MathRevolution wrote:
[GMAT math practice question]
(number properties) \(m * n = 2145\), where \(m\) and \(n\) are positive integers. What is the value of \(m + n\)?
1) \(m\) and \(n\) are two-digit integers.
2) \(m\) and \(n\) both have remainder \(3\) when they are divided by \(4\)
=>
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 \(1\) equation (\(mn=2145\)), D is most likely to be the answer.
Condition 1)
\(m * n = 2145 = 3*5*11*13.\)
Since \(m\) and \(n\) are two-digit numbers, there are four cases to consider:
i) \(m=33 (= 3*11)\) and \(n=65 (= 5*13)\)
ii) \(m=39 (=3*13)\) and \(n=55 (=5*11)\)
iii) \(m=65 (=5*13)\) and \(n=33 (= 3*11)\)
iv) \(m=55 (=5*11)\) and \(n=39 (=3*13)\)
So, there are two possible values of \(m + n\), which are \(98\) and \(94.\)
Condition 1) is not sufficient since it does not yield a unique answer.
Condition 2)
If m = 39 and n = 55, then m + n = 94.
If m = 3 and n = 715(=5*11*13), then m + n = 718.
Condition 2) is not sufficient since it does not yield a unique answer.
Conditions 1) & 2)
\(m * n = 2145 = 3*5*11*13.\)
Condition 1) gives rise to the following four cases for the values of \(m\) and \(n\):
i) \(m=33 (= 3*11)\) and \(n=65 (= 5*13)\)
ii) \(m=39 (=3*13)\) and \(n=55 (=5*11)\)
iii) \(m=65 (=5*13)\) and \(n=33 (= 3*11)\)
iv) \(m=55 (=5*11)\) and \(n=39 (=3*13)\)
Of these, only
\(m=39\) and \(n=55\), and \(m=55\) and \(n=39\) satisfy condition 2)
So, we have a unique answer \(m+n=94\).
Thus, both conditions together are sufficient.
Therefore, C is the answer.
Answer: C
Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
_________________