2019/020) Express $sin ^7x$ as $a\sin7x+b\sin5x+c\sin3x+d\sin\,x$.

We have $\sin\,x = \frac{e^{ix}-e^{-ix}}{2i}$
To avoid fraction we have
$2i\sin\,x = e^{ix}-e^{-ix}$
Take power 7
$-128i\sin^7 x = (e^{ix}-e^{-ix})^7 = {7 \choose 0} e^{7ix} -   {7 \choose 1} e^{5ix} +  {7 \choose 2} e^{3ix} -  {7 \choose 3} e^{ix}$
                                                         $ +  {7 \choose  4} e^{-ix} -  {7 \choose 5} e^{-3ix} +  {7 \choose 6} e^{-5ix}  -  {7 \choose 7} e^{-7ix}$
$= {7 \choose 0} e^{7ix} -   {7 \choose 1} e^{5ix} +  {7 \choose 2} e^{3ix} -  {7 \choose 3} e^{ix}$
   $ +  {7 \choose  3} e^{-ix} -  {7 \choose 2} e^{-3ix} +  {7 \choose 1} e^{-5ix}  -  {7 \choose 0} e^{-7ix}$
$= {7 \choose 0} (e^{7ix} - e^ {-7ix}) -   {7 \choose 1} (e^{5ix} -  e^{-5ix})+  {7 \choose 2} (e^{3ix} - e^{-3ix}) -  {7 \choose 3} (e^{ix} - e^{-ix})$

so $-64\sin^7 x =  {7 \choose 0} \frac{(e^{7ix} - e^ {-7ix})}{2i} -   {7 \choose 1} \frac{(e^{5ix}-  e^{-5ix})}{2i} +  {7 \choose 2} \frac{(e^{3ix} + e^{-3ix})}{2i} -  {7 \choose 3}\frac{ (e^{ix} + e^{-ix})}{2i}$
 $= 1 \sin\, 7x  -   7 \sin\, 5x +  21 \sin\, 3x - 35 \sin\,x$
or $-64\sin^7 x = 1 \sin\, 7x  -   7 \sin\, 5x +  21 \sin\, 3x - 35 \sin\,x$

Hence
$\sin^7 x = - \frac{1}{64} \sin\, 7x  +\frac{7}{64}\sin\, 5x -  \frac{21}{64}\sin\, 3x + \frac{35}{64} \sin\,x$

2019/019) Solve for positive x $4^x+6^x = 9^x$

we have $4=2^2$, $6= 2 * 3$, $9=3^2$
So we get
$2^{2x}+( 2 * 3)^x = 3^{2x}$
divding by $2^{2x}$ we get
$1 + (\frac{3}{2})^x =   (\frac{3}{2})^{2x}$
Putting $(\frac{3}{2})^x = y$ we get
$1 + y = y^2$
so y = golden ratio $\phi = \frac{1+\sqrt{5}}{2}$
Giving
$(\frac{3}{2})^x = \frac{1+\sqrt{5}}{2}$
taking log on both sides we get
$x= \frac{\log \frac{1+\sqrt{5}}{2}}{\log \frac{3}{2}}$

2019/018) $x^2 + bx + a$ leaves same remainder when divided by x + 2 or x - a (where a ≠ -2). Show that a + b = 2.

We have
$f(x) = x^2 + bx +a$
Same remainder when divided by x + 2 or (x-a)

So $f(-2) = 4- 2b + a = f(a) =  a^2 +ab +a$
Or 4 - 2b = a^2 + ab
Or 2b+ ab = 4 - a^2
Or b(2+a) = (2-a)(2+a)
As a is not -2 so we have b = 2-a or a + b = 2

Alternatively as same remainder when divided by x+ 2 and x -a and it is degree 2 polynomial

We have $f(x) = x^2 + bx + a = (x+2)(x-a) +c $ where  c is a constant
Or  $f(x) = x^2 + bx + a = x^2+(2-a)x + c - 2a$
Comparing coefficient of x we have b= 2- a or a+b =2 

2019/017) Each of the numbers $x_1,x_2,\cdots,x_{101}$ is $\pm1$. what is the smallest positive value of $\sum_{1 \le x_i\lt x_j\le 101 }x_ix_j$

