+0  
 
+1
59
7
avatar+336 

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
 #1
avatar+336 
+1

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
 #2
avatar+210 
0

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
 #3
avatar+336 
0

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

nchang  Apr 1, 2020
 #4
avatar+210 
0

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
 #5
avatar+336 
0

\(N(N-1)\) 

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

 Apr 1, 2020
 #6
avatar+210 
0

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
 #7
avatar+210 
0

Or don't be an idiot like me and use modulo to find it easily. Your call.

Impasta  Apr 1, 2020

15 Online Users

avatar