Largest nth Power which has n Digits
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Theorem
The largest $n$th power which has $n$ digits is $9^{21}$:
- $9^{21} = 109 \, 418 \, 989 \, 131 \, 512 \, 359 \, 209$
Proof
The $n$th power of $10$ has $n + 1$ digits.
Hence the $n$th power of $m$ such that $m > 10$ has more than $n$ digits.
The $11$th power of $8$ has $10$ digits:
- $8^{10} = 8 \, 589 \, 934 \, 592$
and so $8^n$ where $n > 10$ has fewer than $n$ digits.
Hence the $n$th power of $m$ such that $m < 9$ and $n < 21$ has fewer than $n$ digits.
Finally we note that:
- $9^{22} = 984 \, 770 \, 902 \, 183 \, 611 \, 232 \, 881$
has $21$ digits.
Hence the $n$th power of $9$ such that $n > 21$ has fewer than $n$ digits.
Hence the result.
$\blacksquare$
Historical Note
David Wells, in his Curious and Interesting Numbers, 2nd ed. of $1997$, attributes this result to a certain Friedlander, but no further information has so far been unearthed.
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $109,418,989,131,512,359,209$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $109,418,989,131,512,359,209$