MathRevolution wrote:
[GMAT math practice question]
If a positive integer \(t\) is not divisible by \(5\), how many possible different remainders can \(t^4\) have when it is divided by \(5\)?
A. one
B. two
C. three
D. four
E. five
Since t is not divisible by 5, then t can be expressed as 5k + 1, 5k + 2, 5k + 3, or 5k + 4 for some integer k.
If t = 5k + 1, then t^4 = (5k + 1)^4 = 5m + 1 (note: in the expansion of (5k + 1)^4, all terms is a multiple of 5 except the last term 1^4 = 1, so we can express it as 5m + 1 for some integer m). We see that the remainder is 1 when t^4 is divided by 5.
If t = 5k + 2, then t^4 = (5k + 2)^4 = 5m + 2^4 = 5m + 16. Since 16/5 = 3 R 1. We see that the remainder is 1 when t^4 is divided by 5.
If t = 5k + 3, then t^4 = (5k + 3)^4 = 5m + 3^4 = 5m + 81. Since 81/5 = 16 R 1. We see that the remainder is 1 when t^4 is divided by 5.
If t = 5k + 4, then t^4 = (5k + 4)^4 = 5m + 4^4 = 5m + 256. Since 256/5 = 51 R 1. We see that the remainder is 1 when t^4 is divided by 5.
Therefore, the remainder is always 1 when t^4 is divided by 5 provided that t is not divisible by 5.
Answer: A
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