 3$

or $x^2+y^2 + z^2\ge 2* 3- 3$ (from (1)

or $x^2+y^2+z^2 \ge 3$ 

2021/087)Given: $x>0,\, n\in\mathbb{N}$ Prove: $(1+x)\times\left(1+x^2 \right)\times\cdots\times\left(1+x^n \right)\geq\left(1+x^{\large{\frac{n+1}{2}}} \right)^n$

We have $x^a + x^b >= 2x^{\frac{a+b}{2}}$ by AM GM inequality

adding $1 + x^{a+b}$ on both sides we get

$(1+x^a)(1+x^b) >= 1 + 2x^{\frac{a+b}{2}} + x^{a+b}= ( 1+ x^{\frac{a+b}{2}})^2$

Putting $b = n+ 1 -a$ we get

$(1+x^a)(1+x^(n+1- a) >= (1+ x^{\frac{n+1}{2}})^2$

For n even taking a from 1 to $\frac{n}{2}$ we get n/2 expressions and multiplying them out we get the result

For n odd we have n-1 ( running a from 1 to $\frac{n-1}{2}$ we get $\frac{n-1}{2}$ terms and as middle term is $(1+x^{\frac{n+1}{2}})$ we get thr result

2021/086) Find the limit of: $lim_{x\to +\infty}\frac{ \lfloor x\rfloor}{x}$

We have

 $\frac{ \lfloor x\rfloor}{x} = \frac{ x- y}{x}$ where y is fractional part of x and hence $0 \le y \lt 1$

Hence 

$  \frac{ \lfloor x\rfloor}{x} =   1- \frac{y}{x} \ge 1 - \frac{1}{x}$

as x goes to infinitte RHS goes to 1 so the required value is between 1 and 1 so it is 1 

2021/085) Let $P(x)=x^3−2x+1$ and $Q(x)=x^3−4x^2+4x−1$. Show that if P(r)=0, then $Q(r^2)=0$

 We have

$P(r) = r^3 - 2r + 1= 0\cdots(1)$

and $Q(r^2) = r^6 - 4r^4 + 4r^2 -1 \cdots(2)$

as $r^3 = 2r  - 1$ so $r^6 = (2r-1)^2 = 4r^2 - 4r + 1$
Putting in (1)  

$Q(r^2) = 4r^2 -4r + 1 - 4r^4 + 4r^2 - 1 = -4r^4 + 8r^2 - 4r = -4r(r^3 - 2r + 1) = 0$ using (1)

2021/084) a,b,c are positive real numbers such that $a^2+b^2= c^2 $ and $ab=c$ find the value of $\frac{(a+b+c)(a+b-c)(a-b+c)(c-a+b)}{c^2}$

 We have

$a^2+b^=c^2\cdots(1)$

And $ab = c\cdots(2)$

Now

$\frac{(a+b+c)(a+b-c)}{c} = \frac{(a+b^2) - c^2 }{c} = \frac{a^2+b^2+2ab-c^2}{c} = \frac{2ab}{c}$ (uing (1)

$ = \frac{2c}{c}$ (sing (2)

$ = 2$

Or

$\frac{(a+b+c)(a+b-c)}{c^2} = 2$

similarly

$\frac{(a-b+c)(-a+b+c}{c} = 2$

multiplying we get $\frac{(a+b+c)(a+b-c)(a-b+c)(c-a+b)}{c^2}= 4$

2021/083) show that for no positive integers a, b,c can all the 3 expresssion $a^2+b+c, b^2+a + c , c^2 + a +b$ be perfect squares

 Proof:

The samllest square above $a^2$ is $a^2+2a+1$. so we must have for $a^2+b+c$ to be a prefect square

$a^2 + b+ c \ge (a+1)^2$

Or $b+c \ge 2a + 1\cdots(1)$

Similarly $c+a \ge 2b + 1\cdots(2)$

And $a + b  \ge 2c + 1\cdots(3)$

Adding above 3 equations we must have $2(a+b+c) >= 2(a + b+ c) + 3$ or $0 \ge  3$ which is contradiction

So above is impossible  or No solution exists 

2021/082) If,for all sets A, $A \cup B = A $ then prove that $B = \emptyset$

This is true.

To prove the same we have $A \cup B = A $ iff $B \subseteq A$

Let us take 2 sets $A_1,A_2$ which are disjoint and because it is true for every set $A_1 \cup B = A_1 $ so $B \subseteq A_1$

and $A_2 \cup B = A_2 $ so $B \subseteq A_2$

So from above 2 we have

$B \subseteq A_1 \cap A_2$

Because $A_1,A_2$ are disjoint sets so we have $A_1 \cap A_2= \emptyset$

So $B = \emptyset$

 

2021/081) A 5-digit number (in base 10) has digits k, k + 1, k + 2, 3k, k + 3 in that order, from left to right. If this number is $m^2$ for some natural number m, find the sum of the digits of m.

Because there is digit 3k so k <= 3

takiing k = 1 we have last 2 digits = 34 so it cannot be a square

taking k = 2 last 2 digits are 65 so it cannot be a square

taking k = 3 we get $34596= 186^2$ or m = 186 so sum of digits = 1 + 8 + 6 = 15 

2020/080) Solve in integer x $5^x - 3^x = 16$

Working mod 3 we have $(-1)^x\equiv 1 \pmod 3$

So x is even say 2n

So we have $^{2n} - 3^{2n} = 16$

Or $(5^n+3^n)(5^n-3^n) =16$

On  LHS both are even and unequal and 1st term is larget and 16 need to be factored into 2 even numbers 8 * 2

So $5^n +3^n = 8$

$5^n - 3^n = 2$

Adding we get $ 2* 5^n = 10$ or n = 1 so x = 2

  

  

2021/079) Solve in natural numbers $(x+4)(x+24) = p^n$ when p is a prime number

We have rhs is power of a prime number so each of the factors on left that ix x+ 4 and x + 20 both are power of same prime number

As x + 4 < x + 24 so we have x+ 4 is a factor of x + 20

Or x + 4 is a factor of x+24 - (x+4) = 20

As  x is natural number so x+ 4 > 4 so we need to consider only the factors of 20 which are greater than 4 that is 5, 10, 20

x+ 4 = 5 => x = 1 x + 20 = 25 power of x + 4 and p = 5 n = 3

x+4 = 10 => x= 6 x + 24 = 30 is not power of x + 4 

Similarly x + 4 = 20 does not give a soltion

  so only solution x = 1, p = 5 , n = 3 

2021/078) What are two numbers whose HCF is 20 and LCM is 300?

Because HCF is 20 so the 2 numbers are 20m,20n where m or n = 1 or they are co-primes. without loss of generality let us assume that that m < n.

So we have LCM = 20mn = 300 or mn= 15

as mn = 15 and m and n are co-primes so m= 1 , n= 15 or m =3 nd n = 5 giving the 2 numbers (20,300) , (60,100) 

 

2021/077) Let $a>b>c$ be the real roots of the equation $\sqrt{2014}x^3-4029x^2+2=0$. Find $b(a+c)$.

To avoid radicals let $\sqrt{2014}=p$

So we get $px^3-(2p^2+1)x^2 +2 = 0$

Or factoring we get $(px-1)(x^2-2px-2)$ = 0

So one root is $x= \frac{1}{p}$ and  other two roots are roots of $x^2-2px-2=0$

For the equation $x^2-2px-2=0$ sum of the roots is 2p and product is -2. so one root has to be -ve and

The positive root shall be above 2p

So $b=\frac{1}{p}\cdots(1)$

And c is the -ve root and $a> 2p$

a,c are roots of $x^2-2px-2=0$ so $a+c = 2p\cdots(2)$

Hence $b(a+c) = 2$ using (1) and (2) 

2021/076) Evaluate $\sum^{89}_{n=0}\frac{1}{1+\tan^3(k^\circ)}$

We have $\sum^{89}_{n=0}\frac{1}{1+\tan^3(k^\circ)}$
 $= \sum^{89}_{n=0}\frac{\cos^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$
$=\frac{\cos^3(0^\circ)}{\cos^3(0^\circ)+\sin ^3(0^\circ)} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)} + \frac{\sin^3(45^\circ)}{\cos^3(45^\circ)+\sin ^3(45^\circ)}+ \sum^{89}_{n=46}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$ splitting into 4 parts
$=\frac{1}{1+0} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)} + \frac{\sin^3(45^\circ)}{\sin^3(45^\circ)+\sin ^3(45^\circ)}+ \sum^{89}_{n=46}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$ as $\sin\,45^\circ = \cos \, 45^\circ$
$=\frac{1}{1+0} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)} + \frac{1}{2}+ \sum^{89}_{n=46}\frac{\cos^3(90^\circ- k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$ putting values and using  $\sin\,x^\circ = \cos \,(90^\circ-x)$
$=\frac{3}{2} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)} +  \sum^{89}_{n=46}\frac{\cos^3(90^\circ- k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$
$=\frac{3}{2} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)} +  \sum^{46}_{n=89}\frac{\cos^3(90^\circ- k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$ reversing order of sum of cos terms
$=\frac{3}{2} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)} +  \sum^{44}_{n=1}\frac{\cos^3( k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$
$=\frac{3}{2} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)+\cos^3( k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$
$=\frac{3}{2} + \sum^{44}_{n=1} 1= 1.5 + 44 = 45.5 $

2021/075) If $x^2-x-1$ divides $ax^{17}+bx^{16} + 1$ find a-b

If $x^2-x-1$ divides $ax^{17}+bx^{16} + 1$ then $x^2=x+1$ =>  $ax^{17}+bx^{16} + 1 = 0$
The above is so because x =t which is root of $x^2= x+1$ is also a root of $ax^{17}+bx^{16} + 1= 0$

We have $x^2=x+1$
Putting $y=\frac{1}{x}$ we get $\frac{1}{y^2} =  \frac{1}{y} +1$
or $y^2 = 1 - y$
 $=>y^4 = (1-y)^2 = 1-2y +y^2 = (1-2y) + (1-y) = 2-3y$
$=>y^8 = (2-3y)^2 = 4-12y +9y^2 = (4-12y) + 9(1-y) =13-21y$
$=>y^{16} = (13-21y)^2 = 169-546y +441y^2 = (169-546y) + 441(1-y) = 610-987y$
$=>y^{17} = 610y-987y^2 = 610y - 987(1-y) = 1597y - 987$
Putting back $x=\frac{1}{y}$ we get
$\frac{1}{x^{17}} =  \frac{1597}{x} - 987$
Or $987x^{17} - 1597x^{16} +1 $ = 0

comparing with above we get a = 987, b = -  1597 so a - b = 2584

2021/074) A number n has sum of digits 100, while 44n has sum of digits 800. Find the sum of the digits of 3n.

