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If \(n\) is an integer greater than 0, what is the remainder

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by Jay@ManhattanReview » Wed Oct 23, 2019 9:12 pm
BTGmoderatorLU wrote:Source: Princeton Review

If \(n\) is an integer greater than 0, what is the remainder when \(9^{12n+3}\) is divided by 10?

A. 0
B. 1
C. 2
D. 7
E. 9

The OA is E
When an integer is divided by 10, the remainder is the units digit of the integer. So, we must find out the units digit of \(9^{12n+3}\).

Note that 9^1 = 9; 9^2 = 81; 9^3 = ...9; 9^4 = ...1; 9^5 = ...9; ...

So, if the exponent is an even number, the units digit is 1; and when the exponent is an odd number, the units digit is 9. We see that the exponent 12n + 3 is an odd number; thus, the units digit of \(9^{12n+3}\) must be 9.

The correct answer: E

Hope this helps!

-Jay
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by Scott@TargetTestPrep » Tue Oct 29, 2019 6:43 pm
BTGmoderatorLU wrote:Source: Princeton Review

If \(n\) is an integer greater than 0, what is the remainder when \(9^{12n+3}\) is divided by 10?

A. 0
B. 1
C. 2
D. 7
E. 9

The OA is E

The remainder when a number is divided by 10 is the units digit of the number. Thus, we need to determine the units digit of 9^(12n + 3) or 9^(12n) * 9^3.

We see that when 9 is raised to an odd power the units digit is 9 and when raised to an even power the units digit is 1.

Thus, the units digit of 9^(12n) * 9^3 is 9 x 1 = 9.

Answer: E

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