I observed this awkward repetition while making a strategy for the game "odd-eve". Can someone tell me why this happens?

Let \({ a }_{ 1 }\) be the sum of digits of \(x\) in base 10.

Let \({ a }_{ n}\) be the sum of digits of \({ a }_{ n-1}\) in base 10, for integers \(n>1\).

Let \(g\left( x \right) =\lim _{ n\rightarrow \infty }{ { a }_{ n } } \)

I observed that:-

\[g\left( 2 \right) =2\] \[g\left( 4 \right) =4\] \[g\left( 8 \right) =8\] \[g\left( 16\right) =7\] \[g\left( 32 \right) =5\] \[g\left( 64 \right) =1\]

then

\[g\left( 128 \right) =2\] \[g\left( 256\right) =4\] \[g\left( 512 \right) =8\] \[g\left( 1024 \right) =7\] \[g\left( 2048 \right) =5\] \[g\left( 4096 \right) =1\]

and this pattern is repeating again and again.

\[g\left( 8192 \right) =2\] \[g\left( 16384 \right) =4\]

and so on.........

So, The pattern for powers of 2 is \(2,4,8,7,5,1,2,4,8,7,5,1\)

and

The pattern for powers of 3 is \(3,9,9,9,9,9,9,9,9,9,9,9\)

The pattern for powers of 4 is \(4,7,1,4,7,1,4,7,1,4,7,1\)

The pattern for powers of 5 is \(5,7,8,4,2,1,5,7,8,4,2,..\)

Why does it work?

Does it work for powers of all integers?

Does it work in all bases?

## Comments

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TopNewestHint: \( \bmod 9 \), divisibility rule. – Pi Han Goh · 2 years, 3 months ago

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– Raghav Vaidyanathan · 2 years, 3 months ago

One line. That beautiful number \(9\)! +1Log in to reply

I got it. The sum of digits mod 9 is same as the number mod 9.

And 2^n mod 9 is repeating every few terms. – Archit Boobna · 2 years, 3 months ago

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