The number can contain only the digits 1,2 besides 0. 1 * 44 = 44 and there is no overflow( if in the number the sum is one then it becomes 8, and if it is 2 the the sum of digits is 16) so the sum of digits is 8 times. if any digit is 3 to 9 then sum of digits less than 8 times so this shall give a lesser sum

Again 1 may be preceded/succeeded  by 0 or 1 as 11 * 44 = 484 . but 2 has to be    preceded/succeeded  by 0 as 21 * 44 = 924 and 12 * 44 = 538 and the sum of digits become less

So the number shall have 0 1 and 2 meeting above conditions so that sum of digits 100 and when we multiply by 3 (that is 3n) the digits shall be 0,3,6 and there is no overflow and sum of digits 300.

2021/073)Consider polynomials P(x) of degree at most 3 each of whose coefficients is an element of { 0,1,2,3,4,5,6,7,8,9}. How many such polynomials satisfy P(-1) = 9.

Because the polynomial is of degrees at most 3 with above coefficients we have 

$P(x) = ax^3+bx^2 + cx +d$ where a, b ,c, d are  element of { 0,1,2,3,4,5,6,7,8,9}.

now $P(-1) = -a + b -c + d = (b+d) - (a+c)$

now maximum sum of b+d can be 18 and as $b+d - (a+c) = 9$ so we have

$a+c =k $ and $b+d = 9+k$ and hence $0 <=<=9$

as the digits are from 0 to 9 for the sum to be k <= 9 one of the numbers can be x = 0 to k and another number can be k -x so there are k+1 possibilities

for the sum to be  $ k >=9$ one of the numbers a should be $k-9$ to $9$ and another number c shall be $k-9$ and there are $(9+1) - (k-9)$ or $19-k$ choices.

for the sum to be  $ k >=9= 9 + m$ there are  $19-(9+m)$ or $10-m$ choices.

For the value  $b+d - (a+c) = 9$ a + c can be k and b+d can be k+9 so there are (k+1)(10-k) ways 

now the value k can be from 0 to 9 so we have sum

$\sum_{k=0}^9(k+1)(10-k) =\sum_{n=1}^{10}n(11-n) = 11\sum_{n=1}^{10}n - \sum_{n=1}^{10}n^2 = 11 * \frac{10*11}{2} - \frac{10 * 11 * 21}{6} = 605- 385 = 220 $

So number of polynomials = 220  

2021/072) if in a triangle $a^2+b^2= \frac{19}{9}c^2$ evaluate $\frac{\cot\,C}{\cot\, A + \cot \, B}$

We are given
$a^2+b^2= \frac{19}{9}c^2$
Using the above and law of cosine we get
$2ab\cos\, C = a^2+b^2-c^2 = \frac{19}{9}c^2 - c^2 = \frac{10}{9}c^2$
or $ab\cos\, C = \frac{5c^2}{9}\cdots(1)$

Further
$\cot\, A +\cot\, B= \frac{\cos\, A}{\sin\, A} +  \frac{\cos\, B}{\sin\, B}$
$= \frac{\cos\, A\sin\, B +  \sin\,A\cos\, B}{\sin \, A\sin\, B}$
$= \frac{\sin(A+B)}{\sin \, A\sin\, B}$
$= \frac{\sin(\pi-C)}{\sin \, A\sin\, B}$ as $A+B+C=\pi$
$= \frac{\sin\,C}{\sin \, A\sin\, B}$

Hence $\frac{\cot\,C}{\cot\, A + \cot \, B} = \frac{\cos\, C \sin\, A\sin\, B}{\sin ^2C}$
$=\frac{ab\cos\,C}{c^2}$ (using law of sin)
$=\frac{5}{9}$ 

2021/071) For the triangle with angles A,B,C, the following trigonometric equality holds. $\sin^2B+\sin^2C−\sin^2A=\sin\,B\sin\,C$ Find the measure of the angle A.

Using law of sin's $\sin A = ka, \sin B= kb, \sin C = kc$

We get

$b^2+c^2 - a^2 = bc$

Or $a^2 = b^2 + c^2 + bc\cdots(1)$

By law of cos

$a^2 = b^2 + c^2 - 2bc \cos A \cdots(2)$

from (1) and (2)

$2 \cos A = - 1$ or $\cos A = \frac{-1}{2}$ or $A = 12^circ$

2021/070) Evaluate $\dfrac{1}{1-\cos \dfrac{\pi}{9}}+\dfrac{1}{1-\cos \dfrac{5\pi}{9}}+\dfrac{1}{1-\cos \dfrac{7\pi}{9}}$.

We know that $\cos \frac{\pi}{9},\cos \frac{5\pi}{9},\cos \frac{7\pi}{9}$ are different and they are

roots of equation $\cos 3x = \cos \frac{\pi}{3} = \frac{1}{2}$

or $4\cos^3 x - 3\cos\,x =\frac{1}{2}$

or

so $\cos\frac{\pi}{9}, \cos\frac{5\pi}{9}, \cos\frac{7\pi}{9}$ are roots of equation

$x^3 - \frac{3}{4}x - \frac{1}{8}= 0$

let $x_1= \cos\frac{\pi}{9}, x_2 = \cos\frac{5\pi}{9}, x_3=\cos\frac{7\pi}{9}$

Now $x_1,x_2,x_3$ are roots of equation

$f(x) = x^3 - \frac{3}{4}x - \frac{1}{8}= 0\cdots(1)$

by Vieta's formula we have

$x_1 + x_2 + x_3 = 0\cdots(2)$

$x_1 x_2 + x_2x_3 + x_3 x_1 = \frac{-3}{4}\cdots(3)$

Further $f(1) = (1-x_1)(1-x_2)(1-x_3) =  1- \frac{1}{4} - \frac{1}{8} = \frac{1}{8}\cdots(3)$

And we need to evaluate $\frac{1}{1-x_1 } + \frac{1}{1-x_2} + \frac{1}{1-x_3}$

Now

$\frac{1}{1-x_1 } + \frac{1}{1-x_2} + \frac{1}{1-x_3}$

$= \frac{(1-x_2)(1-x_3) + (1-x_1)(1-x_3) + (1-x_1)(1-x_2)}{(1-x_1)(1-x_2)(1-x_3)}$

$= \frac{1-x_2 - x_3 + x_2x_3 + 1-x_1 - x_3  + x_1x_3  + 1-x_1 - x_2 + x_1x_2}{(1-x_1)(1-x_2)(1-x_3)}$

$= \frac{3 - 2(x_1 + x_2 + x_3)  + (x_2x_3 + x_3x_1  + x_1x_2)}{(1-x_1)(1-x_2)(1-x_3)}$

$= \frac{3 - 2 * 0 + \frac{-3}{4}}{\frac{1}{8}}$ putting the values using (2) , (3) and (4)

$= 18$

Hence $\frac{1}{1-\cos \frac{\pi}{9}} + \frac{1}{1-\cos \frac{5\pi}{9}} + \frac{1}{1-\frac{7\pi}{9}}= 18$

 

2021/069) Prove that there are no integers $a,\,b,\,c$ and $d$ such that the polynomial $ax^3+bx^2+cx+d$ equals 1 at $x=19$ and 2 at $x=62$.

We have  $f(62)−f(19)=a(62^3−19^3)+b(62^2−19^2)+c(62−19)=1 $

Or  $(62−19)(a(62^2+62∗19+19^2)+b(62+19)+c)=1$ 

LHS is a multiple of 43 and RHS is 1 so this does not have integer solution

2021/068) Prove that the polynomial equation $x^8-x^7+x^2-x+15=0$ has no real solution.

 We have $x^8-x^7 + x^2 -x + 15 = x^7(x-1) + x(x-1) + 15$

Each term is positive for $x > 1$ so LHS is greater than 0 so no solution for $ x > 1$

For x = 1 LHS = 15 so x = 1 is not a solution

Further $x^8-x^7 + x^2 -x + 15 = (15- x) + x^2(1-x^5) + x^8 $

Each term is positive for $x < 1$ so LHS is greater than 0 so no solution for $ x < 1$

Hence no real solution

2021/067) Let a,b,c be three distinct integers and P be a polynomial with integer coefficients. Show that in this case the conditions P(a)=b,P(b)=c,P(c)=a cannot be satisfied simultaneously.

The polynomial is P(x)

we have m-n divides  P(m) - p(n)

Let the given condition is true

So $a-b | P(a) - p(b)$

or $a-b | b- c$

Similarly

$b - c | c- a$

And $ c- a | a - b$

From above 3 have

$a-b | b-c | c- a | a-b$

So all are same hence a = b= c which is a contradiction.

Or they are -1/+1 and this is also a contradiction 

so condition can not be satisfied simultaneously 

2021/066) Prove that two triangles with sides $a,\,b,\,c$ and $a_1,\,b_1,\,c_1$ are similar if and only if $\sqrt{aa_1}+\sqrt{bb_1}+\sqrt{cc_1}=\sqrt{(a+b+c)(a_1+b_1+c_1)}$.

we have

$\sqrt{aa_1}+\sqrt{bb_1}+\sqrt{cc_1}=\sqrt{(a+b+c)(a_1+b_1+c_1)}$.

