Let the number be N.
Therefore N = 5k + 3

On squaring both sides, we get
N² = ( 5k + 3)²
= 5( 5k² + 6k +1 ) + 4
? On dividing x² by 5, the remainder is 4

Let the divisor be x and quotient be y.
Then, the number = xy + 24
Twice the number = 2xy + 48
Now 2xy is completely divisible by x.

On dividing 48 by x remainder is x
? x = 48 - 11
= 37

We know that, a number is divisible by 11 when the sum of the difference between the sum of its digit at even places and sum of its digit at odd places is either 0 or the difference is divisible by 11.

So number is 58129745812974
Sum of digit at odd places = 4 + 9 + 1 + 5 + 7 + 2 + 8 = 36
Sum of digit at even places = 7 + 2 + 8 + 4 + 9 + 1 + 5 = 36
So, required difference = 36 - 36 = 0
? number is divisible by 11.

Number between 0 and 11 which are multiple of 2 or 3
= 11/2 + 11/3 - 11/6 = 5 + 3 - 1
= 7

Number between 0 and -11 which are multiple of 2 or 3
= 11/2 + 11/3 - 11/6 = 5 + 3 - 1
= 7

? Number be 15, including 0

We know that, a number is divisible by 11 when the sum of the difference between the sum of its digit at even places and sum of its digit at odd places is either 0 or the difference is divisible by 11.

For 93455120
Sum of even places = 3 + 5 + 1 + 0 = 9
Sum of odd places = 9 + 4 + 5 + 2 = 20

There difference = 20 - 9 = 11
which is divisible by 11 so 93455120 is divisible by 11.

We know that, ( x^m - a^m ) is divisible by ( x + a ), for even value of m.
? 17²⁰⁰ - 1²⁰⁰ is divisible by 17 +1
? 17²⁰⁰ - 1 is divisible by 18.

Hence when 17²⁰⁰ divided by 18 the remainder is 1.

Remainder of 4¹⁰⁰⁰/7 = (4²)⁵⁰⁰/7 = (16)⁵⁰⁰/7
= 2⁵⁰⁰/7
= Remainder of 4 x 8¹⁶⁶/7
= 4

We know that when m is a odd number ( x^m + a^m ) is divisible by ( x + a ).
? Each one is divisible by (41 + 43).
So Common factor = (41 + 43).

(9¹⁹ + 6 ) / 8 = [( 8 + 1 )¹⁹ + 6] / 8
? Required remainder = ( 1¹⁹ + 6 ) / 8

? Remainder = 7

19¹⁰⁰ = ( 20 -1 )¹⁰⁰
? ( -1 )¹⁰⁰ / 20
? 1 / 20
? Required Remainder = 1