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by ankush123251 » Thu Dec 15, 2011 11:22 pm
The question is not visible completely...can u post it
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by GmatMathPro » Fri Dec 16, 2011 2:37 am
735^4-735^3+735^2-735 = 735^3(735-1)+735(735-1) = 735^3(734)+735(734) = 734(735^3+735) = 734(735)(735^2+1).

So p divides 734(735)(735^2+1), which means p divides one of those factors.

We also know p divides 735^3+736, which is the same as 735^3+735+1 = 735(735^2+1)+1. This tells us that p divides a number that is one more than a multiple of 735 and one more than a multiple of 735^2+1, which means p CANNOT divide 735 or 735^2+1. Thus, p must divide 734.

Therefore the remainder when 734 is divided by p must be 0
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by batmannavneet » Fri Dec 16, 2011 6:38 am
Let 735 = a, which means 734 = a-1 and 736 = a+1
Therefore, (a^4 - a^3 + a^2 - a) is divisible by p
which means, a^2(a^2 + 1) - a(a^2 + 1) is divisible by p
or, a(a-1)(a^2+1) is divisible by p

We also know that a(a^2+1) + 1 is divisible by p, which means that a(a^2+1) is not divisible by p

Therefore, a-1 must be divisible by p
or 734 must be divisible by p.

Remainder = 0
knight247 wrote:OA is A