$\equiv (\sqrt{aa_1}+\sqrt{bb_1}+\sqrt{cc_1})^2=(a+b+c)(a_1+b_1+c_1)$

$\equiv aa_1+bb_1+cc_1+2\sqrt{aa_1bb_1} + 2\sqrt{bb_1cc_1} + 2\sqrt{cc_1aa_1} =  aa_1+ab_1 + ac_1 + ba_1 + bb_1 + bc_1 + ca_1 + cb_1 + cc_1$

$\equiv 2\sqrt{aa_1bb_1} + 2\sqrt{bb_1cc_1} + 2\sqrt{cc_1aa_1} = ab_1 + ac_1 + ba_1 + bc_1 + ca_1 + cb_1$

$\equiv  ab_1 + ac_1 + ba_1 + bc_1 + ca_1 + cb_1-2(\sqrt{aa_1bb_1} + 2\sqrt{bb_1cc_1} + 2\sqrt{cc_1aa_1}) = 0$

$\equiv  (\sqrt{ab_1} - \sqrt{a_1b})^2 +  (\sqrt{ac_1} - \sqrt{a_1c})^2 +  (\sqrt{bc_1} - \sqrt{b_1c})^2 = 0$

The above is true iff $ab_1 = a_1b$, $ac_1 = a_1c$, $bc_1 = b_1c$

giving $\frac{a}{a_1} = \frac{b}{b_1} = \frac{c}{c_1}$ or the 2 triangles are similar

2021/065) Solve the biquadratic equation $(x-9)(x-7)(x+3)(x+5)= 1792$

Because (-7) -(-9) = 2 and and 5-3 = 3 let us take y to be mean of x- 7 and x + 3 that is x - 2

So get get

$(y-7)(y-5)(y+5)(y+7) = 1792$

or $(y-7)(y+7)(y-5)(y+5) = 1792$

or $(y^2-49)(y^2-25) = 1792$

or $y^4 - 74 y^2 + 1225 = 1792$

or $y^4-74 y^2 - 567=0$

or $(y^2-81)(y^2 + 7) = 0$ so $y^2=81$ as $y^2 \ge 0$

giving $y = \pm 9$ or $ x \in \{ 11,7\}$ the solution set  

2021/064) Solve in positive integer a,b $a^3-b^3 = ab + 11$

This type of problem is best solved by finding the bound and checking values in the bound

We have a=b is not a solution because LHS = 0 and RHS= 11

So $(a-b) >=1$ as $a > b$ because RHS is positive

Now $a^3-b^3 = (a-b)(a^2 + b^2 + ab)$

As and b are positive so AM, GM inequality gives   $a^2 + b^2 >= 2ab$

So we have $a^3 - b^3 >=  (a-b)(3ab)$

Or $a^3-b^3 >= 3ab$

So from the given relation $3ab < = ab +11 $ or $ 2ab <=11$ or $ab < 6 $ so $b<=2$ as   $a<=b$

b = 1 gives $a^3 = a + 12$ a =3 gives LHS = 8 RHS = 15

a = 4 gives LHS = 64 RHS = 16 so no integer solution LHS is greater than RHS 

b = 2 gives $a^3 = 2a + 11$ a =3 given LHS = 8, RHS = 17

a = 4 gives LHS = 64 RHS = 19 

So no solution

Hence there is no solution to this equation 

2021/063) Solve $x+\lfloor 2x\rfloor +\lfloor 3x\rfloor=7$

 Because RHS is integer so LHS is integer

As $\lfloor 2x \rfloor$ and $\lfloor 3x \rfloor$ are integers so x is integer so   $\lfloor 2x \rfloor = 2x $ and  $\lfloor 3x \rfloor = 3x$

so x + 2x + 3x = 6x = 7 so $x = \frac{7}{6}$ which is not integer so NO solution 

2021/062) Solve in integers x,y such that $2^x-3^y=7$

x and y have to be non -ve integers because otherwise we shall get a fraction

We have $3^y \equiv 0 \pmod 3$ for  y positive and and 1 for y zero

So let us take the 2 cases separately

$y = 0 => 2^x = 8$ or x = 3 so a solutions (x=3,y=0)

Now let us take y positive so we get

$2^x \equiv 1 \pmod 3$ so x has to be even as $2^x \equiv 1 \pmod 3$ for even x and -1 for odd x

so x = 2n.

Working in mod 4 we get $3^y \equiv 1 \mod 4$ so y has to be even as  $3^x \equiv -1 \pmod 4$ for even x and -1 for odd x

So we get y = 2m

Now $2^(2n) - 3^(3m) = 7$ or factoring we get

$(2^n+3^m)(2^n- 3^m) 7$ and as 7 is prime we have

$2^n+3^m = 7$ and $2^n-3^m = 1$

or solving we get $2^{nd}$ solution (n= 2, m = 1) or (x-4, y=2)

So 2 solutions are $(x,y) = \{(3,0), (4,2)\}$  

 

2021/061) Prove that the number 9999999+1999000 is composite.

Let x = 1000 to keep it simple

So we get

$9999999+1999000  = 10^7-1 + 1999 * 1000$

$= 10 * 10^ 6 - 1+ (2000-1)* 1000$

$= 10 x^2 - 1 + (2x-1) x$ putting 1000 = x

$=12x^2 - x - 1$

$= (4x+1)(3x-1)$

$= 4001 * 2999$ putting back x = 1000 

Hence it is composite 

2021/060) Find a polynomial of degree 3 with real coefficients such that each of its roots is equal to the square of one root of the polynomial $P(x)=x^3+9x^2+9x+9$.

Because the square of the roots shall be roots of the required polynomial so we must have $\sqrt x$ as roots of P(X)

So $(\sqrt x)^3 + 9 x + 9 (\sqrt x) + 9 = 0$

Or $(\sqrt x)(x + 9) = - 9(x+1)$

Or squaring $x(x+9)^2 = 81(x+1)^2$

Or $x(x^2 + 18 x + 81) = 81 x^2 + 162x + 81$

Or $x^3 - 63x^2 - 81 x - 81 = 0$

This is the required equation

2021/059) Let $p$ be a prime number. Find the fractional part of $\dfrac{(p+1)!}{p^2}$.

 Because p is prime by wilson theorem $p | (p-1)!+1$

Or $p^2| (p! + p)$

or $p^2| (p! + p) (p+1)$

Or  $p^2| (p+1)! + p(p+1)$

or $(p+1)!  \equiv  -p(p+1) \pmod p^2$

So fractional part of $\frac{(p+1)!}{p^2}$ is same as fractional part of $\frac{-p(p+1)}{p^2}$ or is $\frac{p-1}{p}$

2021/058) FInd integers a and b such that $5\frac{3}{a} * b\frac{1}{2} = 19$

it is easier to work for b rather than a  as the term $b\frac{1}{2}$ is between 2 integers

Now $ 5 < 5\frac{3}{a} < 6$ 

Now $ 5 * 4 = 20 > 19$ and $ 6 * 3= 18 < 19$ so $b\frac{1}{2}$ is between 3 and 4 and hence b = 3

$b\frac{1}{2} = \frac{7}{2}$

So  $5\frac{3}{a} = 19 * \frac{2}{7} = \frac{38}{7}= 5 \frac{3}{7}$

Or a = 3 

2021/057) Given $a^2+b^2=c^2+d^2=1$ and ac+bd=0.Compute the value of ab+cd

Because $a^2 + b^2 = 1$ we can take $a = \sin\alpha$ and $b = \cos \alpha $

Similarly $c=\sin \beta$ and $d = \cos\beta$

$ac + bd = \sin\alpha \sin  \beta + \cos \alpha \cos \beta = 0$

or $\cos (\alpha - \beta) = 0\cdots (1)$  (using formula of cos of difference)

$ab + cd = \sin \alpha \cos \alpha + \sin \beta \cos \beta = \frac{1}{2}(\sin 2 \alpha + \sin  2 \beta)$ (using formula for sin of twice angle)  

$= \frac{1}{2} \sin (\alpha + \beta) \cos (\alpha - \beta)$ using sum of sines

$= \frac{1}{2} \sin (\alpha + \beta) * 0$ using (1)

= 0 

2021/056) Let S(n) be the sum of digits of a number. Find the number of 3 digits numbers whose $S(S(n)) = 2$

Sum of all the digits of a 3 digit number that is $S(n)$  shall be between 1 and 27.

because $(S(n)) = 2 $ and $S(n)$ is between 1 and 27 so we have $(n) \in \{ 2,11,20\}$ 

Now the number of 2 digit numbers(with leading zero)  sum = n is n+1 for n $<=9$

This is so because 1st digit can be 0 to n and 2nd digit can be 2-n ( n = 0 or 1) or 11 -n ( n= 2 to 9)

The number of 1/2 digit numbers sum = n is 19-n for 9 < n < 19 is 19-n. 

This is so because the 1st digit can go from  n -9 to 9 or 9-(n-9) 

One digit number is allowed ( that is 2 digit number with leading zero)  because we are considering the tens digit of the 3 digit number that can be zero.

Now we shall use the above 2 do the counting

The 3 digit numbers that have sum 3 are 101,110,200 that is 3 numbers

let us count the number of 3 digit that have a sum 11

1st digit is 1 so sum of other 2 digit 10 so number of numbers = 19-10 = 9

1st digit is 2 to 9  so sum of other 2 digit 9 down to 2  number of numbers = $10 + 9 + \cdots 3 = \frac{(10 + 3)*8}{2} = 52$

So the number of numbers that have sum 11 is 9 + 52 = 61

let us count the number of 3 digit that have a sum 20

1st digit can be from 2 to 9 and that shall give the sum of other 2 digits of the number can be from 18 down to 11. the number of numbers can be 19-k with k from 18 to 11

so the sum = $1+2 + \cdots 8 = \frac{8 * 9}{2} = 36$

So total number of numbers with sum 20 = 36

So total number of numbers  = 3 + 61+ 36 = 100

 

2021/055) How many divisors of the number $30^{2003}$ are not divisor of $20^{2000}$?

