How many whole numbers smaller than 100 can you get as a sum of nine consecutive integers?

Question

A) 13 B) 12 C) 11 D) 10 E) 9

ans.?

Answer 1

Let x = starting integer

Then the next consecutive 8 can be represented by x+1, x+2, x+3,x+4, x+5, x+6,x+7, x+8

The sum of those 9 consecutive integers is 9x + 36

Note that the sum from 1 to n is equal to n(n+1)/2.

So, 1+2+3+4+5+6+7+8 = 8(9)/2 = 36.

That sum must be between 0 and 100 including the 0 (because 0 is a whole number).

0 ≤ 9x + 36 < 100

-36 ≤ 9x < 64

-4 ≤ x < 64/9 (or 7 1/9)

So the starting integers are -4, -3, -2, -1, 0, 1, 2, 3, 4 ,5, 6, 7.

Note that if you start at -5:

-5 + (-4) + (-3) + (-2) + (-1) + (0) + 1 + 2 + 3 = -9 which is not a whole number.

If you start at 8:

8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 108 which is a whole number but greater than 100.

Check a starting integer of -4 and a starting integer of 7 to make sure the result is a whole number less than 100.