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rpfinley
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This is my first real post and one of the problems I had a 05 Oct 2006, 18:02
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This is my first real post and one of the problems I had a question on how to solve was:

Whats the remainder of 7^381 divided by 5?



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Nsentra
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This is my first real post and one of the problems I had a 05 Oct 2006, 19:59
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rpfinley
This is my first real post and one of the problems I had a question on how to solve was:

Whats the remainder of 7^381 divided by 5?
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Is it the same as 7^1 -> units digit is 7, so remainder will be 2?
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This is my first real post and one of the problems I had a 05 Oct 2006, 20:29
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i got 2 too.


7's power ending digit has the pattern of 7,9,3,1, and it repeats.
you will find 7^381 has the ending digit of 7. hence the remainder is 2.
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cicerone
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This is my first real post and one of the problems I had a Updated on: 25 Sep 2008, 01:29
Originally posted by cicerone on 05 Oct 2006, 22:12.
Last edited by cicerone on 25 Sep 2008, 01:29, edited 1 time in total.
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It must be 2.

7^381/5

(49^190 x 7)/5

Now 49/5 the remainder is 4, but take it as -1

So [(-1)^190x7]/5 =7/5 .

So remainder is 2
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This is my first real post and one of the problems I had a 06 Oct 2006, 07:07
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cicerone
It must be 2.

7^381/5

(49^190 x 7)/5

Now 49/5 the remainder is 4, but take it as -1

So [(-1)^190x7]/5 =7/5 .

So remainder is 2
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ciceron, I don't understand that tactic behind "Now 49/5 the remainder is 4, but take it as -1" what do you mean? Why couldn't we take it as 1^190 then?
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GMAT 1: 740 Q48 V42
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This is my first real post and one of the problems I had a 06 Oct 2006, 08:13
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I don't have a quicker way of doing this, but
Remainder for
7 = 2
7^2 = 4
7^3 = 3
7^4 = 1
7^5 = 2

So, for all powers of 7 in the form 7^(4x+1), where x is an integer, the remainder will be 2.
381 = (4*95)+1

Ans: 2
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Thanks for the replies, I wish I had an answer but the explanations make perfect sense to me.
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jyotsnasarabu
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This is my first real post and one of the problems I had a 06 Oct 2006, 10:03
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rpfinley
This is my first real post and one of the problems I had a question on how to solve was:

Whats the remainder of 7^381 divided by 5?
Show more


7*7*7*7=7^4=the last digit of the result is 1

380/4=95 with remainder as zero

7*7^380/5=7*1/5=2

hence the remainder is 2 since 7^380=(7^95)^4 and 7^4 always results in 1 as the last digit.

Hope its not very confusing an answer



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