 Let us factorise $30^{2003}$ and $20^{2000}$

$30^{2003}= 3^{2003} * 2 ^{2003} * 5^{2003}$

as 2,3,5 are pairwise co-primes and in factor each can come 0 to 2003 times that is 2004 ways 

so number of factors = $(2003+1) * (2003 + 1) * (2003 +1)= 2004^3$

Now

$20^{2000}= 2^{2000} * (2 * 5) ^{2000}= 2^ {4000} * 5^{2000}$

$GCD(30^{2003},20^{2000}) = 2^{2003} * 5 ^{2000}$

Any number that divides $30^{2003}$ and $20^{2000}$  must divide $GCD(30^{2003},20^{2000})$

So number of numbers that divide $30^{2003}$ and $20^{2000}$ = $(2003+1)(2000+1) = 2004 * 2001$

So number of numbers that divide $30^{2003}$ and does not divide $20^{2000}$ = $2004^3 - 2004 * 2001 = 2004(2004^2-2021) = 8044086060$ 

8044086060 is the number of divisors of $30^{2003}$ that are not divisor of  $20^{2000}$ 

2021/054)Find all integer solutions of the equation $\lfloor \frac{x}{1!}\rfloor + \lfloor \frac{x}{2!}\rfloor +\cdots + \lfloor \frac{x}{10!}\rfloor = 1001$

 Let us define

$f(x) =  \lfloor \frac{x}{1!}\rfloor + \lfloor \frac{x}{2!}\rfloor +\cdots + \lfloor \frac{x}{10!}\rfloor $

So we have 

$f(x) > \lfloor \frac{x}{1!}\rfloor $ for $x>=2$

as $f(x) = 1001$ so 

$x < 1001$

As $7!>1000$ so we have   $\lfloor \frac{x}{n!}\rfloor =0 $ for $x< 1000$ and $n>6$

Further 

$f(kn!+i)= kf(n!) + f(i)\cdots(1)$ for k = 1 to n and i is less than n!

This is so because for m < n 

$\lfloor \frac{kn!+l}{m!} \rfloor =  \lfloor\frac{kn!}{m!}\rfloor  + \lfloor \frac{l}{m!} \rfloor $  

additionally  

$f(kn!+i)= f(kn!) + f(i)\cdots(2)$ for any $k$ and $ i \le n!\cdots(2)$

and $f((n+1)!) = (n+1)f(n!) + 1$

to use the facts let us calculate f(k!) for k = 1 to 6.

f(1) = 1, f(2) = 3, f(6) = 10, f(24) = 41, f(120) = 206

and f(720) = 1237

so the value of  x is less than 720 so let us look at next lowest factorial that is 120

f(120) = 206

$\lfloor \frac{1000}{206}\rfloor = 4$ 

using (1) $f(120*4) = 206 * 4 = 824$

Or f(480) =  824

So we have to account for 1001 - 824 = 177

now f(24) = 41  

$\lfloor \frac{177}{41}\rfloor = 4$ 

so we f(96) = 41 * 4 = 164 

so $f(576) = f(480+96) = f(120 * 4 + 96)

= f(480) + f(96) = 824 + 164 = 988$

Now we have to account for remaining 13 and f(3!) = f(6) = 10

so 6 goes one more time and we ahve

$f(582) = f(576 + 6) = f(24 * 24 + 6) = f(576) + f(6) = 988 + 10 = 998$ using (2)

now we need to account for 3 and as f(2!) = f(2) =3 so we get

E$f(584) = f(582+2) = f(97 * 6) + 2 = f(682) + f(2) = 988 + 3 = 1001$      

so x = 584

2021/053) Prove that $n^{n-1}-1$ is divisible by $(n-1)^2$ for integer n

Replacing n by k+ 1 we need to prove

$(k+1)^k-1$ is divisible by $k^2$

If k is less than 2 it is obvious so let us check for k greater than 2

We have $(k+1)^k - 1 =\sum_{n=0}^k {k \choose n} k^n - 1$

$= 1 + k . k + \sum_{n=2}^k {k \choose n} k^n -1 $

$ =  k . k + \sum_{n=2}^k {k \choose n} k^n$

Each term is divisible by $k^2$ and hence the expression 

2021/052) Let f(x) be a polynomial in x with integer coefficients and suppose that for 5 distinct integers $a_1,a_2,a_3,a_4,a_5$ one has $f(a_1)= f(a_2)= f(a-3) = f(a_4) = f(a_5) = 2 $ Show that there is no integer b such that $f(b) = 9$

Because  $f(a_1)= f(a_2)= f(a-3) = f(a_4) = f(a_5) = 2 $

So defining  $g(x) = f(x) - 2$  

We get $g(a_1)= g(a_2)= g(a-3) = g(a_4) = g(a_5) = 0 $

So $g(x) =  (x-a_1)(x-a_2)(x-a_3)x-a_4)(x-a_5)h(x)$ for some function h(x) of x

So g(x) is product of atleast 5 different numbers 

Because $f(b) = 9$ so $g(b) = 7$ 

For g(x) to be 7 at some b 7 should be product of 5 different numbers but is is not product of have at least 5 different factors but 7  has at most 3(as -7 * -1 * 1) so it is not possible for any b such that $g(b) = 7$ or $f(b)  = 9$

2021/051) Prove that $\frac{5^{125}-1}{5^{25}-1}$ is a composite number

Let $f(x) = x^4 + x^3 + x^2 + x+ 1$
\
we get $f(x) = (x^2 + 1)^2 + x^3 - x^2 + x$

$= (x^3 + 3x + 1)^2 - 2* 3x(x^2+1) - 9x^2 + x^3 - x^2 + x$

$= (x^2 + 3x + 1) ^2 - (5x^3 + 10x^2 + 5x)$

$= (x^2 + 3x +1^2 - 5x(x+1)^2$

$\frac{5^{125}-1}{5^{25}-1} = f(5^{25})$

$f(5^{25}) =  (5^{50} + 3 * 5^{25} +1)^2 - 5 * 5^{25}(5^{25}+1)^2$

$= (5^{50} + 3 * 5^{25} +1)^2 - (5^{13}(5^{25}+1))^2$

$= (5^{50} + 3 * 5^{25} +1 + (5^{13}(5^{25}+1))(5^{50} + 3 * 5^{25} +1 - (5^{13}(5^{25}+1))$

product of 2 numbers neither is 1 so composite 

2021/050) Solve the equation $4x^6-6x^2 + 2\sqrt{2}=0$

Because it has $x^6$ and $x^2$ term we feel that it could be transformed to a trigonometric equation By  putting $x^2=a\cos\,y$

We get $4a^3\cos^3y - 6a\cos\,y + 2\sqrt{2}=0$

Comparing with $4cos^3z - 3\cos\,z$ the coefficients being proportional we get

$\frac{4a^3}{4}= \frac{6a}{3}$ or $a =\sqrt{2}$

Putting $x^2=\sqrt{2}\cos\,y$

we get $4 * 2 \sqrt{2}\cos^3 y - 6\sqrt{2} \cos\,y + 2\sqrt{2} = 0$

or  $4 \sqrt{2}\cos^3 y - 3 \cos\,y + 1 = 0$

or $\cos\,3y = -1 =\cos\, \pi$

so 3 solutions $3y = \pi$ or $3y= 3\pi$ or $3y = 5\pi$

or $x^2 = \sqrt{2} \cos\,\frac{\pi}{3}$   or  $x^2 = \sqrt{2} \cos\,\pi$ or  $x^2 = \sqrt{2} \cos\,\frac{5\pi}{3}$

giving solution $ x = \pm \frac{1}{\sqrt[4]{2}}$(double roots)  or $x= \pm  \sqrt[4]{2}i$

Alternative solution

because there is a $\sqrt{2}$ term and $x^2$ and $x^6$ so we can get rid of $\sqrt{2}$ by putting $x = a \sqrt{2}$ to give

$8a^3\sqrt{2} - 6a\sqrt{2} + 2 \sqrt{2} = 0$

or $8a^3-6a+2= 0$

or $4a^3-3a+1 = 0$

by inspection we get a = -1 is a solution so $a+1$ us a factor

we get $4a^3 - 3a + 1 = 4a^2(a+1) - 4a^2 - 3a + 1 = 4a^2(a+1) - 4a(a+1) + a + 1 = (a+1)(4a^2-4a+1) = (a+1)(2a-1)^2$

giving solution a = -1 or $a= \frac{1}{2}$ (double root)

or $x^2 = - \sqrt{2}$ or  $x^2 = \frac{1}{\sqrt{2}}$ (double root)

 giving solution  $x= \pm  \sqrt[4]{2}i$ or $ x = \pm \frac{1}{\sqrt[4]{2}}$(double roots)

2021/049) Solve the equation $x^3-3x= \sqrt{(x+2)}$

 As RHS is not negative so we have $x^2>=3$

Trying the lowest integer $>\sqrt{3}$ that is 2 we get this satisfies the equation.

Now squaring both sides we get $x^2(x^2-3)^2 = x + 2$

at $x >2$ LHS grows faster than RHS so x = 2 is the only solution

2021/048) Maximise abcd given $ab+cd = 4\cdots(1)$ $ac + bd=8\cdots(2)$

Because abcd is maximum when abcd is positive as we have ab , cd are of same sign and as sum is positive so ab and cd both are positive.

Similarly ac and bd are positive. . so all 4 are positive or all 4 are -ve. with out loss of generality we can assume all 4 are positive.