Let the given sum be S

We have $(\sum_{1 \le  n \le 101 }x_n)^2= \sum_{1 \le  n \le 101 }x_n^2 + 2 \sum_{1 \le  x_i\lt x_j\le 101 }x_ix_j  $

so $2S = (\sum_{1 \le  n \le 101 }x_n)^2 - \sum_{1 \le  n \le 101 }x_n^2$

 as each $x_i$ is $pm1$ so $x_i^2 = 1$

so $\sum_{1 \le  n \le 101 }x_n^2 = 101$
so $2S = (\sum_{1 \le  n \le 101 }x_n)^2 - 101$
 For S to be positive we must have
$(\sum_{1 \le  n \le 101 }x_n)^2 \gt   101$ and odd
for S to be smallest we must have
$(\sum_{1 \le  n \le 101 }x_n)^2 $ smallest number greater than 101 and it is true when $(\sum_{1 \le  n \le 101 }x_n)^2 = 121$
this is possible and this give S = 10.

2019/016) Let the rational number $\frac{p}{q}$ be closest to but not equal to$\frac{22}{7}$ among all rational numbers with denominator less than 100. What is the value of $p-3q$

Solution

We have $\lvert \frac{p}{q}-\frac{22}{7}\rvert$ as close to zero
Or $\lvert \frac{7 p- 22q}{7q}\rvert$
Above value is lowest when 7q is close to the highest and 7p-22q is close to the lowest( as both may not be at the same point)
q highest is when q = 99  this gives p = 311 and 7p-22q = 1 so this is lowest when q is highest
so p = 311, q = 99 and p - 3q = 311 - 3 * 99 = 14

2019/015) Let a and b be positive real numbers such that $a+b=1$. Prove that $a^ab^b +a^bb^a <=1$

without loss of generality we can assume $a>=b$
We have
$1= a+ b = a^{a+b} + b^{a+b}$
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

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

2019/014) Find the values of n such that $n^4+4$ is a prime

We have $n^4+4 = n^4+4n^2 + 4 - 4n^2 = (n^2+2)^2 -(2n)^2 = (n^2+2n+2)(n^2-2n +2)$
$n^4+4$ is a prime iff $n^2+2n+2$ is a prime and $n^2-2n+2=1$
$n^2-2n+2=1=>(n-1)^2 = 0$  or n = 1
And for n= 1 $n^2+2n+2=5$ which is a prime
we could also compute $n^4+4= 1 + 4 =5$
So 1 is the only choice  for n

2019/013) The length of perimeter of a right $\triangle$ is 60 inches and the length of altitude perpendicular to the hypotenuse is 12 inches. find the sides of the $\triangle$

Let the sides be a,b,c where c is the hypotenuse. without loss of generality we assume $a>=b$

We have
$a+b+c= 60$
Or $60-c = a+b\cdots(1)$
$a^2 + b^2 = c^2\cdots(2)$
Area of the triangle $\frac{1}{2}ab = \frac{1}{2}12c$ or
$ab= 12c\cdots(3)$
From (1)
$(60-c)^2 = ( a + b)^2 = a^2 + 2ab + b^2 = (a^2+b^2) + 2 * 12c= c^2 + 24c$ (using (2) and (3))
Or $3600 - 120c + c^2  = c^2 + 24c$
Or $144c = 3600$ or $c=25$
From (1) $a+b= 35\cdots(4)$
From (3) $ab= 300\cdots(5)$
So we have $(a-b)^2 = (a+b)^2 - 4ab = 1225 - 1200 = 25$ (using (4) and (5))
So $a-b = 5\cdots(6)$
From (4) and (6) we have a = 20, b = 15,
So sides of triangle $15,20,25$

2019/012) if a,b,c are sides of a triangle with the the property that $a^n,b^n,c^n$ are sides of a triangle for each positive integer n then prove that the triangle is isosceles

Without loss of generality let us assume that $a > b>c$

So $a^n < b^n + c^n$
Or $1 < (\frac{b}{a})^n + (\frac{c}{a})^n$
If $a>b > c$ then $(\frac{b}{a})^n = 0$ and $(\frac{c}{a})^n =0$ when n goes to infinite
So $1 < 0$ which is contradiction
So b has to be same as a for the sum to be 1 in the limit
Or the triangle is isosceles

Proved

2019/011) Given a number that does not end in 0 or 5, prove that it has a multiple that consists of only 6's and 7's.

