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15 Nov 2018, 05:21
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
70% (01:27) correct
30%
(01:25)
wrong
based on 105
sessions
History
Date
Time
Result
Not Attempted Yet
If m and n are integers, and m is a multiple of 5, is mn a multiple of 135 ?
(1) m is a multiple of 3.
(2) n is a multiple of 9.
(1) m is a multiple of 3.
(2) n is a multiple of 9.
Updated on: 15 Nov 2018, 05:47
Originally posted by EgmatQuantExpert on 15 Nov 2018, 05:41.
Last edited by EgmatQuantExpert on 15 Nov 2018, 05:47, edited 2 times in total.
Last edited by EgmatQuantExpert on 15 Nov 2018, 05:47, edited 2 times in total.
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Solution
Given:
- • m and n are integers
• m is a multiple of 5
To find:
- • Whether mn is a multiple of 135 or not
Approach and Working:
- • For mn to be a multiple of \(135 = 3^3 * 5\) , either m or n or mn must be a multiple of \(3^3 = 27\)
Analysing Statement 1
- • m is a multiple of 3
Implies, mn need not be a multiple of 27
Therefore, Statement 1 is not sufficient
Analysing Statement 2
- • n is a multiple of 9, implies that mn need not be a multiple of 27
Therefore, Statement 2 is not sufficient
Combining Both Statements
- • From statement 1, we have m is a multiple of 3
• From statement 2, we have n is a multiple of 9
• Combining both we get, mn is a multiple of 27
Therefore, both statements together are sufficient
Hence, the correct answer is Option C.
Answer: C
15 Nov 2018, 05:44
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Bunuel
Since 135 = 100+35 = 5*(20 + 7) = 5*27=5*(3^3), we need to know if mn contains 3^3 in its factorization.
We'll look for statements that give us this information, a Logical approach.
(1) but are there more or less than three 3s in the factorization of m?
Insufficient.
(2) This gives two out of the three 3s that we need. If there is another one, the answer is YES, if not and if m has no 3s then NO
Insufficient.
Together we have (at least) one 3 from m and (at least) two 3s from n, which gives us the 3^3 we need.
Sufficient.
(C) is our answer.
