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Senior Manager
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For what value of 'n' will the remainder of 351^n and 352^n [#permalink]
 31 Oct 2003, 09:30
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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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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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(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
3 is the answer
(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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Thanks Stolyar!! Your approach sounds better.
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