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M23-12
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16 Sep 2014, 01:18
16 Sep 2014, 01:18
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E
Difficulty:
75%
(hard)
Question Stats:
54% (02:16) correct
46%
(02:03)
wrong
based on 141
sessions
If \(n\) is a positive integer, is \(n^2-n\) divisible by 12 ?
(1) \(n\) is divisible by 11.
(2) \(n\) is divisible by 19.
(1) \(n\) is divisible by 11.
(2) \(n\) is divisible by 19.
Re M23-12
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16 Sep 2014, 01:18
16 Sep 2014, 01:18
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Official Solution:
If \(n\) is a positive integer, is \(n^2-n\) divisible by 12 ?
(1) \(n\) is divisible by 11.
If \(n=11\), then \(n^2-n=n(n-1)=11*10\) and the answer is NO.
If \(n=11*12\), then \(n^2-n=(a \ multiple \ of \ 12) - (a \ multiple \ of \ 12) =(a \ multiple \ of \ 12) \) and the answer is YES.
Not sufficient.
(2) \(n\) is divisible by 19.
If \(n=19\), then \(n^2-n=n(n-1)=19*18\) and the answer is NO.
If \(n=19*12\), then \(n^2-n=(a \ multiple \ of \ 12) - (a \ multiple \ of \ 12) =(a \ multiple \ of \ 12) \) and the answer is YES.
Not sufficient.
(1)+(2) From the above, we have that \(n\) is divisible by \(11*19\):
If \(n=11*19\), then \(n^2-n=n(n-1)=11*19*(209-1)=11*19*208\) and the answer is NO (since \(11*19*208\) is not divisible by 3, it's not divisible by \(12=3*4\) either).
If \(n=11*19*12\), then \(n^2-n=(a \ multiple \ of \ 12) - (a \ multiple \ of \ 12) =(a \ multiple \ of \ 12) \) and the answer is YES.
Not sufficient.
Answer: E
If \(n\) is a positive integer, is \(n^2-n\) divisible by 12 ?
(1) \(n\) is divisible by 11.
If \(n=11\), then \(n^2-n=n(n-1)=11*10\) and the answer is NO.
If \(n=11*12\), then \(n^2-n=(a \ multiple \ of \ 12) - (a \ multiple \ of \ 12) =(a \ multiple \ of \ 12) \) and the answer is YES.
Not sufficient.
(2) \(n\) is divisible by 19.
If \(n=19\), then \(n^2-n=n(n-1)=19*18\) and the answer is NO.
If \(n=19*12\), then \(n^2-n=(a \ multiple \ of \ 12) - (a \ multiple \ of \ 12) =(a \ multiple \ of \ 12) \) and the answer is YES.
Not sufficient.
(1)+(2) From the above, we have that \(n\) is divisible by \(11*19\):
If \(n=11*19\), then \(n^2-n=n(n-1)=11*19*(209-1)=11*19*208\) and the answer is NO (since \(11*19*208\) is not divisible by 3, it's not divisible by \(12=3*4\) either).
If \(n=11*19*12\), then \(n^2-n=(a \ multiple \ of \ 12) - (a \ multiple \ of \ 12) =(a \ multiple \ of \ 12) \) and the answer is YES.
Not sufficient.
Answer: E
Re: M23-12
[#permalink]
25 Nov 2016, 00:48
25 Nov 2016, 00:48
Can you explain how you know that n(n−1)=11∗12∗(11∗12−1) is divisible by 12?
