The intent of the problem seems to be as follows:
How many integers between 10! and 10!+20, inclusive, are divisible by 3?
6
7
8
9
10
Two number property rules:
(multiple of k) + (multiple of k) = (multiple of k).
(multiple of k) + (non-multiple of k) = (non-multiple of k).
Since 10! = 10*9*8*7*6*5*4*3*2, 10! is a multiple of 3.
To yield subsequent multiples of 3, we must add to 10! consecutive multiples of 3
Options between 10! and 10! + 20, inclusive:
10!
10! + 3
10! + 6
10! + 9
10! + 12
10! + 15
10! + 18.
Total options = 7.
The correct answer is
B.
Last edited by
GMATGuruNY on Fri Sep 05, 2014 11:36 pm, edited 1 time in total.
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