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\[x ^{n+1}-(x+1)^{n}=2001\] in positive integers x,n.
@Yahoo! @Australopithecus <-- Lucy?
@unkabogable
are u familiar with modular arithmetic ?
I never did well on the AMC anyway...
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\[x^{n+1}-(x+1)^n+1=2002\]LHS is divisible by \(x\) so .. \(x\) must be a divisor of \(2002\)
I got it
taking mod \(3\) and mod \(x+1\) will be helpful... only solution is \(x=13\) and \(n=2\)
yes
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