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Two number property rules:How many integers between 10! and 10!+20, inclusive, are divisible by 3?
6
7
8
9
10
There's a nice rule that says: If M is divisible by k, and N is divisible by k, then (M + N) is divisible by k.[email protected] wrote:How many integers are divisible by 3 between 10! and 10!+20 inclusive?
A) 6
B) 7
C) 8
D) 9
E) 10
HI Guru/Brent,GMATGuruNY wrote:The intent of the problem seems to be as follows:
Two number property rules:How many integers between 10! and 10!+20, inclusive, are divisible by 3?
A)6
B)7
C)8
D)9
E)10
(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.
Since 3 is a factor of 10!, 10! is divisible by 3. From this point, we can keep adding 3 to 10! to obtain integers that are divisible by 3, so the integers between 10! and 10!+20 inclusive that are divisible by 3 are:[email protected] wrote:How many integers are divisible by 3 between 10! and 10!+20 inclusive?
6
7
8
9
10
