Just for fun!-IX

Number Theory Level 2

Find the number of trailing zeros in ${ 9 }^{ 999 }+1$

The answer is 1.

1 solution

Aaaaa Bbbbb
Aug 18, 2015

It is very easy to see that: $S=9^{999}+1=(10-1)^{999}+1$ According to newton: $S=S_{1}+10-1+1=S_{1}+10$ So the number of trailing zeros is: $\boxed{1}$

Can you explain what $S_ 1$ is ? Why can't $S _ 1 + 10$ end with 2 zeros?

Calvin Lin Staff - 5 years, 10 months ago

Because all previous items that all have at least two trailing zeros, so when their sum add 10 will get a number with one trailing zero.

aaaaa bbbbb - 5 years, 10 months ago

Right, you should make that more explicit in your solution. Explain that you are expanding it out via the binomial theorem, and also explain where you got the +10 from.

In particular, it should have been $+ {999 \choose 1 } \times 10 \times 1^{998}$ , which would have yielded a +90 instead of a +10.

Calvin Lin Staff - 5 years, 9 months ago

Just for fun!-IX

The answer is 1.

1 solution

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