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What is the remainder when 1+n+n^2 +…+ n^8 is divided by

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What is the remainder when 1+n+n^2 +…+ n^8 is divided by

by Max@Math Revolution » Thu Aug 15, 2019 5:28 pm

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[GMAT math practice question]

What is the remainder when 1+n+n^2 +...+ n^8 is divided by 5?

1) The remainder when n is divided by 5 is 3
2) n is less than 5
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Source: — Data Sufficiency |
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by Max@Math Revolution » Sun Aug 18, 2019 5:13 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 conditions if necessary.

The easiest way to solve remainder questions is to plug in numbers.
The units digits of 3^n for n = 1, 2, 3, 4, ... are 3, 9, 7, 1, 3, 9, 7, 1, ...
So, the units digits of 3^n have period 4:
They form the cycle 3 -> 9 -> 7 -> 1.
Thus, if n has remainder 3 when it is divided by 5, 1+n+n^2 +...+ n^8 has the same remainder as 1 + 3 + 9 + 7 + 1 + 3 + 9 + 7 + 1 = 21 when it is divided by 5. It has a remainder of 1 when it is divided by 5.
Condition 1) is sufficient.

Condition 2)
If n = 1, then 1+n+n^2 +...+ n^8 = 9, which has remainder 4 when it is divided by 5.
If n = 3, then 1+n+n^2 +...+ n^8 has remainder 1 when it is divided by 5.
Since condition 2) doesn't yield a unique solution, it is not sufficient.

Therefore, A is the answer.
Answer: A
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