2026/077)The set M consists of all 7-digit positive integer numbers that contain (in decimal notation) each of the digits 1,3,4,6,7,8 and 9 exactly once. (a) Find the smallest positive difference d of two numbers from M. (b) How many pairs (x,y) with x and y from M are there for which x−y=d?

 

It is 9 . this can be checked 1346798 - 1346789,

This should be divisible by 9 as 2 permutations of a number as they contain same digits so both have same remainder and hence the difference should be divisible by 9 and hence 9 is the answer.

What are the numbers that give a difference 9. The unit digit of larger number shall be 1 less than the tens digit.

The number must end with (2 digits) 43,76,87,98 end smaller number ends with 34,67,78,89

That is 4 sets of number

They must have 5 digits same that can be from rest 5 digits they can be in 120 ways(5 digits can be permuted in 120 ways

So number of pair of numbers 120 * 4 = 480

2026/076) Find multiple of 29 having last 2 digits 47

Because last 2 digits are 47 so the number is of the form 100n+47

Because it is multiple of 29 so it is of the form 29m

So 29m = 100n + 47

Or $29m - 100n = 47\cdots(1)$

Because GCD(29,100) =1 this has got  solution

Let us find 1 in form of 100a + 29b

We have $100 = 29*3 + 13\cdots(2)$

$29 = 2 * 13 + 3\cdots(3)$

$13 =  3 *4+1 \cdots(4)$

From (4)

$1 = 13 - 3 * 4$

$= 13 - 4(29 - 2 *13) = 9 * 13 - 4 * 29$  (from (3))

$= 9*(100-3*29) - 4* 29$ (from (2))

$= 9 *100 - 31 *29\cdots(5)$

 

From (1) and (5) we get

 

$47 = 29 + 18$

 

Or $47 = 29 +  18(9 * 100 - 31 *29)$

 

Or $47 = 29 + 100*(162)  - 29 *558$

 

$= 100 *162 - 29 * 557$

 

$-29 * 557 = 47 - 100 *162$

 

Here we need to make the multiple of 100 positive this shall make multiple of 29 positive

 

We need to add 29 * 100 * 6 (600 is least multiple of 100 above 557 ) to LHS and 29* 6 * 100 on RHS both 17400 to get  

 

$29 * (600 - 557) = 47 + 100 *( 29 * 6 - 162$

 

Or $29 * 43 = 100 * 12 + 47= 1247$

 

This is the smallest multiple of 29 having last 2 digits 47 and we have

 

$29 * (43 +100n) = (12 + 29n) *100 + 47 = 1247 + 2900n$ are all positive numbers when $n >=0$  

 

  

2026/075) The equation $x^3+px+q =0$ where $q\ne 0$ has one root raciprocal of another route , Show that $p+q^2=1$

let roots be $m,\frac{1}{m},n$

Using  Vieta's formula we have

$m +\frac{1}{m} + n=0\cdots(1)$

$ m * \frac{1}{m} + m * n + n *\frac{1}{m} =  p \cdots(2)$

We have sum of product of roots $m* \frac{1}{m} *n$ = - q \cdots(3)$

From(3) $n = -q\cdots(4)$

From (1) $n = - (m + \frac{1}{m})\cdots(5)$ 

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

Or $1+ n * (-n) = p$ using (5)$

Or $1 - n^2 = p$

Or $1 = p + n^2$ 

Or $1 = p +q^2$ using (4)

Proved   

2026/074) What is the largest What is the largest integer n with no repeated digits that is relatively prime to 6?

 Note : 2 numbers a and b are relatively prime (also called coprimes if GCD(a,b) =1

 Let us take the largest no 9876543210. This number is divisible by both 2 and 3. Any permutation shall be divisible by 3 so one digit has to be removed. The one digit number should not be divisible by 3.

The digit that came be removed are 1,2,4,5,7,8.