This question I have solved at https://mathhelpboards.com/challenge-questions-puzzles-28/multiple-consists-only-6s-7s-26532.html#post116812

I am providing a structured solution for the same. (not same as in the link above)

There are 2 cases:
1) The number is odd
2) The number is even

 Now we provide solution for both cases to make the solution complete

1) The number is odd say k

The number is not ending with 5 so the number is not divisible by 5

We take numbers $a_1=1, a_2=11, a_3= 111$  so on upto a number with k 1's that is $a_k = 1111....1111$ (k times) that is $a_n$ is a number with n 1's.
Divide the k numbers by k
We have k remainder 0 and so either one remainder is zero or 2 remainder are same (as there are maximum k remainders 0 to k-1)
We assert that one of the remainders is zero.

If that is not the case then 2 remainders are same say for $a_n$ and $a_m$ with $n > m $ so $a_n-a_m = a_{n-m} * 10^m$ divided by k is zero so $a_{n-m}$ divided by k is zero which is a contradiction

so one of the remainders is zero let it be $a_m$ . which is divisible by k. multiply the number bu 6 or 7 so that all the digits are 6 or 7 and it is a multiple of k which is true

case 2: The number is even that is $k*2^n$ Where k  is odd and m is $>0$

Before that we prove the following
We have $10^n \equiv 2^n \pmod {2^{n+1}}\cdots(1)$

Proof:

$10^n = 5^n * 2^n =  \frac{5^n-1}{2} *2^{n+1} + 2^n\equiv  = 2^n \pmod {2^{n+1}}   $ as $\frac{5^n-1}{2}$ is integer

Now 6 is divisible by 2 and 76 by ${2^2}$

If a number is divisible by $2^n$ then either it divisible by $2^{n+1}$ or remainder when divided by $2^{n+1}$ is $2^n$

If it is divisible by $2^{n+1}$ then  add $6*10^{n+1}$ to it to make is divisible by$ 2^{n+1}$ as both terms are divisible by $2^{n+1}$

If it is not divisible by $2^{n+1}$ that is remainder is $2^n$ then  add $7*10^{n+1}$ to it to make is divisible by$ 2^{n+1}$ as both terms leave a remainder $2^n$ and sum zero  divisible by $2^{n+1}$

So in both cases we have a number consisting  of 7 and 6 (n digits) which is divisible by $2^n$

Note: there my be a smaller number satisfying the criteria with m digits (say y) may be $m < n$

So let is consider the number y with p digits ( digits are 6 or 7) which is divisible by $2^n$

Now we choose k numbers $x_1=1 $, $x_2 = 10^p + 1$,$x_3= 10^{2p}+10^p + 1$, $x_4=10^{3p} + 10^{2p} + 10^p + 1 $  so on $x_{k} = 10^{(k-1)p} + 10^{(k-2)p} \cdots + 1 $

By using same argument as in case 1 we can show that

So one of the $x$ above is divisible by k. multiplying by  y we get a number consisting of 7 and 6 which is a multiple of given number

Proved




-

2019/010) Given A, B, C, integers satisfying $A \log\, 16 + B \log\, 18 + C \log\, 24 = 0$ find minimum value of $A^2 + B^2+C^2$,

We are given
$A \log\, 16 + B \log\, 18 + C \log\, 24 = 0$
or
$A \log\, 2^4 + B \log\, 2* 3^2 + C \log\, 2^3*3 = 0$
or
$A (4\log\, 2) + B (\log\, 2 + 2\log\,  3) + C (3 \log\, 2 + \log\, 3) = 0$
or $\log\, 2( 4 A+ B + 3C) + \log\, 3(2B + C = 0$
as $\log\,2$ and $\log\,3$ are irrational so we must have

$4A + B + 3C = 0\cdots(1)$
$2B + C = 0\cdots(2)$
From (2)
$C = - 2B\cdots(3)$
And putting in (1) we get $4A= 5B\cdots(4)$

