2026/078) Find the remainder when we divide $3^{33}-2$ by 18

We need to factor 18 into product of co-primes $18 = 9 *2 $

Now let us find  $3^{33}-2$ mod 9 and $3^{33}-2$ mod 2 then we shall combine both

Let us proceed one by one 

As $3^{33}$ is  divisible by $3^2=9$ we have 

  $3^{33} \equiv 0 \pmod 9$

So $3^{33} -2 \equiv -2 \pmod 9$ 

Adding 9 on RHS to make is positive

so $3^{33} -2 \equiv 7 \pmod 9$ 

$3^{33} -2$ is odd

So $3^{33} -2 \equiv 1 \pmod 2$ 

To find the remainder when divided by 18 we to find the number $x \lt 18$ such that 

$x \equiv 7 \pmod 9$ 

$x \equiv 1 \pmod 2$ 

This can be solved using Chinese Remainder Theorem but as 2 and 9 are small numbers we can solve

Simply by taking numbers which are 7 mod 9 and checking it is 1 mod 2 or odd.

The number should be less than 18. the 2 numbers are 7 and 16 and 7 is odd

So $3^{33} -2 \equiv 7 \pmod {18}$ 

 

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} $