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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
Joined: 16 Aug 2015
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If a positive integer t is not divisible by 5, how many possible diffe  [#permalink]

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27 Mar 2019, 01:04
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Difficulty:

65% (hard)

Question Stats:

39% (01:33) correct 61% (01:36) wrong based on 46 sessions

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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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"Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Expert Joined: 02 Aug 2009 Posts: 7573 Re: If a positive integer t is not divisible by 5, how many possible diffe [#permalink] ### Show Tags 27 Mar 2019, 07:35 1 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 _________________ CEO Joined: 18 Aug 2017 Posts: 3008 Location: India Concentration: Sustainability, Marketing GPA: 4 WE: Marketing (Energy and Utilities) Re: If a positive integer t is not divisible by 5, how many possible diffe [#permalink] ### Show Tags 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 _________________ If you liked my solution then please give Kudos. Kudos encourage active discussions. Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7230 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If a positive integer t is not divisible by 5, how many possible diffe [#permalink] ### Show Tags 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
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Re: If a positive integer t is not divisible by 5, how many possible diffe  [#permalink]

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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.

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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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