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A power-full number is defined as a number with $n$ amount of non-zero digits $D_x$. The number takes the form $D_n$&$...$&$D_2$&$D_1$. The digits in a power-full number satisfies the rule
$D_{x+1}$&$D_x=k_x^{x+1}$ where $k_x$ are positive integers. This rule applies for $1\le x\lt n$.

Essentially a power-full number has paired digits that equal some perfect power corresponding to its position in the chain + 1.

Given that the magnitude of a power-full number and $n$ determines how powerful it is (the bigger the better), what is the most powerful number?

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  • $\begingroup$ Feel free to help with formatting if there is an easier way to convey this puzzle $\endgroup$ – Adam May 21 at 13:35
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    $\begingroup$ Do you mean kx are positive integers or is there a single k? $\endgroup$ – Jonathan Allan May 21 at 13:50
  • $\begingroup$ Oops, fixed! @JonathanAllan $\endgroup$ – Adam May 21 at 14:00
  • $\begingroup$ My edits were to make the spelling consistent; now you have both 'power-full' and 'powerful' in your title and body. Pedants may think the definitions are different ... $\endgroup$ – Glorfindel May 21 at 14:27
  • $\begingroup$ @Veedrac $1\le x\lt n$ shouldn't include $n$. I thought it would be more confusing to have $1\le x\le n-1$ $\endgroup$ – Adam May 21 at 15:47
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I think the most powerful number is

1649, $n=4$ (or, if zeros are allowed: 01649, $n=5$)

It's powerful since

49 is a square, 64 a third power, 16 a fourth power (and 01 a fifth power).

In order to have a more powerful number

with $n=6$, it needs to start with two digits forming a sixth power, so either 01 or 64. However, there is no two-digit fifth power starting with 1 or 4; we have only 01 and 32.
Another $n=5$ powerful number without zeros should start with 32, but there's no two-digit fourth power starting with 2, so that's impossible too.
Another $n=4$ powerful number could start with 81, but there are no two-digit third powers starting with one. Similarly, there are no other two-digit third powers starting with 6 than 64, so no improvement can be made there either.

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    $\begingroup$ To complete the demonstration, you should add why you can't have the same n but the leading digit being the number of letters of the 5th word of your answer. (the reason is similar to what you already said) $\endgroup$ – Evargalo May 21 at 15:09
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    $\begingroup$ @Evargalo you're right, thanks; updated $\endgroup$ – Glorfindel May 21 at 15:13

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