Question

$$\mathsf{\large a_n = \underbrace{1\ldots 1}_{3^n\text{digits}}}$$ $$\mathsf{\text{Prove,}\quad a_n\equiv 0\mod{3a_{n-1}}}$$

Answer 1

I don't like your questions, they take too long..:-)

Answer 2

I tired to convert \(a_n\) into \(a_{n-1}\) but ended up with Geometric Sum.

Neither do I :/

Answer 3

How long now?

Answer 4

Must be close...

Answer 5

I don't think this is a rigorous proof, but here goes...
\(a_n\) has 3 times as many digits as \(a_{n-1}\). so we could write \(a_n\) as:\[a_n=a_{n-1}a_{n-1}a_{n-1}\]therefore dividing \(a_n\) by \(a_{n-1}\) would give us a number that has the form:\[1000...1000...1000...\]and this number is always exactly divisible by 3 because its digit sum (3) is divisible by 3.
therefore:

\(a_n=0\) mod \(3a_{n-1}\)

Answer 6

111111111=/=(111)*(111)*(111)?

Answer 7

sorry, when I write:\[a_n=a_{n-1}a_{n-1}a_{n-1}\]I meant that \(a_n\) can be written as the digits of \(a_{n-1}\) repeated one after the other 3 times.

Answer 8

i.e. the digits of \(a_n\) can be partitioned into three identical groups, each of which has all the digits of \(a_{n-1}\)

Answer 9

\[a_n = 10^{3^n - 3^{n-1}}a_{n-1} + 10^{3^n-2\cdot3^{n-1}}a_{n-1} + 10^{3^n -3\cdot 3^{n-1}}a_{n-1}\]
\[\large a_n = 10^{3^n}a_{n-1}\left(\left(\frac{1}{10}\right)^{3^{n-1}}+\left(\frac{1}{10}\right)^{2\cdot 3^{n-1}}+\left(\frac{1}{10}\right)^{3\cdot 3^{n-1}}\right)\]
\[\large a_n = 10^{3^n - 3^{n-1}}a_{n-1} \left(\frac{\frac{9}{10}}{\frac{10^3-1}{10^3}}\right)\]

Thanks asnaseer, your approach is really good. and thanks estudier for trying. :-)_

Answer 10

yw