We are given

$ab+cd = 4\cdots(1)$

$ac + bd=8\cdots(2)$

From  (1) we have using AM GM inequality $abcd <= 4$ and 

$abcd = 4$ when

$ab = cd  = 2\cdots(3)$

The value shall be 4 in case we find some abcd satisfying (2) and (3)

From (3) we have $a=\frac{2}{b}$ and $c=\frac{2}{d}$

Putting in (2) we get $ac + db =\frac{4}{bd} + db = 8$

or $(bd)^2 - 8bd + 4 =0$

So $bd = 4 + \sqrt{12}$ one solution

taking a = 1 we get

$a =1, b= 2 , c = 4- \sqrt{12} , d = 2 + \sqrt{3}$ satisfy (1) and (2) hence   maximum value of abcd = 4

 

2021/047) Find all natural numbers n such that $n^2+7| n^3 + 3$

We have $n^3+3 = n(n^2+7) - (7n-3)$

Because $n^2+7| n^3 + 3$ so $n^2+ 7 | 7n - 3$

As 7n - 3 is positive so we have

$n^2 + 7 \le 7n- 3$

Or $n^2 - 7n + 4 \le 0$ giving $ n < 7$

By trying the values of n from 1 to 6 we get  1

Solution set $\{2,5\}$ 

2021/046) Find all 2 digit numbers having 6 factors

we shall be using the following formula for number of factors of a number.

If the number N is of the form $p_1^{q_1}p_2^{q_2} \cdots p_n^{q_n}$ when 

$p_1,p_2,p_3\cdots p_n$ are prime numbers and

 $q_1,q_2\cdots q_n$ are the powers of the prime numbers then the number of factors 

 are $(q_1+1)(q_2+1)\cdots(q_n+1)$  

now let us factor 6 is different ways 6 = 6 = 2 * 3

if we take 6 then the number is $p^5$ and for this to be between 10 and 100 this is $2^5= 32$ as $3^5 = 243$ is bigger

If we take 2* 3 then the number is $p_1p_2^2$ and for this to be between 10 and 100 

Taking $p_1=2$ we get $2* 3^2= 18$, $2 * 5^2 = 50$, $ 2 * 7^2 = 98$.

Taking $p_1=3$ we get $3* 2^2= 12$, $3 * 5^2 = 75$

Taking $p_1=5$ we get $5* 2^2= 20$, $5 * 3^2 = 45$

Taking $p_1=7$ we get $7* 2^2= 28$, $7 * 3^2 = 63$

Taking $p_1=11$ we get $11* 2^2= 44$, $11 * 3^2 = 99$

Taking $p_1= 13$ we get $13* 2^2= 52$

Taking $p_1= 17$ we get $17* 2^2= 68$

Taking $p_1= 19$ we get $19* 2^2= 76$

Taking $p_1= 23$ we get $23* 2^2= 92$

So solution set 

$\{12,18,20,28, 32,44,45,50,52,63,68,75,76,92,98,99\}$

2021/045) Let $p>5$ be the prime number. Prove that the expression $p^4-10p^2+9$ is divisible by 1920?

 we need to show that $p^4 - 10p^2 +9)$ is divisible by 1920.

Let us now factor $p^4 - 10p^2 +9$

$p^4 - 10p^2 +9 = (p^2-1)(p^2-9)$

$= (p+1)(p-1)(p+3)(p-3)$

$= (p-3)(p-1)(p+1)(p+3)$

as p is greater than 5 and a prime so p is odd so let p = 2n + 1

so we get above expression

$=(2n-2)(2n)(2n+2)(2n+4) = 16(n-1)n(n+1)(n+2)$

$(n-1)n(n+1)(n+2)$ being product of 4 consecutive numbers is divisible by 24 so $16(n-1)n(n+1)(n+2)$ is divisible by 16 * 24 = 384

Further $(n-1)n(n+1)(n+2)= \frac{(n-1)n(n+1)(n+2)(n+3}{n+3}$ 

$(n-1)n(n+1)(n+2)(n+3)$ being product of 5 consecutive numbers is divisible by 5 but is p is prime so p is not

 divisible by 5 or 2n+1 is not divisible by 5 or 2n +6 is not divisible 5 or n +3 is not divisible by 5 so $(n-1)n(n+1)(n+2)$ is not divisible by 5

Hence $16(n-1)n(n+1)(n+2)$ is divisible by 5 and is it is divisible by 384 so divisible by 5 * 384 or 1920

so $p^4 - 10p^2 +9$ is divisible by 1920 hence proved.

2021/044) Prove that $13^{99} - 19^{93}$ is divisible by 162

we have $162= 2 * 81 =2 * 3^4$

So we need to show that the given expression is divisible by 2 and 81

Divisible by 2 is simple and both terms are odd so the difference is even so divisible by 2

Now let us find the value of expression mod 81

To find the same let us evaluate each term mod 81

$13^{99}= (12+1)^{99} = \sum_{n=0}^{99}{99 \choose n}12^n 1^{99-n}$

for n = 4 to 99 there is power of $12^n$ so each of the terms is divisible by $3^4$ or 81

so we have $13^{99} \equiv 1 + 99 * 12 + \frac{99 *98}{2} * 12^2 +   \frac{99 *98 * 96 }{6} * 12^3 \pmod {81}cdots(1)$

out of above the 3rd term onwards divisible by 81( as it contains 99 so one 9 is there so we require another 9 that comes from $12^2$ so taking mod 81 all the terms from 3rd term onwards is zero

So

$13^{99} \equiv 1 + 99 * 12 =  1+ (81 * 18) * 12 \equiv 1 + 18 * 12 \equiv 217 \equiv 55 \pmod {81}$

Now

$19^{93}= (18+1)^{93} = \sum_{n=0}^{93}{93 \choose n}18^n 1^{93-n}= 1 + 18 * 92 + \sum_{n=2}^{93}{93 \choose n}18^n 1^{93-n}$

except 1st 2 terms all are divisible by 81 so we have

$19^{93}\equiv 1 + 18 * 93 \equiv 1 + 18 * 12 \equiv 217 \equiv 55  \pmod {81}\cdots(2)$

from (1) and (2)

$13^{99} - 19 ^{93} \equiv 0 \pmod {81}$

as $13^{99} - 19 ^{93} \equiv 0 \pmod {2}$

from above 2 we get $13^{99} - 19 ^{93} \equiv 0 \pmod {162}$  

2021/043) Find the largest integer n for which $\sum_{k=1}^\infty \frac{1}{k^n}$ diverges

Ans n = 1.

To prove that same we show that if n = 1 then this diverges and if n =2 this converges

To prove that it diverges for n = 1

Let the sum be S

We have $S = \sum_{k=1}^\infty \frac{1}{k^n}$

Or 

$S = 1 +  \sum_{k=1}^\infty (\sum_{i=2^{k-1}+1}^{2^{k}} \frac{1}{i^n})$

 $S = 1 + \sum_{k=1}^\infty s_k\cdots(1)$

where $s_k = \sum_{i=2^{k-1} +1}^{2^{k}} \frac{1}{i^n}$

Basically we have broken the sum to smaller groups with each group $k^{th}$ group containing the sum of reciprocal of $2^{k-1} +1$ to $2^{k}$ we have separated 1 and 1/2 so that k can start from 1 

Now let us compute $s_k = \sum_{i=2^{k-1} +1}^{2^{k}} \frac{1}{i^n}$

The above has got $2^{k-1}$ terms 

and for $i < 2^k$ we have $ \frac{1}{i} > \frac{ 1}{2^k}$

Hence $s_k >= 2^{k-1} * \frac{ 1}{2^k}$

or $s_k >= \frac{ 1}{2}$

Putting in (1) we get

$S = 1 + \sum_{k=1}^\infty s_k = 1 + \sum_{k=1}^\infty \frac{1}{2} $

Hence it diverges 

Now let us prove that it converges for n =2

For $n >=2$

We have $n^2 > n(n-1)$ or $\frac{1}{n^2} < \frac{1}{n(n-1)}$

We have $n^2 > n(n-1)$ or $\frac{1}{n^2} < \frac{1}{n(n-1)}$

as $\frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}$

so $\frac{1}{n^2} <  \frac{1}{n-1} - \frac{1}{n}$

So $\sum_{k=1}^\infty \frac{1}{k^n}$

$= 1 + \sum_{k=2}^\infty \frac{1}{k^n}$

$<  1 + \sum_{k=2}^\infty(\frac{1}{k-1} - \frac{1}{k}$

$<= 1 + (1- \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3} ) + \cdots$

$ < 2$

 

2021/042) For real numbers a and b that satisfy $a^3+12a^2+49a + 69=0$ and $b^3 -9b^2+ 28b -31 = 0$ find a+b.

We have
 $a^3+12a^2+49a + 69=0$
=> $(a+4)^3 + (a+5) = 0$
or $(a+4)^3 + (a+4) + 1= 0\cdots(1)$

$b^3 -9b^2+ 28b -31 = 0$
$=>(b-3)^3 + (b-4) = 0$
$=>(b-3)^3 + (b-3) -1 = 0\cdots(2)$

if we define $f(x) = x^3+x$ knowing that f(x) is odd function
$(f(a+4) = -1 $ and $-f(b-3) = 1$

so $(a+4) = - (b- 3)$

or $a + b = - 1$

2021/041) The equation $(x+a)(x+b)=9$ has a root a+b . Prove that $ab \le 1$

 Solution 

Because a+b is a root so $(x+a)(x+b)=9$ shall be satisfied when x = a + b.
Putting x = a + b we get
$(2a+b)(2b+a) = 9$
or $2a^2 + 2b^ 2 + 5ab = 9$
or $2(a^2+b^2) + 5ab = 9$
or $2(a^2+b^2-2ab) + 9ab = 9$
or $2(a- b)^2 = 9(1-ab)$
now LHS is >=0 so is RHS so $9(1-ab) \ge 0$ or $1-ab \ge 0$ or $ab \le 1$

2021/040) Prove that there are infinitely many positive integers n such that n(n+1) can be expressed as a sum of two squares in atleast two different ways.

 Let the number be $z = n(n+1)$

So $z = n^2 + n$

If we choose n to be a square say $m^2$ then we have z already a sum of 2 squares

We have $z = (m^2)^2 + m^2= m^2(m^2+1)$

If we have $m^2$ as sum of 2 squares say $x^2+y^2$ then 

we have

$z = (p^2 + q^2) (m^2 + 1) = p^2m^2 + q^2m^2 + p^2 + q^2$

$= (p^2m^2 + q^2 - 2pqm) + (q^2m^2 +p^2 + 2pqm)$

$= (pm - q)^2 + (qm +p)^2$

This is another way of representation

As (p,q,m) are sides of a Pythagorean triple and there are infinite Pythagorean triples so there are infinite independent values.

