The number 6 ²ⁿ - 1, where n is any positive integer, is always divisible by?

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)

Let the number be x and (15-x)
Then, x² + (15 - x)²= 113
? x² - 15x + 56 = 0
? (x-7) (x-8) = 0
? x = 8 or x = 7
So, the numbers are 7 and 8

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.

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

Let the value for ? be N
So N x 9 x 7 x 3 = a number with unit digit 1.
Clearly, the minimum value of N is 9.

Let second number be 3N, then, first one is 6N and the third one is 2N.
From question 3N + 6N + 2N = 132
? 11N = 132
? N = 12

? Second number = 3N = 3 x 12 = 36