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22 Nov 2018, 02:34
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C
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Difficulty:
35%
(medium)
Question Stats:
72% (01:41) correct
28%
(01:24)
wrong
based on 251
sessions
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Are positive integers m and n both multiples of 3?
(1) Two-digit number mn, where m is the tens digit and n is the units digit, is a multiple of 9
(2) n is a multiple of 3
M36-129
(1) Two-digit number mn, where m is the tens digit and n is the units digit, is a multiple of 9
(2) n is a multiple of 3
M36-129
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16 Nov 2021, 09:25
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Bunuel
Official Solution:
Are positive integers \(m\) and \(n\) both multiples of 3?
(1) Two-digit number \(mn\), where \(m\) is the tens digit and \(n\) is the units digit, is a multiple of 9
An integer is a multiple of 9 if the sum of its digit is a multiple of 9. So, \(m+n\) must be a multiple of 9.
If two-digit number \(mn\) is 36, 63 or 99, then the answer is YES;
If two-digit number \(mn\) is 18, 27, 45, 54, 72 or 81, then the answer is NO.
(2) \(n\) is a multiple of 3
No info about \(m\). Not sufficient.
(1)+(2) If \(m\) is not a multiple of 3, then \(m+n=(not \ a \ multiple \ of \ 3) + (a \ multiple \ of \ 3) = (not \ a \ multiple \ of \ 3) \) but we know that \(m+n\) is a multiple of 9, so it's also must be a multiple of 3. Thus, \(m\) must also be a multiple of 3. Sufficient.
Answer: C
General Discussion
Updated on: 17 Jan 2019, 01:32
Originally posted by Archit3110 on 22 Nov 2018, 02:58.
Last edited by Archit3110 on 17 Jan 2019, 01:32, edited 1 time in total.
Last edited by Archit3110 on 17 Jan 2019, 01:32, edited 1 time in total.
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Bunuel
frm 1: mn can be 18,36 so yes and no that mn are multiples of 3 insufficient
frm 2: no info of m is insufficient
frm 1& 2: we can have n as 3,6,9 units digit and mn to be multiple of 9 can be 63,36... sufficient
IMO C is correct