To keep the number as large as possible the lower digit need to be removed 

Let us remove 1 the smallest digit

So we get 987654320

This is not divisible by 3.But it is divisible by 2

For the number to be odd (that is not divisible by 2) we need an odd digit in unit place

So we get 987654203

We next try to find a larger number if it exists

1st six digits 987654 cannot be changed as if we change we shall have a smaller number

So only change possible is is in 3 digits must contain 0 3(removing them shall not affect as it shall remain divisible by 3) and either 1 or 2 (either can be removed to make non multiple of 3)

203. larger numbers are 230( discarded as it is even), 301, 302( discarded as even)

So the number is 301 and the original required number is 987654301

 

2026/073) How do you find the remainder when $x^10$ is divided by $x^4+x^3+x^2+x+1$

This we can do it by long division,

Here we realize that  

$(x^4+x^3+x^2+x+1)(x-1)= x^5-1$

And $x^{10}-1= (x^5+1)(x^5-1)$

Or   $x^{10}-1= (x^5+1)(x^5-1)= (x^5+1)(x-1)(x^4+x^3+x^2+x+1)$

Or  $x^{10}= (x^5+1)(x-1)(x^4+x^3+x^2+x+1) + 1$

Hence remainder is 1 

 

2026/072) What is the value of $\sin^3 10^\circ +\sin^3 50^\circ −\sin^3 70^\circ$ equal to?

We have Formula for $\sin 3t $

$\sin 3t = 3 \sin t-4 \sin^3 t$

Putting  $10^\circ$, $10^\circ$,$70^\circ$ we get

 $\sin 30^\circ = 3\sin 10^\circ -4 \sin^3 10^\circ$

or   $\frac{1}{2} = 3 \sin 10^\circ -4 \sin^3 10^\circ\cdots(1)$

  $\sin 150^\circ = 3 \sin 50^\circ -4 \sin^3 50^\circ$

or   $\frac{1}{2} = 3 \sin 50^\circ -4 \sin^3 50^\circ\cdots(2)$

$\sin 210^\circ = 3 \sin 70^\circ -4 \sin^3 70^\circ$ 

or   $\frac{-1}{2} = 3 \sin 70^\circ -4 \sin^3 70^\circ\cdots(3)$

Adding (1) , (2) and subtracting (3) we get

$\frac{3}{2} = 3(\sin 10^\circ + \sin 50^\circ - \sin ^70^\circ) + 4(\sin^3 10^\circ +\sin^3 50^\circ −\sin^3 70^\circ) $ 

Or

$\sin^3 10^\circ +\sin^3 50^\circ −\sin^3 70^\circ = \frac{1}{4}(\frac{3}{2} -   3(\sin 10^\circ + \sin 50^\circ - \sin 70^\circ)  \cdots(1)$

Now we need to evaluate  

$\sin 10^\circ + \sin 50^\circ - \sin 70^\circ$

 Using $\sin A + \sin B = 2 \sin\frac{A+B}{2}\cos \frac{A-B}{2}$ we get

 $\sin 50^\circ + \sin 10^\circ = 2 \sin 30^\circ \cos 20^\circ$

$2 * |frac{1}{2} \cos 20^\circ$

$ \cos 20^\circ = \sin 70^\circ $

Or = $\sin 10^\circ + \sin 50^\circ - \sin 70^\circ = 0$

putting in (1) we get

 $\sin^3 10^\circ +\sin^3 50^\circ −\sin^3 70^\circ = -\frac{3}{8} $

 

 

 

 

2026/071) if $k_1,k_2,\cdots k_{15}$ are the roots of the equation $x^{15} - 2x^{14} + 3 x^{13} + \cdots + 15x-16= 0$ evaluate $(1+k_1)(1+k_2)\cdots(1+k_{15})$

 Because $k_1,k_2,\cdots k_{15}$ are the root

Hence $(x-k_1)(x-k_2)\cdots(x-k_{15}) = x^{15} - 2x^{14} + 3 x^{13} + \cdots + 15x-16$ as RHS is a 

degree 15 polynomial and coefficient is 1

Putting $x= -1 $ we get
 

$(-1-k_1)(-1-k_2)\cdots(-1-k_{15}) = -1 - 2 -3 - \cdots - 15-16= -136$
 

hence
 

$(-1)^{15}(1+k_1)(1+k_2),\cdots(-1+k_{15}) = -136$
 

or $(1+k_1)(1+k_2)\cdots(1+k_{15}) = 136$

2026/070) Simplify $\sqrt{8+\sqrt{7}} - \sqrt{8-\sqrt{7}}$

Because $8 + 2 \sqrt(7)$ has $\sqrt(7)$ as one of the terms so the square root is of the form $a + b \sqrt(7)$ where a and b are rational numbers