2021/039) if $\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$ where a,b,c are positive integers with no common factor then prove that a+b is a perfect square

 We have 

$\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$

or $bc + ac = ab$

or $(a-c)(b-c)= c^2$

Now let a prime divide c 

There are two cases 

p divides a-c and b-c in which case p divides a,b,c so they have a common factor p which cannot be true

or $p^2$ divides a-c or b-c so there exists m and n(either of them can be 1) such that

$c= mn$ and $(a-c) = m^2$ and $(b-c) = n^2$   

so $a = m^2 + c$  and $b = n^2 +c $

so $a+b= m^2 + n^2 + 2c = m^2 + n^2 + 2mn = (m+n)^2$

or $a+b$ is a perfect square

2021/038) if $n^2+7n+4$ is divisible by 11 what is the remainder when divided by 121

 Working in mod 11 we have

$n^2+7n + 4 \equiv n^2 + (7-11)n + 4$

$\equiv n^2 - 4n + 4 \equiv (n-2)^2$

so if $n^2 + 7n + 4$ is divisible by 11 then n-2 is divisible by 11

so $n = 11k + 2$

So  $n^2+7n+4 = (11k+2)^2 + 7(11k+2) + 4$

= $121k^2 + 121k + 22$

so remainder when divided  by 121 is 22 

2021/037) Solve the system of equations

$a + b + c = 7\cdots(1)$

$a^2 + b^2 + c^2 = 21\cdots(2)$

$abc= 8 \cdots(3)$

Solution 

We are given 

$a + b + c = 7\cdots(1)$

$a^2 + b^2 + c^2 = 21\cdots(2)$

$abc= 8 \cdots(3)$

From (1) and (2) we get

$(a+b+c)^2 - (a^2+b^2+ c^2) = 7^2 - 21$

Or $2(ab+bc+ca) = 28$

Or $ab+bc+ca = 14\cdots(4)$

a , b, c are roots of equation

$(x-a)(x-b)(x-c) = 0$

Or $x^3-x^2(a+b+c) + x(ab + bc + ca) - abc= 0$

Putting the values from (1), (3) and (4) we get 

$x^3 - 7x^2 + 14x - 8 = 0$

We know from above that the value becomes zero when x = 1 so x =1 is a factor and (dividing by x- 1) we get

$(x-1)(x^2 - 6x + 8) = 0$

Or $(x-1)(x-2)(x-4) = 0$

So roots are 1,2, 4 and so we have

$\{a,b,c\} = \{1,2,4\}$ 

2021/036) Simplify : $\sqrt{4 -4a + a^2} + \sqrt{4 + 4a + a^2}$, if $a < -2 $

We have $\sqrt{4 -4a + a^2} = \sqrt{(a-2)^2}= | a- 2 | $

as $a < -2 $ so  $\sqrt{4 -4a + a^2} = 2 - a$

Further we have $\sqrt{4 + 4a + a^2} = \sqrt{(a+2)^2}= | a + 2 | $

as $a < -2 $ so  $\sqrt{4 -4a + a^2} = - 2 - a$

adding we get $\sqrt{4 -4a + a^2}  +  \sqrt{4 + 4a + a^2}= - 2a$ 

2021/035) Let $x,\,y,\,z$ be integers such that $(x-y)^2+(y-z)^2+(z-x)^2=xyz$, prove that $x^3+y^3+z^3$ is divisible by $x+y+z+6$.

 We know $x^3+y^3+z^3 - 3xyz = \frac{1}{2}(x+y+z)((x-y)^2 + (y-z)^2 + (z-x)^2)$

Hence  $x^3+y^3+z^3 = 3xyz + \frac{1}{2}(x+y+z)((x-y)^2 + (y-z)^2 + (z-x)^2)$

Hence $x^3+y^3+z^3 = 3xyz + \frac{1}{2}(x+y+z)(xyz)$ (putting the value from given condition)

Or $x^3+y^3+z^3 = xyz( 3 + \frac{1}{2}(x+y+z))$

Or $x^3+y^3+z^3 = \frac{xyz}{2}( 6 + x+y+z)$

If we can prove that xyz is even then  we are through

As (x-y), (y-z) and (z-x) sum to give zero so atleast one of them is even. So xyz is even from the given condition so  $\frac{xyz}{2}$ is an integer and hence $x^3+y^3+z^3$ is multiple of $(6 + x+y+z)$

2021/034)How do you prove if a, b, and c are integers such that $a|b$ and $c|b$ and a and c are co-primes , then $ac | b$

 Now  $a|b$  so there exists integer d such that $$b = ad\cdots(1)$$

Further  c and a are co-primes so as ber Bezout's identity

There exists integer x and y such that

$cx + ay = 1$

Multiply both sides by d to get

$cx + ayd = d$

or $cx + by = d$ using (1)

as c divides b so there exists integer p such that

$b= pc$

So $cx + pcy = d$

Or $c(x+py) = d$

So from (1) $b = ad = ac(x+py)$ hence ac is factor of b

2021/033) Solve the system of equations

$xy = z - x - y\cdots(1)$

$yz = x - y - z\cdots(2)$

$zx = y - x - z\cdots(3)$

From (1) we have

Solution

x + y + xy = z

or xy + x + y + 1 = z + 1

or $(x+1)(y+1) = z+1\cdots(4)$

similarly from (2) and (3) get

 $(y+1)(z+1) = x+1\cdots(5)$

 $(z+1)(x+1) = y+1\cdots(6)$

 multiplying (4),(5),(56) we get

 $(x+1)^2(y+1)^2(z+1)^2 = (x+1)(y+1)(z+1)$

 so $(x+1)(y+1)(z+1) = 1$ or $(x+1)(y+1)(z+1) = 0$

  $(x+1)(y+1)(z+1) = 0$ gives x+ 1 = 0 or y + 1 = 0 or z + 1 = 0

  x+1 = 0 using (4) and (6)  y+ 1 = z+1 =0

 Similarly y + 1 = 0 gives  x+ 1 = z+1 =0

    and  z + 1 = 0 gives x+1 =  + 1 =0 

    so solution x = -1, y  = -1, z = -1

  product 1 along with (1), (2), (3) give x+1 = y+ 1 = z + 1 = 1 ( that is x= 1, y = 1, z = 1)

  or x + 1 = y+ 1 = -1 and z + 1 = 1( x=y=-2,z = 0)

 or y + 1 = z+ 1 = -1 and x + 1 = 1( y=z = -2,x = 0)

 or x + 1 = z+ 1 = -1 and y + 1 = 1( x=z=-2,y = 0) 

2021/032) Find non -ve integer x and y satisfying $x^3 = y^3 + 2y + 1$

We have $(y+2)^3 = y^3 + 6y^2+12y + 8 > y^3 + 2y + 1$

and $y^3 < y^3 + 2y + 1$

so x = y + 1

or $(y+1)^3 = y^2 + 3y^2 + 3y + 1 = y^2 + 2y + 1$

or  $2y^2 + y = 0$ giving y = 0(other root -ve and so not admissible)  and x = 1

2021/031) Given $(4-a)(4-b)(4-c)(4-d)(4-e)=12$ a,b,c,d,e are negative or positive integers which have different values Find the sum of a+b+c+d+e

First let us factor 12 we have 12 = 2 * 2 * 3.

The above has 3 factors and given expression has 5 factors so
12 = 1 * 1 * 2 * 2 * 3
In the above we ave 5 factors but not all 5 are different and to make them different we can have (knowing that -ve numbers are allowed
12 = 1 * (-1) * 2 * (-2) * 3
The (4-a,4-b,4-c,4-d,4-e) can be chosen as permutation of(1,-1,2,-2,3) giving a sum 3 and so a+b+c+d+e = 20 -3 or 17.

2021/030)Find prime p and positive integers x,y such that $p^x = y^4 + 4$

We get factoring RHS 

$p^x = y^4 + 4 = (y^4 + 4y^2 + 4) - 4y^2$

$= (y^2 - 2y +2)(y^2+ 2y2)$

as RHS is a power of p so each factor is power of p

and clearly $ (y^2 - 2y +2) < (y^2+ 2y+2)$

so $(y^2 - 2y +2) |  (y^2+ 2y+2)\cdots(1)$

Or $(y^2 - 2y +2) |  (y^2 - 2y +2) -  (y^2+ 2y+2)$

Or $(y^2 - 2y +2) |  (y^2 - 2y+2) -  (y^2+ 2y+2)$

Or $(y^2- 2y+2 | 4y\cdots(2)$

Hence $(y^2- 2y+2 | 4y^2\cdots(3)$

From (1) and (3) we get

$(y^2 - 2y +2) |  4(y^2+ 2y+2)- 4y^2$

or $(y^2 - 2y +2) |  8y+8 $

$(y^2 - 2y +2) |  8(y + 1)\cdots(4) $

As y and y+ 1 are co-primes from 2) ad (4)

$(y^2 - 2y +2) |  8\cdots(4) $

 

So we have 4 cases $y^2-2y+2$ = 1 or 2 or 4 or 8

Let us take the csses one by 1

case 1)

$y^2-2y+2 = 1$

or $y^2-2y+1= 0$ or $y= 1$

this gives $p^x = 1^4 + 4 = 5$ giving $p=5,x = 1$

 Case 2)

 

$y^2-2y+2 = 2$

or $y^2-2y = 0$ or $y= 0$ or 2 

$y=2$ gives $p^x = 20$ which s not a power of a prime

Case 3)

 

$y^2-2y+2 = 4$

or $y^2 - 2y - 2= 0$ this does not have integer solution

Case 4)

 

$y^2-2y + 2 = 8$

or $y^2 - 2y - 6= 0$ and this does not have integer solution

So the solution set $p=5,x=1,y=1$

 

2021/029) If (a,b,c) is a Pythagorean triple prove that $60 | abc$

 we have 60 = 3 * 4 * 5.