Lowest A, B , - C(all positive)  shall give lowest $A^2+B^2+C^2$
From (4) we get
$A=5, B= 4$ hence $C = -8$ from (3)

So lowest $A^2+B^2+C^2= 25 + 16 + 64 = 105$
(A= -5, B = -4, C= 8) gives same value

2019/009) Find all non-negative integers x,y satisfying $(xy-7)^2 = x^2 + y^2$

We have adding 2xy on both sides

$(xy-7)^2 + 2xy = x^2 +y^2 + 2xy = (x+y)^2$
or $(x^2y^2 -14xy + 49) + 2xy = (x+y)^2$
or $(x^2y^2 -12xy + 49)  = (x+y)^2$
$(x^2y^2 -12xy +36 ) + 13 = (x+y)^2$
or $(xy-6)^2 + 13 = (x+y)^2$
or $13 = (x+y)^2 - (xy-6)^2 = (x+y+xy - 6)(x + y -xy + 6)$
or $13 = ((x+1)(y+1) -7)(5 - (x-1)(y-1))$
this is product of 1 and 13 so we have
$(x+1)(y+1) -7 = 13$ and $5 - (x-1)(y-1) = 1$ giving $x+1)(y+1) = 20$ and $(x-1)(y-1) = 6$
giving (x,y) = (3,4) or (4,3)

Or 
$(x+1)(y+1) -7 = 1$ and $5 - (x-1)(y-1) = 13$ giving $x+1)(y+1) = 8$ and $(x-1)(y-1) = -8$
giving (x,y) = (0,7) or (7,0)

So solution set $(x,y) = (0,7)\, or \,(7,0)\, or\, (3,4)\, or\, (4,3)$

2019/008) Show that $n^2(n^2-1)(n^2-4)$ is divisible by 360

Proof:
We have $n^2(n^2-1)(n^2-4)= n^2(n-1)(n+1)(n-2)(n+2) = (n-2)(n-1)n^2(n+1)(n+2)$
$=5! * {n\choose 5}$

Now $5!=120$ Additionally one of (n-2),(n-1),n is divisible by 3.
If n is divisible by 3 then $n^2$ is divisible by 9 or 2 number (n-2) and (n+1) or (n-1) and (n+2) are divisible by 3.
So the product by $3^2$ so the number is divisible by $LCM(120,9)$ or 360
Or alternatively there is an additional 3 so the product is divisible by 120 *3 or 320 

2019/007) Show that if $34x = 43 y$ then x + y is not a prime

if $34 x = 43 y$
then $34x + 34 y = 77 y$
or $34(x+y) = 77 y$
as 77 and 34 are co primes so 77 is divisor of x +y and hence x+y is a multiple of 77. even if it is 77 it is not a prime
so x + y is not a prime

2019/006) if $a^{x-1} = bc, b^{y-1} = ca, c^{z-1} = ab$ show that $xy + yz + zx = xyz$

From $a^{(x-1)} = bc$ we have $a^x= abc$ or $a=(abc)^\frac{1}{x}\cdots(1)$

similarly $b=(abc)^\frac{1}{y}\cdots(2)$
$c=(abc)^\frac{1}{z}\cdots(3)$

multiplying (1) (2) and (3) we get $abc = (abc)^{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}$
or

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

multiplying both sides by xyz we get the result 

2019/005) Prove If $gcd(a,b)=1$ then $gcd(a,b^2)=1$

We have $ax + by = 1$  as per bezout identity 

So$ b = 1 * b = b* (ax + by)$ 
$= bax + b^ 2 y$ 
So $by = baxy + b^2y^2$ 
Or $by + ax = baxy + ax + b^2 y^2$ 
Or $1 = ax(1+by) + b^2 y^2$ 

As we can put 1 as linear combination of $a$ and $b^2$ so $gcd(a,b^2) = 1$

2019/004) If $\frac{a}{b+ c} + \frac{b}{c+a} + \frac{c}{a+b} = 1$ then prove that $\frac{a^2}{b+ c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} = 0$

