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If a positive integer t is not divisible by 5, how many possible diffe

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Math Revolution GMAT Instructor
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If a positive integer t is not divisible by 5, how many possible diffe  [#permalink]

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New post 27 Mar 2019, 01:04
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[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

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Re: If a positive integer t is not divisible by 5, how many possible diffe  [#permalink]

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New post 27 Mar 2019, 07:35
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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


100% incorrect means some lack of understanding on remainders.

Remainders will never be negative. They can be added, multiplied, subtracted as per the terms.

If t is not divisible by 5, the remainder could be 1, 2 , 3 or 4.
Now we can replace t in \(t^4\) by th epossible remainders.
1) If t gives remainder 1, \(t^4\) will give \(1^4\) or 1 as the remainder.
2) If t gives remainder 2, \(t^4\) will give \(2^4\) or 16 as the remainder, but 16=3*5+1, so 1 is the remainder again.
3) If t gives remainder 3, \(t^4\) will give \(3^4\) or 81 as the remainder, but 81=16*5+1, so 1 is the remainder again.
4) If t gives remainder 4, \(t^4\) will give \(4^4\) or 6 will be the last digit, so 1 is the remainder again.

Thus, in all cases, 1 is the remainder.

A
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Re: If a positive integer t is not divisible by 5, how many possible diffe  [#permalink]

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New post 27 Mar 2019, 08:20
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


i solved question by substituting value of t as 6
so when t^4 /5 ; we get 1 as remainder
similarly check with t=7,8.
IMO A
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Re: If a positive integer t is not divisible by 5, how many possible diffe  [#permalink]

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New post 29 Mar 2019, 00:17
=>

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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Re: If a positive integer t is not divisible by 5, how many possible diffe  [#permalink]

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New post 02 Apr 2019, 19:11
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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Re: If a positive integer t is not divisible by 5, how many possible diffe   [#permalink] 02 Apr 2019, 19:11
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