If m and n are positive integers, is m+n divisible by 15?

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[GMAT math practice question]

If m and n are positive integers, is m+n divisible by 15?

1) m is divisible by 9 and n is divisible by 15.
2) mn is divisible by 225.
Source: — Data Sufficiency |

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m and n

by GMATGuruNY » Thu Mar 21, 2019 2:06 am

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Max@Math Revolution wrote:[GMAT math practice question]

If m and n are positive integers, is m+n divisible by 15?

1) m is divisible by 9 and n is divisible by 15.
2) mn is divisible by 225.
Rule:
MULTIPLE OF X + MULTIPLE OF X = MULTIPLE OF X
MULTIPLE OF X + NON-MULTIPLE OF X = NON-MULTIPLE OF X


225 = 9*25

Statements combined:
Case 1: m=9 and n=25*15, with the result that m is a multiple of 9, n is a multiple of 15, and mn = 9*25*15 = 225*15 = multiple of 225
In this case, m+n = NON-MULTIPLE OF 15 + MULTIPLE OF 15 = NON-MULTIPLE OF 15, so the answer to the question stem is NO.

Case 2: m=15*9 and n=25*15, with the result that m is a multiple of 9, n is a multiple of 15, and mn = 15*9*25*15 = 15*225*15 = multiple of 225
In this case, m+n = MULTIPLE OF 15 + MULTIPLE OF 15 = MULTIPLE OF 15, so the answer to the question stem is YES.

Since the answer is NO in Case 1 but YES in Case 2, the two statements combined are INSUFFICIENT.

The correct answer is E.
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by Max@Math Revolution » Sun Mar 24, 2019 5:02 pm

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=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 2 variables (m and n) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
If m = 45 and n = 45, then m + n = 90 is divisible by 15 and the answer is 'yes'.
If m = 9 and n = 225, then m + n = 234 is not divisible by 15 and the answer is 'no'.
Thus, both conditions together are not sufficient, since they do not yield a unique solution.

Therefore, E is the answer.
Answer: E

In cases where 3 or more additional equations are required, such as for original conditions with "3 variables", or "4 variables and 1 equation", or "5 variables and 2 equations", conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.