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What is the remainder, after division by 100, of \(7^{10}\)?

Expert replies
Source: — Problem Solving |

by Jay@ManhattanReview » Tue Apr 16, 2019 2:00 am
swerve wrote:What is the remainder, after division by 100, of \(7^{10}\)?

A. 1
B. 7
C. 43
D. 49
E. 70

The OA is D

Source: GMAT Prep
The remainder, after division by 100, of \(7^{10}\) would be the last two digits. For example, 243 divided by 100 would leave a remainder 43.

Let's observe the units digits of 7^(positive integer)

"¢ 7^1 = 7; units digit = 7;
"¢ 7^2 = 49; units digit = 9;
"¢ 7^3 = 243; units digit = 3;
"¢ 7^4 = _ _ _1; units digit = 1;

"¢ 7^5 = _ _ _7; units digit = 7;
"¢ 7^6 = _ _ _9; units digit = 9;
"¢ 7^7 = _ _ _3; units digit = 3;
"¢ 7^8 = _ _ _1; units digit = 1;

We see that the units digits repeat after every four cycles.

Thus, 7^10 = 7^(2*4 + 2)

The units digit of 7^(2*4 + 2) would be the units digit of 7^2, which is 9. The only option eligible is D.

The correct answer: D

Hope this helps!

-Jay
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by Scott@TargetTestPrep » Tue Apr 16, 2019 4:51 pm
swerve wrote:What is the remainder, after division by 100, of \(7^{10}\)?

A. 1
B. 7
C. 43
D. 49
E. 70

The OA is D

Source: GMAT Prep
7^10 = (7^2)^5 = 49^5 = 49^2 x 49^2 x 49 = 2401 x 2401 x 49. Notice that 2401/100 = 24 R 1, so (2401 x 2401 x 49)/100 has the same remainder when (1 x 1 x 49)/100. However, since 1 x 1 x 49 = 49, so the remainder must be 49.

Alternate Solution:

Let's find the units digit of 7^10. We look at the pattern of the units digits of powers of 7. When writing out the pattern, notice that we are concerned ONLY with the units digit of 7 raised to each power.

7^1 = 7

7^2 = 9

7^3 = 3

7^4 = 1

7^5 = 7

The pattern of the units digit of powers of 7 repeats every 4 exponents. The pattern is 7-9-3-1. In this pattern, all positive exponents that are multiples of 4 will produce 1 as its units digit. Thus:

7^8 has a units digit of 1, 7^9 has a units digit of 7 and so 7^10 has a units digit of 9. The only answer choice with a units digit of 9 is D.

Answer: D

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