The sum of any seven consecutive numbers is divisible by

Solution: 7

Let x be the first integer. The seven consecutive integers are x, x+1, x+2, x+3, x+4, x+5, x+6.

Their sum is:

x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) + (x+6) = 7x + 21 = 7(x+3)

Because the sum equals 7 times the integer (x+3), it is always divisible by 7 for any integer x.

• Quick check: For x = 0, the sum is 0 + 1 + 2 + 3 + 4 + 5 + 6 = 21 = 7 × 3.

Therefore the sum of any seven consecutive integers is always divisible by 7.