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

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

by Max@Math Revolution » Thu Feb 13, 2020 12:24 am
[GMAT math practice question]

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

1) n is divisible by 2.
2) n is divisible by 3.
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Source: — Data Sufficiency |
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Re: What is the remainder of 9^n-1 when it is divided by 10?

by deloitte247 » Sat Feb 15, 2020 6:27 pm
$$Find\ the\ remainder\ of\ \frac{9^{n-1}}{10}$$
Statement 1: n is divisible by 2.
This implies 'n' is an even number and n-1 is an odd number and 9 raised to the power of any odd number yield an old number.
Given that n is always even;
$$If\ n=2,\ then\ 9^{n-1}=9^{2-1}=9^1=9$$
$$If\ n=4,\ then\ 9^{n-1}=9^{4-1}=9^3=729$$
$$If\ n=6,\ then\ 9^{n-1}=9^{6-1}=9^5=59,049$$
$$If\ n=8,\ then\ 9^{n-1}=9^{8-1}=9^7=4,782,969$$
$$When\ n\ is\ even,\ in\ all\ cases\ of\ 9^{n-1},\ the\ last\ digit\ is\ always\ 9$$
$$Hence,\ \frac{9^{n-1}}{10}\ will\ have\ a\ remainder\ 9,\ thus,\ statement\ 1is\ SUFFICIENT.$$

Statement 1: n is divisible by 3.
This means 'n' can either be even or odd because 6 is even and it is divisible by 3; 9 is odd and it is divisible by 3. So if 'n' is odd, then 'n' will be even.
$$If\ n-1=2,\ then\ 9^2=81$$
$$If\ n-1=4,\ then\ 9^4=6561$$
$$If\ n-1=6,\ then\ 9^6=531,441$$
$$If\ n\ is\ odd,\ then\ \frac{9^{n-1}}{10}will\ have\ remainder\ 1$$
$$If\ n\ is\ even,\ then\ \frac{9^{n-1}}{10}will\ have\ remainder\ 9$$
The information is not enough to arrive at a definite demainder. Hence, statement 2 is NOT SUFFICIENT.

Since statement 1 alone is SUFFICIENT, the correct answer choice is option A
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Re: What is the remainder of 9^n-1 when it is divided by 10?

by Max@Math Revolution » Sun Feb 16, 2020 5:48 pm
=>

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.

9^1 = 9, 9^2 = 81 ~ 1, 9^3 = 729 ~ 9, 9^4 ~ 1, …
The odd number powers of 9 have the units digit 9 and the even number powers of 9 have the units digits 1.

Condition 1) tells us that n is an even number and 9^n – 1 ~ 1 – 1 = 0.
Thus condition 1) is sufficient.

Condition 2)
If n = 3, then we have 9^3 – 1 ~ 9 – 1 = 8. However, if n = 6, then we have 9^6 – 1 ~ 1 – 1 = 0. Therefore, condition 2) does not yield a unique solution.
Condition 2) is not sufficient.

Therefore, A is the answer.
Answer: A

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations,” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A, B, C, or E.
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