[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
If a positive integer t is not divisible by 5, how many poss
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- Max@Math Revolution
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Since t is NOT divisible by 5, we need only test t = 1, 2, 3 and 4Max@Math Revolution 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
If t = 1, then t� = 1, and 1 divided by 5 leaves remainder 1
If t = 2, then t� = 16, and 16 divided by 5 leaves remainder 1
If t = 3, then t� = 81, and 81 divided by 5 leaves remainder 1
If t = 4, then t� = 256, and 256 divided by 5 leaves remainder 1
So, there's only ONE possible remainder
Answer: A
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If t has remainder 1 when it is divided by 5, t^4=~1^4 has remainder 1 when it is divided by 5.
If t has a remainder 2 when it is divided by 5, t^4=~2^4=~16 has remainder 1 when it is divided by 5.
If t has remainder 3 when it is divided by 5, t^4=~3^4=~81 has remainder 1 when it is divided by 5.
If t has remainder 4 when it is divided by 5, t^4=~4^4=~256 has remainder 1 when it is divided by 5.
t^4 has the unique remainder, which is 1, for all values of t.
Therefore, the answer is A.
Answer: A
If t has remainder 1 when it is divided by 5, t^4=~1^4 has remainder 1 when it is divided by 5.
If t has a remainder 2 when it is divided by 5, t^4=~2^4=~16 has remainder 1 when it is divided by 5.
If t has remainder 3 when it is divided by 5, t^4=~3^4=~81 has remainder 1 when it is divided by 5.
If t has remainder 4 when it is divided by 5, t^4=~4^4=~256 has remainder 1 when it is divided by 5.
t^4 has the unique remainder, which is 1, for all values of t.
Therefore, the answer is A.
Answer: A
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If 't' is not divisible by 5, remainder will be either 1,2,3 or 4.
Checking for all possible remainders, we can replace 't' in 't^4' with all possible remainder.
$$If\ t=1,\ then\ t^4=1^4=1\ as\ the\ remainder.$$
$$If\ t=2,\ then\ t^4=2^4=16\ as\ the\ remainder.$$
But, 16= (3*5) + 1. So, 1 is the remainder
$$If\ t=3,\ then\ t^4=3^4=81\ as\ the\ remainder\ but\ \left(16\cdot5\right)+1=81.$$
So, the remainder is 1.
$$If\ t=4,\ then\ t^4=4^4=256\ as\ the\ remainder\ but\ \left(51\cdot5\right)+1=256$$
The remainder is also 1.
Therefore, in all case, 1 (one) is the remainder.
Hence, option A is correct
Checking for all possible remainders, we can replace 't' in 't^4' with all possible remainder.
$$If\ t=1,\ then\ t^4=1^4=1\ as\ the\ remainder.$$
$$If\ t=2,\ then\ t^4=2^4=16\ as\ the\ remainder.$$
But, 16= (3*5) + 1. So, 1 is the remainder
$$If\ t=3,\ then\ t^4=3^4=81\ as\ the\ remainder\ but\ \left(16\cdot5\right)+1=81.$$
So, the remainder is 1.
$$If\ t=4,\ then\ t^4=4^4=256\ as\ the\ remainder\ but\ \left(51\cdot5\right)+1=256$$
The remainder is also 1.
Therefore, in all case, 1 (one) is the remainder.
Hence, option A is correct