Proof by Binomial theorem

Question

I am supposed to prove using Binomial theorem that number $$11^{10}-1$$ ends with at least with two zeroes.

My solution so far: $$11^{10}-1=(10+1)^{10}-1$$ $$\sum _{i=1}^{10} \binom{10}{i}10^{10-i}1^{i}=10^{2}\left [ \sum _{i=0}^{8} \binom{10}{i}10^{8-i}..\right ]$$ and from this point I do not know how to continue. Can anyone please help me?

https://math.stackexchange.com/questions/231350/number-of-consecutive-zeros-at-the-end-of-11100-1 — lab bhattacharjee, Oct 18 '19 at 14:11

score 4 · Accepted Answer · answered Oct 18 '19 at 13:16

4

You have$$11^{10}=(1+10)^{10}=1+\overbrace{10\times10+\binom{10}2\times10^2+\cdots}^{\text{sum of multiples of }100}$$

answered Oct 18 '19 at 13:16

José Carlos Santos

427,504

okay, so I will rewrite it, but how can it help me? – Peter F. Oct 18 '19 at 13:25
If $11^{10}$ is $1$ plus a multiple of $100$, then $11^{10}-1$ is a multiple of $100$. – José Carlos Santos Oct 18 '19 at 13:27
okay, so I do not use anything what I wrote before? Is it just that simple? – Peter F. Oct 18 '19 at 13:29
Why do you ask that? Do you something in what I wrote that you disagree with? – José Carlos Santos Oct 18 '19 at 13:30
No no, I just thought that this is not enough and that I have to prove it in another way, but as I can see, this is enough, thanks – Peter F. Oct 18 '19 at 13:35

Proof by Binomial theorem

1 Answers1