# perfect numbers and repeated digits

I am looking the way to proceed further in following difficulties. Observe the following:

n = 3 + 3(6) + 3[$6^2$] + 3[$6^3$] ......(1) = 129

Now by writing 129 in base 6, we see the repeated digit 3 up to three times.

If we took one more term 3[$6^3$] in (1), we will get repeated 3 up to 4 times and so on.

Now, my question is, how it is repeated and why this happening like this? is there any reason beside the adding number of terms and having those many threes in base 6.

If we replace 3 by a and 6 by k, for writing n in base k, we get repeated a up to m times, where m is number of terms in (1).

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It is hard to understand your question, Can you please elaborate? Why what is "happening like this"? – Dennis Gulko May 24 '12 at 11:12
This seems to be quite general and to follow directly from the very definition of the decomposition of integers in a given base. – Did May 24 '12 at 11:15
What is the connection to perfect numbers? – Dennis Gulko May 24 '12 at 11:17
I have voted to close. The question is basically asking why $aaa \dots aaa$ is a sequence of $a$'s in base $b$. Answer: because you constructed it that way. – Brandon Carter May 24 '12 at 12:31

Didier's comment should answer the question, but I will expand on it here.

What does 5829 mean (as an ordinary, base-10, number)?

Well, it means $5\times10^3+8\times10^2+2\times10+9$.

And what would 5829 mean in base-$b$ (assume $b\gt9$)?

Well, $5b^3+8b^2+2b+9$.

So, what would aaaa mean in base-$b$?

It would, of course, mean $ab^3+ab^2+ab+a$.

The question is just turning this around and noticing that $a+ab+ab^2+ab^3$, written in base-$b$, comes out aaaa, and, more generally, $a+ab+ab^2+\cdots+ab^k$, written to base-$b$, comes out aaa...a, with a repeated $k+1$ times.

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 you are right. I understand your approach. Thank you and sorry for the late reply. – mr.math May 29 '12 at 5:42