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Man Cat 4 #14-Base 7 remainder

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Manager
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Man Cat 4 #14-Base 7 remainder [#permalink] New post 19 May 2009, 22:57
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If x is a positive integer, what is the remainder when 7^(12x+3) + 3 is divided by 5?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
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Re: Man Cat 4 #14-Base 7 remainder [#permalink] New post 19 May 2009, 23:56
It should be 0 since the unit digit of expression is 0.
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Re: Man Cat 4 #14-Base 7 remainder [#permalink] New post 20 May 2009, 00:09
We know x > 1

we know that 7^1/5 = 2, 7^2/5 = 4, 7^3/5 = 4,7^4/5 = 4. This goes on in a cycle for every 4.

so x = 1 7^15/5( divide 15 by 4 rem = 3) + 3/5 = 3+ 3 = 6/5 = 1
so x = 2 7^27/5( divide 15 by 4 rem = 3) + 3/5 = 3+ 3 = 6/5 = 1
so x = 3 7^39/5( divide 15 by 4 rem = 3) + 3/5 = 3+ 3 = 6/5 = 1

I think Ans. B
Manager
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Re: Man Cat 4 #14-Base 7 remainder [#permalink] New post 20 May 2009, 07:57
Let x=1.

7^15 + 3

7 follows cyclicity of 4 and hence 7^15 will have 3 as Unit digit. So sum will be 3+3 =6

Hence remainder is 1.

Hence B
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Re: Man Cat 4 #14-Base 7 remainder [#permalink] New post 20 May 2009, 09:41
agree with B

7,9,3,1 is the cycle for powers of 7

so 12x+3 gives 15, 27, 39..etc, which produce factors of seven ending in 3

3+3=6 if we divide by 5 the remainder will be 1
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Re: Man Cat 4 #14-Base 7 remainder [#permalink] New post 21 May 2009, 19:40
bigtreezl wrote:
agree with B

7,9,3,1 is the cycle for powers of 7

so 12x+3 gives 15, 27, 39..etc, which produce factors of seven ending in 3

3+3=6 if we divide by 5 the remainder will be 1


Now, I understand. I just have to deal with the last digit. Thanks! :)
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Re: Man Cat 4 #14-Base 7 remainder [#permalink] New post 18 Oct 2009, 16:12
B
bigtreezl has it good at generalization.
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Re: Man Cat 4 #14-Base 7 remainder [#permalink] New post 21 Oct 2009, 10:56
bigtreezl wrote:
agree with B

7,9,3,1 is the cycle for powers of 7

so 12x+3 gives 15, 27, 39..etc, which produce factors of seven ending in 3

3+3=6 if we divide by 5 the remainder will be 1


Well 7^15 = 7^5 (taking last digit into account) will result in 7
7^27 = 7^7 will result in 3
7^9 wil result in 7 .. so we have end results of 7 and 3 both .. isn't it ?
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Re: Man Cat 4 #14-Base 7 remainder   [#permalink] 21 Oct 2009, 10:56
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Man Cat 4 #14-Base 7 remainder

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