Squaring we get

$a^2 + 7b^2 + 2ab \sqrt(7) = 8 + 2 \sqrt(7)$

So comparing rational and irrational parts we get

$a^2 + 7b^2 = 8\cdots(1)$

and $ab = 1\cdots(2)$

So a and b both are positive (-ve shall give -ve square root)

From (2) we get

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

Putting in (1) we get

$a^2 + 7 \frac{1}{a^2} = 8$

or $a^4 - 8a^2 +7=0$

or $(a^2-1)(a^2-7) = 0$

as a is rational so $a^2-7=0$ is ruled out and we have $a^2-1=0$

As a is positive a = 1 and so b = 1 from (2)

So $\sqrt{8+\sqrt{7}} = 1+ \sqrt{7}\cdots(3)$

Now Similarly $\sqrt{8 -\sqrt{7}} = \pm (1- \sqrt{7})$ we need to chooses the proper sign  

We need to take the principal root that is the value

 $\sqrt{8 -\sqrt{7}} = \sqrt{7}-1\cdots(4)$

from (3) and (4)

 $\sqrt{8+\sqrt{7}} - \sqrt{8-\sqrt{7}}= 2$

 

 

2026/069) Find the smallest positive three-digit integer n such that $3^n+4^n$ is divisible by 5.

We need to find $3^n+ 4^n \mod 5$

Let us take the $3^n \mod 5$ and see the behaviour As 5 is prime we have

$3^4 \equiv 1 \mod 5$

and also

$4^4 \equiv 1 \mod 5$

so $3^n + 4^n \mod 5 $ have period  4

Let us consider $f(n) = 3^n + 4^n \pmod 5$

We have

$f(0) = 2$

$f(1) = 2$

$f(2) = 0$

$f(3) = 1$

And $f(n) = f(4k+n)$

We see that $f(2)$ is zero so $f(4k+2)$ is zero

So we need to find the smallest k suck that $4k + 2 > 99$ and get $k = 25$ and smallest number $102$

 

2026/068) What is the smallest positive integer that has exactly 24 factors

Basis

To find the number of positive divisors of an integer, find its prime factorization, add one to each exponent, and multiply those sums together.

That is If $N=p_1^{q_1}p_2^{q_2}\cdots p_n^{q_n}$

it shall have $(q_1+1) (q_2+1)\cdots(q_n+1)$ factors

kindy note that $p_1,p_2\cdots p_n$ are relatively prime and not $q_1+1,q_2+1\cdots q_n+1$ 

Solution

Let is factor 24 in as many ways as we can and put N as product of power of primes and for the number to be lower higher power shall be with lower numer

 24 = 24 this gives number $2^23$

24 = 8 * 3 giving $2^7 *3^2= 1152 $=  this is smaller

can we make it smaller 

24 = 4 * 2 *3 = 4 * 3 * 2 giving $ 2^3 *3^2 * 5 = 360$ 

if we try to make a smaller number we get 3 * 2 * 2 * 2 giving $2^2 * 3 * 5 * 7 = 420$ and it is larger

so the ans is $360$   

 

2026/067) Given that $ \sum_{k=1}^{35}\sin\,5k =\tan(\frac{a}{b}) $ where angles are measured in degrees and a and b are relatively prime positive integers that satisfy $\frac{a}{b} < 90$ evaluate a + b

We shall use the formula for sum of sin of angles in AP

We have

$ \sum_{k=1}^{n}\sin\,kx = \dfrac{\sin\frac{nx}{2}\sin\frac{(n+1)x}{2}}{\sin \frac{x}{2}}$

