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Re: If n = 20! + 17, then n is divisible by which of the

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Bunuel wrote:
Walkabout wrote:
If n = 20! + 17, then n is divisible by which of the following?

I. 15
II. 17
III. 19

(A) None
(B) I only
(C) II only
(D) I and II
(E) II and II


20! is the product of all integers from 1 to 20, inclusive, thus it's divisible by each of the integers 15, 17, and 19.

Next, notice that we can factor out 17 from from 20! + 17, thus 20! + 17 is divisible by 17 but we cannot factor out neither 15 nor 19 from 20! + 17, thus 20! + 17 is not divisible by either of them.

Answer: C.


GENERALLY:
If integers a and b are both multiples of some integer k>1 (divisible by k), then their sum and difference will also be a multiple of k (divisible by k):
Example: a=6 and b=9, both divisible by 3 ---> a+b=15 and a-b=-3, again both divisible by 3.

If out of integers a and b one is a multiple of some integer k>1 and another is not, then their sum and difference will NOT be a multiple of k (divisible by k):
Example: a=6, divisible by 3 and b=5, not divisible by 3 ---> a+b=11 and a-b=1, neither is divisible by 3.

If integers a and b both are NOT multiples of some integer k>1 (divisible by k), then their sum and difference may or may not be a multiple of k (divisible by k):
Example: a=5 and b=4, neither is divisible by 3 ---> a+b=9, is divisible by 3 and a-b=1, is not divisible by 3;
OR: a=6 and b=3, neither is divisible by 5 ---> a+b=9 and a-b=3, neither is divisible by 5;
OR: a=2 and b=2, neither is divisible by 4 ---> a+b=4 and a-b=0, both are divisible by 4.

Hope it helps.


Please tell me for the multiplication cases as well like you explained about addition & subtraction.

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