# Finding the sum of the digits of $3^{1000}$ [closed]

Can we work out the sum of the digits of the number $3^{1000}$ without any program?

I just means the way of pure mathematics.

I'm interested in learning more way to solve this problem.Let's solve it together.

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## closed as off-topic by nbubis, Weapon of Choice, anorton, Algebraic Pavel, SurbAug 12 at 22:58

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We have been asked not to discuss Project Euler problems. –  Gerry Myerson Jul 15 '12 at 8:45
@GerryMyerson We should disregard those people. People who wish to do these problems earnestly will try them on their own before (if ever) looking up the answer. Your reasoning would imply that we should consider request to not answer certain questions because they may be exercises from textbooks which could potentially be someone's homework. –  William Jul 15 '12 at 9:16
Please see meta.math.stackexchange.com/questions/1090/… and some of the links and discussion there. –  Gerry Myerson Jul 15 '12 at 10:59
I think this is an interesting question, and pointedly different from the PE way of doing things. However, I don't think SE is really a social site, and I think if you want to make friends there are better places to go. –  Ben Millwood Jul 15 '12 at 11:16
Project Euler questions should not be answered. –  nbubis Aug 12 at 19:58

## 3 Answers

Hint : $3^{1000} = 9^{500}$ , $9 = 10 - 1$ and Binomial Theorem

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$9 = 3^2$, so $3^{1000} = 9^{500}$. But the binomial theorem doesn't help here, computing the number by repeated squaring is less work than computing all the binomial coefficients. –  Daniel Fischer May 6 at 12:50
Thanks for pointing out Btw –  zscoder May 6 at 12:53
There will be a lot of carry, so the coefficients have very little to do with the decimal digits... –  chubakueno Jun 7 at 19:12

I found some unusual pattern, and maybe this can help $3^{1000}=(3^2)^{500}$ Or, $9^{500}$

Lets, look at few initial terms, $9,81,729,6561,59049,531441..$.

If you sum up the digits of these terms, you get either $9,18,27,36..$ as you go on increasing the power. Overall the sum of digits will always be a multiple of 9 with no fixed pattern.

For eg. 59049 adds upto 27, but 531441 adds upto 18.

But this gets a little fixed after 10th power,because digits of $9^{10}$ adds up to 45. digits of $9^{20}$ adds upto 90 which is $45*2$. So,$9^{30}$ will add upto $45*3=135$ and so on.

Now, $500=50*10$, The sum of digits will be $45*50=2250$

I couldn't arrive to this pattern, without a calculator so may be this is not the correct answer, but that's one way to go about it! :)

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The sum of the digits of $9^{30}$ is $99$, the pattern does not continue. –  Daniel Fischer Jul 9 at 14:17
Yup..you are right. My mistake. –  MonK Jul 10 at 2:35

OEIS identifies this sequence as A00416. There doesn't appear to be a nice closed form.

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