Question

prove that the sum of five consecutive natural number is even

Answer 1

Let the smallest number of the five numbers be x. Then the numbers are — x, x + 1, x + 2, x + 3, and x + 4. We have to prove or disprove that the sum of these consecutive integers is divisible by 5 without leaving a remainder.

Sum of the consecutive integers

= x+x+1+x+2+x+3+x+4

= 5x+10

= 5(x+2)

5x + 10 = 5(x + 2) implies that the sum of the numbers are divisible by 5 where 5 is the divisor, 5x + 10 is the divident for any integer value of x and x + 2 is the quotient and 0 is the remainder.

Thus it is proved that sum of any five consecutive integer is divisible by five without leaving a remainder.

Checking: Let x = 3. Then, the other numbers are 4, 5, 6, and 7. Then

3 + 4 + 5 + 6 + 7 = 25, which is divisible by 5.

Let x = —10. Then the other integers are —11, —12, —13 and —14. Then

-10 + (—11) + (—12) + (—13) + (—14)

= —60, which is divisible by 5.

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