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What is the remainder when 7^n + 2 is divided by 5 (1) when

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What is the remainder when 7^n + 2 is divided by 5 (1) when [#permalink] New post 19 Jun 2008, 03:51
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What is the remainder when 7^n + 2 is divided by 5
(1) when n is divided by 4, the remainder is 1
(2) when n is divided by 3, the remainder is 2

just wanted to confirm if this solution method is optimal
using statement 1
In case of 7^n the unit digit circles as 7,9,3,1
And in each case the unit digit of 7^n would be 7 only, which means 9 will be the unit digit, so in each case remainder is 4
using statement 2
In case of 7^n the unit digit circles as 7,9,3,1
And in each case the unit digit of 7^n can vary and we cannot pin point what is the remainder because multiple values are possible.
Therefore A is sufficient
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Re: DS question [#permalink] New post 19 Jun 2008, 14:22
Quote:
What is the remainder when 7^n + 2 is divided by 5
(1) when n is divided by 4, the remainder is 1
(2) when n is divided by 3, the remainder is 2


I think this approach should work – I mean to study the periodicity of unit digits depending on n. And you are right, the answer is A.

This is more generalised and detailed explanation, just for clarity's sake:

Let’s first examine 7^n+2:
n=0 => 3 => remainder 3
n=1 => 9 => remainder 4
n=2 => 1 => remainder 1
n=3 => 5 => remainder 0
n=4 => 3 => remainder 3


In general:
n=4k => remainder 3
n=4k+1 => remainder 4
n=4k+2 => remainder 1
n=4k+3 => remainder 0

So periodicity with which n gives us different remainders is 4.

Now, first statement gives us that n=4k+1 => we can find the remainder. Thus, suff.

Second statement says that n=3l+2, but this can’t help us, since only representing n as of 4*smth + smth would provide us with unambiguous value for the remainder. So insuff.

As for more efficient solution – I’d like to know it too :)
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Re: DS question [#permalink] New post 19 Jun 2008, 17:09
vdhawan1 wrote:
What is the remainder when 7^n + 2 is divided by 5
(1) when n is divided by 4, the remainder is 1
(2) when n is divided by 3, the remainder is 2


1) n = 4*k+1, k = 0,1,2,3
(7^n+2)(5)=(7*7^4k+2)(5)=(2*49^(2k)+2)(5)=(2*(-1)^2k+2)(5)=4(5) -> suff

2) n = 3*k+2 //k = 0,1,2
(7^n+2)(5) = (49*7^3k+2)(5)=((-1)*(-1)*7^k+2)(5)=(7^k+2)(5) = (2^k+2)(5)
k=0 -> remainder = 3
k=1 -> remainder = 4 -> insufficient

A is the answer
Re: DS question   [#permalink] 19 Jun 2008, 17:09
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