For what value of 'n' will the remainder of 351^n and 352^n : PS Archive
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# For what value of 'n' will the remainder of 351^n and 352^n

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For what value of 'n' will the remainder of 351^n and 352^n [#permalink]

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31 Oct 2003, 08:30
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For what value of 'n' will the remainder of 351^n and 352^n be the same when divided by 7?

(1) 2 (2) 3
(3) 6 (4) 4
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31 Oct 2003, 09:06
The answer should be 3. Do not know if my reasoning is correct or not.

When 351 is divided by 7 it will give the remainder 1 and 352 will give the remainder 2.

Now take any small number that when divided by 7 will give the remainder 1.

Let us assume 8. This will give 1 remainder.

Now 8^2, 8^3, 8^4 will all give the remainder of 1. So for the given question it does not matter which value of n we pick up for 351^n. It will always give the remainder of 1.

So the question is to find out n in such a manner that 352^n will have a remainder of 1 when divide by 7.

To find that out take another number which when divided by 7 will give a remainder 2. Say 9.

Now 9^2, when divided by 7 will give remainder 4. So 2 can not be answer.

9^3 when divided by 7 will give the remainder 1.

Similarly if we check 9^4 and 9^6, they will not give 1 as remainder when divided by 7.

So IMO, the answer is 3.

Can anybody suggets a different method?
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31 Oct 2003, 10:34
(350+1)^n always gives 1 as a remainder (binomial expansion)
search for the same remainder
(350+2)^2 gives a remainder of 4
(350+2)^3 gives a remainder of 8, or in fact 1

(350+2)^4 gives a remainder of 2
(350+2)^5 gives a remainder of 4
(350+2)^6 gives a remainder of 1

6 matches as well
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31 Oct 2003, 11:17
Thanks Stolyar!! Your approach sounds better.
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