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 $360$