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.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
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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