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For positive integers m and n, is 3+3^2+3^3+….+3^{mn+1} di

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by Max@Math Revolution » Thu Jun 27, 2019 1:17 am

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Difficulty—

[GMAT math practice question]

For positive integers m and n, is 3+3^2+3^3+....+3^{mn+1} divisible by 6?

1) m^2 + n^2 has remainder 1 when it is divided by 4.
2) m / n is an integer.
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Source: — Data Sufficiency |
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by Max@Math Revolution » Sun Jun 30, 2019 4:50 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.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify the conditions if necessary.

The statement that 3+3^2+3^3+....+3^{mn} is divisible by 6 is equivalent to the statement that 3+3^2+3^3+....+3^{mn} is divisible by 2, because each of 3, 3^2,3^3, ... and 3^{mn} is divisible by 3. It is also equivalent to the statement that mn is an odd number or (mn+1) is an even number.

Condition 1)
The squares of 1, 2, 3, 4, ... are 1, 4, 9, 16, ..., respectively, and they have remainders of 1, 0, 1, 2, ... , respectively, when they are divided by 4.
Thus, if m^2 + n^2 has a remainder of 2 when it is divided by 4, both m and n are odd integers.
It follows that mn is an odd number, and condition 1) is sufficient.

Condition 2)
If m = 1 and n = 1, then 3^1 + 3^2 = 12 is an even number, and the answer is 'yes'.
If m = 2 and n = 1, then 3^1 + 3^2 + 3^3 = 37 is an odd number, and the answer is 'no'.
Since condition 2) doesn't yield a unique solution, it is not sufficient.

Therefore, A is the answer.
Answer: A
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