Question

show that one and only one out of x, x+4,x+8 is divisible by 5 where x is any integer​

Answer 1

Answer:

Given below.

Step-by-step explanation:

Let x be 1.

∴ sequence = 1, 1+4, 1+8

∴ sequence = 1, 5, 9

Out of these, only one, 5, is divisible by 5.

Let x be 2.

∴ sequence = 2, 2+4, 2+8

∴ sequence = 1, 6, 10

Out of these, only one, 10, is divisible by 5.

Let x be -3.

∴ sequence = -3, -3+4, -3+8

∴ sequence = -3, 1, 5

Out of these, only one, 5, is divisible by 5.

Thus we can see with three examples that only one out of the three is divisible by when x is any integer.

This is because there is a difference of 4 between all the three expressions, and 5 is greater than 4 by one.