# Proof

• Sep 8th 2008, 05:09 PM
fifthrapiers
Proof
I can't find the topic I made.. ugh.

Any way I'm trying to prove $\displaystyle 5^{2008}$ always ends in the digits $\displaystyle 25$.

I almost had it.. something like:

$\displaystyle 5(2x+5) = 10x + 25$ . . . this doesn't quite work. I have to tweek that somehow to show it.
• Sep 8th 2008, 07:29 PM
fifthrapiers
Anyone? I know the proof is similar to what I have. I just can't tweek it to get integers x = 1, 2, ... to satisfy that and include all the powers of 5 to show that you get something (something like 1000 + 100 + 25) and hence will always have 00 + 25 which completes the proof.
• Sep 8th 2008, 11:53 PM
Moo
Hello,

You can prove that $\displaystyle 5^x$ always ends by 25 for $\displaystyle x \ge 2$

Do it by induction if you want to (Tongueout)
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It's ok for x=2

Assume $\displaystyle 5^x$ ends up by 25, that is to say $\displaystyle 5^x=100k+25$, where k is a (positive) integer.

$\displaystyle 5^{x+1}=5 \cdot 5^x=5 \cdot (100k+25)=100 \cdot (5k)+5 \cdot (20+5)=100 \cdot (5k)+100+25$

$\displaystyle 5^{x+1}=100 \cdot (5k+1)+25$, which obviously ends up by 25 :D