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# Help! I really don't know where to start.

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The positive integers $$N$$ and $$N^2$$ both end in the same sequence of four digits $$abcd$$ when written in base 10, where digit a is not zero. Find the three-digit number $$abc$$.

Apr 1, 2020

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I think we need to use modular arithmetic. This is how far I got $$n^2 - n = n(n - 1)\equiv 0\mod{10000}$$.

I just realized that my "n" needs to be capitalized but whatever... :)

Apr 1, 2020
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Notice that the digits before $$abcd$$ are not at all that important. Beyond that, I'm still thinking. Perhaps try modulo arithmetic? d has to be 1, 5, or 6.

Apr 1, 2020
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Yeah, you and I had the same idea! :)

nchang  Apr 1, 2020
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Hmm... Perhaps it won't help us that much... I mean, what can we do next?

Nevermind. We can use modulo, though why you're doing AIME Problem #8 is beyond me.

Impasta  Apr 1, 2020
edited by Impasta  Apr 1, 2020
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$$N(N-1)$$

one must be divisible by 625 and the other must be divisible by 16. This might lead us anywhere?

Apr 1, 2020
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There aren't that many four digit numbers divisible by 625. Calculator bash time. Due to parity, only check 625 x odd numbers.

Impasta  Apr 1, 2020
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Or don't be an idiot like me and use modulo to find it easily. Your call.

Impasta  Apr 1, 2020