Put n = 35 and $ x = 5^\circ$

To get 

 $ \sum_{k=1}^{35}\sin\,5k^\circ = \frac{\sin\dfrac{175}{2}^\circ\sin\,90^\circ}{\sin \dfrac{5}{2}^\circ}$

$ = \dfrac{\sin\dfrac{175}{2}^\circ}{\cos \dfrac{175}{2}^\circ}=\tan  \dfrac{175}{2}^\circ$

From given condition $a=175,b=2,\dfrac{a}{b} = 87.5 <90$ 

So $a+b = 177$

2026/066) Determine all triples (a,b,c) of positive integers with $a^{b^c} = (a^b)^c$

 We are given $a^{b^c} = a^{bc}$

Or $a^{b^c-bc} = 1$
 

This has got 2 sets of solutions
 

1) a = 1 and b and c can be any positive integer
 

2) a = any positive integer other than 1 and
 

$b^c - bc = 0$
 

Or $b(b^{c-1}) = c$
 

As b is not zero so we have $ b^{c-1} = c $
 

if c = 1 we get b any number
 

Or $b= \sqrt[c-1]c$ $c>=2$
 

The root can be taken to be integer if c-1 = 1 or c =2 giving b = 2
 

So solution set $(1,b,c)$ or $(a,2,2)$ or $(a,b,1)$ where a,b,c are any integers

2026/065) If x and y are positive real numbers that satisfy the equation $x+4\sqrt{xy} -2 \sqrt{x} - 4 \sqrt{y} + 4y =3$ evaluate $\frac{\sqrt{x} + 2\sqrt{y} + 2014}{ 4- \sqrt{x} - 2\sqrt{y}}$

Solution 

We are given

$x+4\sqrt{xy} -2 \sqrt{x} - 4 \sqrt{y} + 4y =3$ 

Adding 1 to both sides

$x+4\sqrt{xy} -2 \sqrt{x} - 4 \sqrt{y} + 4y + 1=4$

Or $ (\sqrt{x} +  2\sqrt{y} -1)^2 = 4$ or $ (\sqrt{x} +  2\sqrt{y} -1) = 2$ as both square roots are positive

So  $ (\sqrt{x} +  2\sqrt{y}) = 3$

So  $\frac{\sqrt{x} + 2\sqrt{y} + 2014}{ 4- \sqrt{x} -  2\sqrt{y}}=  \frac{3 + 2014}{ 4- 3} = 2017$
 

2026/064) Given a,b,c,d are roots of the equation $x^4−7x^3+3x^2−21x+1=0$ Evaluate $(a+b+c)(b+c+d)(c+d+a)(d+a+b)$

Because a,b ,c ,d are roots of the equation $x^4−7x^3+3x^2−21x+1=0$

We have by vieta's formula
 

$a+b+c+d = 7\cdots(1)$
 

Let
 

$f(x) = x^4−7x^3+3x^2−21x+1=(x-a)(x-b)(x-c)(x-d)\cdots(2)$
 

From (1) we have
 

$a+b+c = 7-d\cdots(3)$
 

$b+c+d  = 7 -a\cdots(4)$
 

$c+d+a = 7 -b\cdots(5)$
 

 $d+a +b = 7-c\cdots(6)$
 

From (3), (4), (5),(6) we have
 

$(a+b+c)(b+c+d)(c+d+a)(d+a+b)= (7-d)(7-a)(7- b)(7-c) = f(7)\cdots(7)$ from (2)
 

As $f(x) = x^4−7x^3+3x^2−21x+1$
 

So $f(7) = 7^4 - 7 * 7^3 +3 *7^2 - 21 * 7 + 1 = 1$
 

From (7) and above we have
 

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

 

2026/063) Factor $n^4+6n^3+11n^2+6n+1$

Let $f(n) =  n^4+6n^3+11n^2+6n+1$

This does not have a change of sign so there is no positive root

So it can have -ve root if it has real root it has to be -1

Checking using the rational root theorem $(f(-1) = 1 - 6 + 11 -6 +1 = 1$

So -1  is not a root

So it has to be product of 2 quadratic polynomials 

This is a quartic polynomal. And the coefficients are symmetric.

