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# What is the reminder of 7^381 divided by 5

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Senior Manager
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What is the reminder of 7^381 divided by 5 [#permalink]

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10 Oct 2005, 21:16
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What is the reminder of 7^381 divided by 5
Manager
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10 Oct 2005, 21:25
The remainder is 2. We only need to know the last digit of 7^381 to know the remainder.

last digits are
7^0=1
7^1=7
7^2=9
7^3=3
7^4=1
7^5=7

as you may notice, the last digits repeat every 4 powers, ie., last digit of 7^2 = last digit of 7^6, etc.

381 is one more than 380 (which is divisible by 4), thus 7^381 will have the last digit 7, therefore the remainder will be 2
SVP
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10 Oct 2005, 21:28
chets wrote:
The remainder is 2. We only need to know the last digit of 7^381 to know the remainder. last digits are
7^0=1
7^1=7
7^2=9
7^3=3
7^4=1
7^5=7
as you may notice, the last digits repeat every 4 powers, ie., last digit of 7^2 = last digit of 7^6, etc.

381 is one more than 380 (which is divisible by 4), thus 7^381 will have the last digit 7, therefore the remainder will be 2

this is the best approach. first find the pattren, then AC.
Director
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10 Oct 2005, 21:31
There has to be a pattern here. Let's see -
7^1 = 7 (rem - 2 when div by 5)
7^2 = 49 (rem - 4)
7^3 = 343 (rem - 3)
7^4 = 2401 (rem - 1)
7^5 = 16807 (rem - 2)
7^6 = 117649 (rem - 4)
...
I would stop here and believe that there is a pattern in the remainder, else I'll be doing this problem for whole duration of the test (and may still not complete it in the end! )

7^381 = 7^(4(95) + 1)
so this is the 5 th in the sequence pattern and hence has the remainder as 2
Am I correct! Please say yes!
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10 Oct 2005, 22:30
gsr your answer is correct. But Chets approach is more preferred. Reason, it involves less calculation; hence less prone to error and saves time.

Hope you agree with me.
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10 Oct 2005, 22:40
sudhagar, It's true. I agree. But both methods are same and to make it more explainatory (one won't this liberty in the exam!), i listed out all the digits
10 Oct 2005, 22:40
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