For 60 to be factor of abc we need to show that 3 is a factor of a or b or c. 4 is a factor
of a or b or c and 5 is a factor of a or b or c. We shall prove the same for
primitive Pythagorean triple

We now prove the 3 one by one and we shall be proving for primitive triples
1) For primitive Pythagorean triple one of the legs is divisible by 3
Proof:
let no leg be divisible by 3
then we have the 2 sides a and b as $a=3n\pm 1$ and $b=3m\pm 1$
so we have $c^2 = a^2 + b^2= (3n\pm 1)^2 + (3m\pm 1)^2 = (9n^2 \pm 6n + 1) + (9m^2 \pm 6m + 1)$
or $c^2 = 3(3n^2+3m^2 \pm 2n \pm 2m) + 2$
but dividing a square number by 3 does not leave a remainder 2 so this is not possibles. So atleast one side shall be divisible by 3.
2) Exactly one of a, b is divisible by 4
before proving this 1st we need to prove that exactly one of a, b is divisible by 2
because it is primitive so both cannot be divisible by 2 so let us assume that both are odd
so let a = 2x + 1 and b= 2y + 1
so $a^2+b^2 = (2x+1)^2 + (2y+1)^2 = 4x^2 + 4x + 1 + 4y^2 + 4y + 1 = 4(x^2+x+y^2+y) +2$
but this cannot be a perfect square as if it is even then it leaves a remainder 0 when divided by 4.
without loss oe generality let a be odd and b even.
Now a is odd so $a= 2x + 1$
There are 2 cases for b. either b is divisible by 4 or not 5
If b is not divisible by 4 then b is of the form $4y+2$
we have 60 = 3 * 4 * 5.
for 60 to be factor of abc we need to show that 3 is a factor of a or b or c. 4 is a factor of a or b or c and 5 is a factor of a or b or c. We shall prove the same for primitive Pythagorean triple
we now prove the 3 one by one and we shall be proving for primitive triples
1) For primitive Pythagorean triple one of the legs is divisible by 3
Proof:
let no leg be divisible by 3
then we have the 2 sides a and b as $a=3n\pm 1$ and $b=3m\pm 1$
so we have $c^2 = a^2 + b^2= (3n\pm 1)^2 + (3m\pm 1)^2 = (9n^2 \pm 6n + 1) + (9m^2 \pm 6m + 1)$
or $c^2 = 3(3n^2+3m^2 \pm 2n \pm 2m) + 2$
but dividing a square number by 3 does not leave a remainder 2 so this is not possibles. So atleast one side shall be divisible by 3.
2) Exactly one of a, b is divisible by 4
before proving this 1st we need to prove that exactly one of a, b is divisible by 2
because it is primitive so both cannot be divisible by 2 so let us assume that both are odd
so let a = 2x + 1 and b= 2y + 1
so $a^2+b^2 = (2x+1)^2 + (2y+1)^2 = 4x^2 + 4x + 1 + 4y^2 + 4y + 1 = 4(x^2+x+y^2+y) +2$
but this cannot be a perfect square as if it is even then it leaves a remainder 0 when divided by 4.
we have 60 = 3 * 4 * 5.
For 60 to be factor of abc we need to show that 3 is a factor of a or b or c. 4 is a factor
of a or b or c and 5 is a factor of a or b or c. We shall prove the same for
primitive Pythagorean triple
We now prove the 3 one by one and we shall be proving for primitive triples
1) For primitive Pythagorean triple one of the legs is divisible by 3
Proof:
let no leg be divisible by 3
then we have the 2 sides a and b as $a=3n\pm 1$ and $b=3m\pm 1$
so we have $c^2 = a^2 + b^2= (3n\pm 1)^2 + (3m\pm 1)^2 = (9n^2 \pm 6n + 1) + (9m^2 \pm 6m + 1)$
or $c^2 = 3(3n^2+3m^2 \pm 2n \pm 2m) + 2$
but dividing a square number by 3 does not leave a remainder 2 so this is not possibles. So atleast one side shall be divisible by 3.
2) Exactly one of a, b is divisible by 4
before proving this 1st we need to prove that exactly one of a, b is divisible by 2
because it is primitive so both cannot be divisible by 2 so let us assume that both are odd
so let a = 2x + 1 and b= 2y + 1
so $a^2+b^2 = (2x+1)^2 + (2y+1)^2 = 4x^2 + 4x + 1 + 4y^2 + 4y + 1 = 4(x^2+x+y^2+y) +2$
but this cannot be a perfect square as if it is even then it leaves a remainder 0 when divided by 4.
without loss oe generality let a be odd and b even.
Now a is odd so $a= 2x + 1$
There are 2 cases for b. either b is divisible by 4 or not 5
If b is not divisible by 4 then b is of the form $4y+2$
we have 60 = 3 * 4 * 5.
for 60 to be factor of abc we need to show that 3 is a factor of a or b or c. 4 is a factor of a or b or c and 5 is a factor of a or b or c. We shall prove the same for primitive Pythagorean triple
we now prove the 3 one by one and we shall be proving for primitive triples
1) For primitive Pythagorean triple one of the legs is divisible by 3
Proof:
let no leg be divisible by 3
then we have the 2 sides a and b as $a=3n\pm 1$ and $b=3m\pm 1$
so we have $c^2 = a^2 + b^2= (3n\pm 1)^2 + (3m\pm 1)^2 = (9n^2 \pm 6n + 1) + (9m^2 \pm 6m + 1)$
or $c^2 = 3(3n^2+3m^2 \pm 2n \pm 2m) + 2$
but dividing a square number by 3 does not leave a remainder 2 so this is not possibles. So atleast one side shall be divisible by 3.
2) Exactly one of a, b is divisible by 4
before proving this 1st we need to prove that exactly one of a, b is divisible by 2
because it is primitive so both cannot be divisible by 2 so let us assume that both are odd
so let a = 2x + 1 and b= 2y + 1
so $a^2+b^2 = (2x+1)^2 + (2y+1)^2 = 4x^2 + 4x + 1 + 4y^2 + 4y + 1 = 4(x^2+x+y^2+y) +2$
but this cannot be a perfect square as if it is even then it leaves a remainder 0 when divided by 4.
without loss oe generality let a be odd and b even.
Now a is odd so $a= 2x + 1$
There are 2 cases for b. either b is divisible by 4 or not 5
If b is not divisible by 4 then b is of the form $4y+2$
So $a^2 + b^2 = (2x+1)^2 + (4y+2)^2 = 4x^2+4x + 1 + 16y^2 + 16y + 4$
$= 4x(x+1) + 1 + 16y(y+1) + 4$
We have $a^2 + b^2 \equiv 5 \pmod 8$
But because if c is odd say c = 2t + 1
$c^2 = (2t+1) ^2$
Or $c^2 = 4t^2 + 4t + 1 = 4t(t+1) + 1$
So $c^2 \equiv 1 \pmod 8$
Hence $a^2 + b^2 \equiv 5 \pmod 8$ is not possible so b is divisible by 4
3) Either a or b or c is divisible by 5
If either a or b is divisible by 5 then we are through. So let us assume that neither a nor b is divisible by 5 . working in mod 5 we have
$x^2 \in \{0,1,4\} \pmod 5$
as a and b are not divisible by 5 so
$a^2 \in \{1,4\} \pmod 5$
$b^2 \in \{1,4\} \pmod 5$
if we choose $a^2=1 \pmod 5$ and $b^2=4 \pmod 5$ or $a^2=4 \pmod 5$ and $b^2=1 \pmod 5$ then only the sum could be a perfect square and and other combination does not satisfy the criteria working on mod 5 (we get 2 or 3) and this gives $c^2=0 \pmod 5$ or c is divisible by 5.
As we have proved that 3 divides a or b so 3 divides abc, 4 divides a or b so 4 divides abc and 5 divides a or b or c so 5 divides abc. hence 60 is a factor or abc.
Proved

2021/028) Find non -ve integer x and y satisfying $x^3 = y^3 + 2y + 1$

We have $y^3 + 2y + 1 > y^3$

Further $(y+2)^3 = y^3 + 6y^2 + 12y + 8 > y^3 + 2y + 1$ 

So we have $(y+2)^3 > x^3 > y^3$

So $x^3 = (y+1)^3 = y^3 + 3y^2 + 3y +1 = y^3 + 2y + 1$

Or $3y^2 + 3y + 1 = 2y + 1$

Or $3y^2 + y = 0$

Or y = 0 as we are considering non -ve roots

So solution x=1, y = 0

2021/027) Show that for all real numbers $x,\,y,\,z$ such that $x+y+z=0$ and $xy+yz+zx=-3$, the expression $x^3y+y^3z+z^3x$ is a constant.

