Problem Solving - 43

If the sum of four consecutive positive integers a three digit multiple of 50, the mean of the these integers must be one of x possible values, where x=

(A) 7
(B) 8
(C) 9
(D) 10
(E) more than 10

Answer: C

Suppose four integers are a, a+1, a+2 and a+3
Hence a + (a+1) + (a+2) + (a+3) = 4a+6
Now 4a+6 will be a multiple of 50 i.e 4a+6=50m where m can take any value from {2,3...,19}

But a = (50m-6)/4 = (25m-3)/2 must be an integer, so m must be odd.

Thus m can be any odd integer from 3 to 19
3=1+1*2
19=1+9*2

So there are 9 different values for m, a and 4a+6, as well as (4a+6)/4