So if n is a root then $\frac{1}{n}$ is a root. So the coeffcient of$x^2$ and constant should same in both polynomials. However I shall continue as below

$f(n) =  n^4+6n^3+11n^2+6n+1$

$=  (n^4+1) +11n^2+6(n^3+n)$ reordering the terms

$= (n^2+1)^2 -2n^2 +11n^2+6n(n^2+1)$ putting $n^4+1$ in terms of $(n^2+1)$

$=(n^2+1)^2 + 6n(n^2+1) + 9n^2$

$= (n^2+1)^2+ 2(3n)(n^2+1) + (3n)^2$ geting expressing in $a^2+2ab+b^2$ form

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

$= (n^2+3n+1)^2$ putting in standard form  

 

2026/062) Define $f(n)=LCM(1,2,\cdots\,n)$. Determine the smallest positive integer a such that $f(a)=f(a+2)$

This shall require that $a+1$ and $a+2$ do not increase the LCM.

This shall require both $a+1$ and $a +2 $to be composite and neither is a perfect power of a prime. 

Let us explain it for $a+1$ and the same logic holds for $s+ 2$

If $a+1$ is prime then we have not encountered the same in any of the numbers and this shall contribute to LCM and it shall increase. LCM should be multiplied by $a+ 1$ 

If  $a+1 = b^k$ where b is a prime then it should be multiplied by $b$ as $b^(k-1)$ must have come previously but not $b^k$

if $a+1=b^kc^m$ that is product of powers of 2 primes then $b^k$ and $c^m$ has already come and hence it shall not contribute to  a higher LCM

Same for $a+2$     

 So we require smallest n such that n+1, n+ 2 are composite and producut of power of at least 2 primes 

Looking at $1,2,3,4,5,6,7,8,9,10,11,12,13,14,15$

$14 = 2 * 7 $so LCM does not increase

$15 = 3 * 5$ so LCM does not increase

For $(8,9)$ both power of primes ,

 For $(9,10)$, $9$ is power of prime

 So $a = 13$ 

Next $a = 19$ as  $20=2 ^2 * 5, 21 = 3 *7$ do not increase meets criteria  

  

2026/061) Five distinct 2-digit numbers are in a geometric progression. Find the middle term.

Without loss of generality we can assume that nmbers are in increasing sequence 

We have 3 powers of 2 as 2 digit numbers . They are 16,32,64. 

If we start at 10 we get 4 numbers in geometric progression when ratio is 2. 

So we need some ratio less than 2 . Common ratio can be fraction as long as we do not get a fraction after multiplying by common ratio . 

Because we need 5 numbers we need to multiply 4 times . So we start with a 4th power of 2 that is 16 and common ratio $\frac{3}{2}$ 16 giving 16, 24,36,54,81 and middle term is 36.

Basically  5 numbers $a^4,a^3b,a^2b^2,ab^3,b^4$ where a is starting number and $\frac{b}{a}$ as common ratio. form  GP.

 

2026/060) Show that $f(n)=n^5+n^4+1$ is not prime for $n>1$

We shall prove a stronger result. we shall show that

$g(n) = n^{3k+2} + n^{3m+1} + 1\cdots(1)$ is not a prime for $n>1$ and $k+m > 0$

To prove it we shall snow that is is divisible by $n^2+n+1$

We have 

$n^2+n+1 = (n-\omega)(n-\omega^2)..\cdots$  where $\omega$ is cube root of 1

And $w^3 = 1\cdots(3)$

And $w^2+w+1=0\cdots(4)$ 

We shall snow that g(n) is divisible by  $ (n-\omega)$ and $ (n-\omega^2)$

 We get putting $\omega$ in (1) for n 

$g(\omega) =   \omega^{3k+2} + \omega^{3m+1} +1$

$= (\omega^3)^k \omega^2 +   (\omega^3)^m \omega + 1$

$=   \omega^2 +   \omega + 1 = 0$ using (3) and (4)

So $g(n)$ is divisible by  $(n-\omega)$

 Similarly $g(n)$ is divisible by $(n-\omega^2)$

So $g(n)$ is divisible by  $n^2+n+1$

Because  $k+m$ is greater than 0 so at least one of them is greater than zero

So $g(n) > n^2+n+1$ and as $n^2+n+1 > 0$ g(n) is product of 2 numbers neither is 1 

Hence g(n) is composite

 Putting k = m = 1 we get $f(n)$

So f(n) is not prime 

 Proved

 

2026/059) Show that the product of two positive integers of the form $a^2+ab+b^2$ has the same form.