We are given
$\frac{a}{b+ c} + \frac{b}{c+a} + \frac{c}{a+b} = 1$
Multiplying both sides by (a+b+c) we get
$\frac{a(a+b+c)}{b+ c} + \frac{b(a+b+c)}{c+a} + \frac{c(a+b+c)}{a+b} = a+b+c$
Or
$\frac{a^2}{b+ c} + a + \frac{b^2}{c+a} + b + \frac{c^2}{a+b} + c= a+b+c$
Or
$\frac{a^2}{b+ c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} = 0$

2019/003) if $a^2=b+c$, $b^2=c+a$, $c^2 = a+b$ find $\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1}$

We have
$a^2=b+c$
Hence $a^2+a= a+b+c$
Hence $a(a+1) = a + b+ c$
or $\frac{1}{a+1} = \frac{a}{a+b+c}\cdots(1)$
Similarly $\frac{1}{b+1} = \frac{b}{a+b+c}\cdots(2)$
and $\frac{1}{c+1} = \frac{c}{a+b+c}\cdots(3)$

adding these 3 we get

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

2019/002) Given $x^5+\frac{1}{x^5}$ Show that $x+\frac{1}{x}=3$ where x is real

let $x + \frac{1}{x} = y$ 
so cube both sides 
$x^3 + \frac{1}{x^3} + 3(x + \frac{1}{x}) = y^3$ 
or $x^3 + \frac{1}{x^3} = y^3-3y$ 
and $x^2+ \frac{1}{x^2} = y^2 -2$ 
multiply both to get 
$x^5 + \frac{1}{x} + x + \frac{1}{x^5} = 123 + y = (y^3-3y)(y^2-2) = y^5 - 5y^3 + 6y$ 
or $y^5 - 5y^3 + 5y - 123 =0$ 
or $(y-3)(u^4 + 3y^3 + 4y^2 + 12y + 41) = 0$ 
gives y = 3 or complex solutions 
so $x+ \frac{1}{x} = 3$ if x is real

2019/001) Find smallest n such that n! ends with 2019 zeroes.

n! ends with k zeroes if is is divisible by $2^k$ and $5^k$ but not with $2^{k+1}$ or $5^{k+1}$.
in the product the power 2

but in n! there are lot many 2 than 5 so it should be divisible by $5^k$  but not $5^{k+1}$.
$\lfloor\frac{n}{5}\rfloor$ numbers shall be divisible by 5
$\lfloor\frac{n}{5^2}\rfloor$numbers shall be divisible by $5^2$
$\lfloor\frac{n}{5^3}\rfloor$ numbers shall be divisible by $5^3$

so number of times 5 appears in product n ! = $p(n) =  \sum_{k=1}^{\inf}\lfloor\frac{n}{5^k}\rfloor$

this shall stop when $5^k > n$ so finite number is elements are to be taken.

We need to find an estimate for n and then correct the same.

Based on the we need to estimate n and then correct the same.

out of 25 consecutive numbers 5 numbers are divisible by 5 and one of those 5 is divisible by one more 5 so out of 25 so product of 25 consecutive numbers so 25 consective number product is divisble by $5^6$

so estimate for n = $\frac{2019*25}{6} = 8412.5$ so take 8410 the highest multiple of  5 less than 8412.

so n = 8410

now p(8410) = 2100

so we have overshot by 2100- 2019 = 81

so n should be reduced by $\frac{81*25}{6} = 337$

giving n -= 8410 - 337 = 8073 or highest multiple of 5 below it 8070.

n = 8070 and p(8070) = 2014

we fall short by 5 and next we take P(8075) = 2016 and P(8080) = 2017, P(8085) = 2018, P(8090) = 2019

so correct n = 8090

TITBIT001) Some interesting property of 2019.

Happy New Year 2019.  I have not solved a  problem here and I shall be starting shortly.Here I provide some interesting property of 2019.

2019 is the smallest number that can be expressed as sum of 3 squares of prime number in 6 different ways

2019 = $7^2+11^2+43^2$
        = $7^2 + 17^2 + 41^2$
        = $13^2+13^2 + 41^2$
        =  $11^2+23^2+37^2$
        = $17^2 + 19^2 + 37^2$
        = $23^2+23^2+31^2$