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What is the remainder when 9^n -1 is divided by 10?

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by Max@Math Revolution » Sun Aug 25, 2019 11:22 pm

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[GMAT math practice question]

What is the remainder when 9^n -1 is divided by 10?

1) n is a multiple of 2
2) n is a multiple of 3
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Source: — Data Sufficiency |
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by Max@Math Revolution » Tue Aug 27, 2019 11:33 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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

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 remainder when 9^n -1 is divided by 10 is the same as the units digit of 9^n -1. This is easily determined from the units digit of 9^n.

9^1 = 9, 9^2 = 81 ~ 1, 9^3 ~ 9, 9^4 ~ 81 ~ 1, ...
So, the units digits of 9n have period 2:
They form the cycle 9 -> 1.
Thus, 9^n has a units digit of 9, if n is an odd number and a units digit if 1, if n is an even number.

Thus, condition 1) is sufficient.

Condition 2) is not sufficient since a multiple of 3 can be either an even or an odd number.

Therefore, A is the answer.
Answer: A
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by deloitte247 » Thu Aug 29, 2019 7:23 pm
Statement 1=> n is a multiple of 2.
This means that 'n' is even when '9' is raised to an even number.
Also, 10 is also an even number and when an even number is divided by even number, we have remainder = 0.
$$e.g\ if\ n=2;\ 9^2-1=80;\ \ \ \ \ \ So,\ \frac{80}{10}=8\ remainder\ 0$$
$$e.g\ if\ n=4;\ 9^4-1=6560;\ \ \ \ \ \ So,\ \frac{6560}{10}=656\ \ remainder\ 0$$
Hence, statement 1 is SUFFICIENT.

Statement 2=> n is a multiple of 3.
This means that 'n' can be either an even or odd number since multiples of 3 can either be even or odd (e.g 3, 6, 9, 12).
So, when 9 is raised to any number (even or odd) in the original expression, we should have a remainder of zero (0) for statement 2 to be sufficient.
$$e.g\ if\ n=3;\ 9^3-1=728;\ \ \ \ \ \ So,\ \frac{728}{10}=72\ \ remainder\ 8$$
$$e.g\ if\ n=6;\ 9^6-1=531441;\ \ \ \ \ \ So,\ \frac{531440}{10}=53144\ \ remainder\ 0$$
From the expression above, with n=odd, the remainder is not zero but with n=even, we have the remainder to be zero. Since the multiple of 3 cannot totally give us a remainder zero (0), then statement 2 is NOT SUFFICIENT.

Therefore, only statement 1 alone is SUFFICIENT. Hence, option A is the current answer.
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