We have (given)

$x+y+z=0 \cdots(1)$

$xy+yz+zx=-3\cdots(2)$

From (1)

$(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy +yz + zx) = 0$

or $x^2 + y^2 + z^2 + 2(-3) = 0$

or $x^2 + y^2 + z^2 = 6\cdots(3)$

Further from (1)

$x+y = -z\cdots(4)$

$y+z = -x\cdots(5)$

$z+x = -y\cdots(6)$

Now let us prove that

$x^3y+ y^3 z + z^3 x = xy^3 + yz^3 + zx^3\cdots(7)$

to prove the same

$x^3y+ y^3 z + z^3 x - (xy^3 + yz^3 + zx^3)$

$= (x^3y - xy^3) + (y^3z - yz^3) + (z^3 x - zx^3)$

$= xy(x^2 - y^2) + yz(y^2 - z^2) + zx(z^2 - x^2)$

$=xy(x+y)(x-y) + yz(y+z)( y-z) + zx(z+x)(z-x)$

$=xy(-z)(x-y) + yz(-x) (y-z) + xz(-y) (z-x)$ using (4), (5), (6)

$= - xyz(x-y) - xyz(y-z) - xyz(z-x)$

= 0

so (7) is true

Now $(x^2 + y^2 + z^2)(xy + yz + zx) = 6 * (-3) $ putting values  from above

Or $x^3y + x^2yz + zx^3 + xy^3 + y^3 z + y^2zx + z^2yx + yz^3 + z^3 x = 18$

or $(x^3y + y^3 z + z^3x) + (xy^3 + yz^3 + zx^3 ) + (x^2yz + xy^2z + xyz^2) = - 18$

or  $(x^3y + y^3 z + z^3x) + (x^3y + y^3z + z^3x ) + (x^2yz + xy^2z + xyz^2) = - 18$ (from (7)

or  $2(x^3y + y^3 z + z^3x) + xyz(x+y+z) = - 18$

or $2(x^3y + y^3 z + z^3x) + xyz. 0 = - 18$ from (1)

or $2(x^3y + y^3 z + z^3x)=  - 18$ from (1)

or $(x^3y + y^3 z + z^3x)  = - 9$

Which is a constant

Hence proved

 

2021/026) Find positive integer n for which $2^n + n | 8^n + n$

We have $8^n + n ^3 = (2^n)^3 + n^3$

So

$2^n + n | 8^n + n^3$

using above and the given condition

$2^n + n |  n^3-n$

Now there are 2 cases

1) $n^3-n = 0$ which gives n = 1 ( as other roots n = -1 and 0 are not positive

2) or $n^3-n > 0$ as for positive integer n >=2 ( n= 1 as already considered) 

as $2^n + n | n^3-n$ so we have $n^3- n \ge 2^n + n$

or $n^3 \ge 2^n + 2n$

we can show by induction that for $n > 10 $ $2^n > n^3$ 

so $2 \le n \ le 9$

by checking out the values from 2 to 9 we get

$ n \in \{2,4,6\}$

so we ahve the solution set

$n \in \{ 1,2,4,6\}$ 

2021/025) Solve the following $\lfloor x\lfloor x\rfloor \rfloor = 10$ in positive x

 Let $\lfloor x \rfloor = n$

So we have $ n \le x < n+1$

So 

$ nx \le \lfloor x \lfloor x \rfloor \rfloor <  x(n+1) $

Ss $n^2 \ le nx $ and $(n+1)^2 > x(n+1)$ we have

$ n^2 \le \lfloor x \lfloor x \rfloor \rfloor <  (n+1)^2 $

Or

$ n^2 \le 10 < (n+1)^2 $

And $ 10 \le nx < 11$

Or n= 3 and putting n = 3 we have $\frac{10}{3} \le x < \frac{11}{3}$

2021/024) if $a^3+b^3+c^3=3abc$ and a,b, c are different then prove that $\frac{(b+c)^2}{3bc} + \frac{(c+a)^2}{3ac}+\frac{(a+b)2}{3ab}=1$

We have $a^3+b^3+c^3=3abc$ so a=b=c or a+b+c = 0

As a,b,c are different so a+b + c = 0

Hence

$\frac{(b+c)^2}{3bc}= \frac{(-a)^2}{3bc}=\frac{a^3}{3abc}$

Similarly we have 

$\frac{(c+a)^2}{3ac}= \frac{b^3}{3abc}$

$\frac{(a+b)^2}{3ab}= \frac{c^3}{3abc}$ 

Adding these 3 we get

$\frac{(b+c)^2}{3bc} + \frac{(c+a)^2}{3ac}+\frac{(a+b)^2}{3ab}= \frac{a^3+b^3+c^3}{3abc} = \frac{3abc}{3abc}=1$

 

2021/023) Find n when $2^n + 105$ is a perfect square

 Let $2^n + 105 = p ^2 $

Let us work mod 3. for a number to be perfect square it as to  1 or zero mod 3. 

So $2^n + 105 \equiv 0 \pmod 3 $ or $2^n + 105 \equiv 0 \pmod 3 $

or $2^n \equiv 0 \pmod 3 $ or $2^n \equiv 1 \pmod 3 $

but $2^n \equiv 0 \pmod 3 $ is not possible 

$2^n \equiv 1 \pmod 3 $

Now $2^x \equiv 1 \pmod 3 $ if x is even and -1 if x is odd

so n is even say 2m

so $2^{2m} + 105 = p^2$

or $p^2- 2^{2m} = 105$

or $(p+2^m)(p - 2^m ) = 105 = 105 * 1 = 35 * 3 = 15 * 7 = 21 * 5$

Now $(p + 2^m) - (p- 2^m) = 2^{m+1}$ a power of 2

from the above the combination we get

$p+2^m = 35$ $p-2^m= 3$ giving p =19 and m = 4 or n = 8 

Or 

$p+2^m = 15$ $p-2^m= 7$ giving p =11 and m = 2 or n = 4

Or $p+2^m = 21$ $p-2^m= 5$ giving p = 13 and m = 3 or n = 6

so n = 4 or 6 or 8  

2021/022)if $S(n)$ is sum of digits of n find n for which $n + S(n) = 2021$

 As $n+ S(n) = 2021$ so $n < 2021$.

Now highest $S(n)$ for number $n < 2021$ is 28 that is when 

n = 1999.

so $ n >= 2021-28$ or $n>=1993$

Now working mod 9 we have $ n \equiv S(n) \pmod 9$

So $n + S(n) \equiv 2n \pmod 9$

so $2n \equiv 2021 \pmod 9$

or $2n \equiv 5 \pmod 9$

as 5 is odd add 9 to get even

so $2n \equiv 14 \pmod 9$

or $n \equiv 7 \pmod 9$

so we need to check for candidates between 1993 and 2021 which are 7 mod 9 and they are 1996, 2005,2014 out of which 1996 and 2014 satisfy the condition 

2021/021)There are two values of a for which the equation $4x^2 +ax+8x+9 = 0$ has only one solution for x.

What is the sum of these values of a?  (A) -16 (B) -8 (C) 0 (D) 8  

say 8+ a =m

$4x^2+ mx + 9$ has only one solution for x only if $4x^2+ mx + 9$ is a perfect square

or $4x^2+ mx + 9= (2x\pm 3)^3$

Taking $(2x+3)^2$ we get m = 12 or 8+a = 12 or a = 4

Taking $(2x-3)^2$ we get m = -12 or 8+a = -12 or a = -20

so sum of a = -16 so ans (D)

2021/020) If a,b,c are three successive terms of an A.P., then prove that$ a^2+8bc=(2b+c)^2$

Because a, b, c are 3  successive terms of an A.P. so we have

b- a = c- b or a = 2b - c

now LHS = $a^2 + 8bc = (2b-c)^2 + 8bc = 4b^2 - 4bc + c^2 + 8bc = 4b^2 + 4bc + c^2 = (2b+c)^2$

2021/019) If $A = \log_{12} 18$ and $B = \log_{(24)}54$ then evaluate, AB + 5(A - B)?

We have

$A = \frac{\log18}{\log12} = \frac{log\,2 + 2\log\,3}{2\log\, 2 + \log\, 3}$

$A = \frac{\log 54}{\log 24} = \frac{log\,2 + 3\log\,3}{3\log\, 2 + \log\, 3}$

putting $x=\log\, 2$ and $y=\log\,3$ we get

$A= \frac{x+2y}{2x+y}$

and $B= \frac{x+3y}{3x+y}$

So 

AB + 5(A - B)

= $\frac{(x+2y)(x+3y)}{(2x+y)(3x+y)} + 5(\frac{(x+2y)(3x+y) - (2x+y)(x+3y)}{ (2x+y)(3x+y)})$

= $\frac{[(x^2+5xy+6y^2) + 5(x^2 - y^2)]}{ (2x+y)(3x+y)}$

= $\frac{6x^2 + 5xy + y^2}{x^2 + 5xy + y^2} =1 $

.

2012/018) For integer n if 4 is not a factor of n the prove that $5 | 1^n + 2^n + 3^n + 4^n$

4 is not a factor of n so there are 2 cases 

1) n is odd 2) n is even of the form 2(2m + 1) where m in integer.

4 is not a factor of n so there are 2 cases 

1) n is odd 2) n is even of the form 2(2m + 1) where m in integer.

Let us deal both the cases one by one

1) Case 1: n is odd

We have rearranging the terms $1^n + 2^n + 3^n + 4^n = (1^n + 4^n) + (2^n + 3^n) $

if n is odd then we have (a+b) is a factor of $a^n+ b^n$ so we get

(1+4) is a factor of $1^n + 4^n$ or 5 is a factor of $1^n + 4^n$

(2+3) is a factor of $2^n + 3^n$ or 5 is a factor of $2^n + 3^n$ 

adding the above 2 we get $5 | 1^n + 2^n + 3^n + 4^n$

2) case 2: n is even of the form 2(2m + 1)

$1^{2 * (2m+1)} + 2^{2 * ( 2* (2m + 1)}  + 3^{2 * (2m + 1)} + 4 ^ {2* (3m+1)} $

$= (1^2)^{2m+1} + (2^2)^{2m + 1}  + (3^2)^{2m + 1} + (4 ^ 2)^{3m+1} $

$= 1^{2m+1} + 4^{2m + 1}  + 9^{2m + 1} + 16^{3m+1} $

using the arguments is case 1 we know

1 + 4 or 5 is a factor of $1^{2m+1} + 4^{2m + 1}$ 

And 9 + 16 or 25 is a factor of $9^{2m + 1} + 16^{2m+1} $ and hence 5 is a factor of  $1^{2m+1} + 4^{2m + 1}  + 9^{2m + 1} + 16^{3m+1} $ or the original expression.

hence proved

as we have proved both parts so we have proved the same

2021/017) find integer n such that n!+3 is a perfect square

we know that for a perfect square x $x^2 \equiv\, 0 or \, 4 \pmod 4$

and for $n \ge 4$  $n! + 3 \equiv\, 3 \pmod 4$

so for  $n \ge 4 $ there is no solution so only n = 0 to 3 need to be checked

$0!+3 = 4 = 2^2$ is a perfect square

$1!+3 = 4 = 2^2$ is a perfect square

$2!+3 = 5 $ is not a perfect square

$3!+3 = 9 = 3^2$ is a perfect square

so n = 0 or 1 or 3