The Sum of two number is 15 and sum of their square is 113. The numbers are?

Let the two part be x and (24 - x),
Then, 7x + 5(24 - x) = 146
? 7x + 120 - 5x = 146
? 2x = 26
? x = 13
? First part = x = 13

Taking p = 11
2^p - 1 = 2¹¹ - 1
= 2047

Since 2047 is divisible by 23 it is not prime. Thus, required least positive prime number is 11.

When N is a natural number, then there is only one possible case that N, N + 2, N + 4 are prime numbers.

When N = 3, then N, N + 2, N + 4 = 3, 5, 7 all are primes.

6x² + 6x = 6x( x + 1 ) which is clearly divisible by 6 and 12 as x( x + 1 ) is even.

We can check divisibility of 19⁵ + 21⁵ by 10 by adding the unit digit of 9⁵ + 1⁵ which is equal to 9 + 1 = 10.
So it must be divisible by 10.

Now, for divisibility by 20 we add 19 and 21 which is equal to 40. So, it is clear that it is also divisible by 20.

So 19⁵ + 21⁵ is divisible by both 10 and 20

Let the number be x,
Then x - 4 = 21/x
? x² - 4x - 21 = 0
? x² - 7x + 3x - 21 = 0
? x(x - 7) + 3(x - 7) = 0
? (x - 7) (x + 3) = 0
? x = 7 (Neglecting = -3)

6² -1 = 35, which is divisible by both 5 and 7.

A number that ends in 2 , 3 , 7 or 8 is never a perfect square.

Let the second number be N.
Then, first number = 2N and third number = 4N
? 2N + N+ 4N/3 = 56
? 7N = 3 x 56
? N = 3 x 56/7 = 24
So, the smallest number is 24.

Let x/5 = p and let y when divided by 5 gives q as quotient and 1 as remainder.
Then, y = 5q + 1
Now x = 5p and y = 5q + 1
? x + y = 5p + 5q + 1 = 5(p + q) + 1
So, required remainder = 1