Let two integers be $a^2+ab+b^2$ and $c^2+cd+d^2$

We have

$a^2+ab+b^2 = (a-b\omega)(a-b\omega^2)\cdots(1)$ where $\omega$ is complex cube root of 1

And

$c^2+cd+b^2 = (c-d\omega)(c-d\omega^2)\cdots(2)$ 

 Multiplying (1) by (2) we get

 $(a^2+ab+b^2)( c^2+cd+d^2) = (a-b\omega)(a-b\omega^2)(c-d\omega)(c-d\omega^2)$

Now as $\omega$ is cube root of 1 so we have

$\omega^2+\omega+1 = 0$

Or 

$\omega^2 = - (1+\omega)\cdot(3)$

Also

 $\omega = - (1+\omega^2)\cdot(4)$ 

Now get us calculate $(a-b\omega)(c-d\omega)$

$(a-b\omega)(c-d\omega) = ac -\omega(ad + bc) + bd\omega^2$

$= ac -\omega(ad+bc) -bd(1+\omega)$ from (3)

$= (ac-bd) -\omega(ad+bc+bd)\cdots(5)$

 

Now get us calculate $(a-b\omega^2)(c-d\omega^2)$

$(a-b\omega^2)(c-d\omega^2) = ac -\omega^2(ad + bc) + bd\omega^4$

$= ac -\omega^2(ad+bc) +bd(\omega)$ as $\omega^3=1$ 

$= (ac -\omega(ad+bc) +bd(1+\omega^2)$ from (4)

 $= (ac-bd) -\omega^2(ad+bc+bd)\cdots(6)$

Now   $(a^2+ab+b^2)( c^2+cd+d^2) = (a-b\omega)(a-b\omega^2)(c-d\omega)(c-d\omega^2)$

$=(a-b\omega)(c-d\omega)(a-b\omega^2)(c-d\omega^2)$

$=((ac-bd) -\omega(ad+bc+bd))((ac-bd) -\omega^2(ad+bc+bd))$

$=((ac-bd)^3 +(ac-bd)(ad+bc+db)+(ad+bc+db)^3$                               using $x^3+y^3=(x-\omega y)(x-\omega^2 y)$

= $m^2+mn + n^2$ where  $m= ac-bd$ and $n = ad+bc+bd$

Hence proved  

  

 

2026/058) Find the number of ordered triples of positive integers $(a,b,c)$ such that $6a+10b+15c=3000$

We are given

$6a+10b+15c=3000\cdots(1)$ 

We have rearranging the terms let us get a in terms of others 

$6a = 3000 - 10b-15c = 5(600-2b-2c)$

As the RHS is divisible by 5 and GCD(5,6) = 1 so a is divisible by 5 and so

$a=5x\cdots(2)$ for some x

Again to get b in terms of others

 $10b = 3000 - 6a-15c = 3(1000-2a-5c)$

As the RHS is divisible by 3 and GCD(3,10) = 1 so b is divisible by 3 and so

$b=3y\cdots(3)$ for some y

 Again to get c

 $15c = 3000 - 6a-10b = 2(1500-3a-5b)$

As the RHS is divisible by 2 and GCD(2,15) = 1 so c is divisible by 2 and so

$c=3z\cdots(4)$ for some z

Putting the values a,b,c from (2),(3),(4) respectively in (1)
 

$30x + 30y + 30z = 3000$

Or $x + y + z = 100$

We need to find x,y,z all 3 positive integers sum is 100

Let us assume that there are 100 stones in a line . There are 99 gaps. we can make them into 3 parts of each part containing above one by putting 2 sticks in 99 gaps. this can be done in $99 \choose 2$ ways