# Dudeney's Property of 2592

## Theorem

$2592 = 2^5 \times 9^2$

It is the only number $n$ that has the property that:

$n = \sqbrk {abcd} = a^b \times c^d$

where $\sqbrk {abcd}$ denotes the decimal representation of $n$.

## Proof

First we verify that $2592$ does indeed satisfy the given property.

 $\ds 2592$ $=$ $\ds 2^5 \times 3^4$ Prime Decomposition of $2592$ $\ds$ $=$ $\ds 2^5 \times \paren {3^2}^2$ $\ds$ $=$ $\ds 2^5 \times 9^2$

$\Box$

It remains to be shown that this is the only such number.

Because $\sqbrk {abcd} = a^b \times c^d$, neither $a^b$ nor $c^d$ can have more than $4$ digits.

Hence, for each digit, the highest power is:

 $\ds 1^9$ $=$ $\ds 1$ $\ds 2^9$ $=$ $\ds 512$ $\ds 3^8$ $=$ $\ds 6561$ $\ds 4^6$ $=$ $\ds 4096$ $\ds 5^5$ $=$ $\ds 3125$ $\ds 6^5$ $=$ $\ds 7776$ $\ds 7^4$ $=$ $\ds 2401$ $\ds 8^4$ $=$ $\ds 4096$ $\ds 9^4$ $=$ $\ds 6561$

Neither $a$ or $c$ can be zero, or that would make $\sqbrk {abcd} = 0$.

Suppose $a = 1$ or $b = 0$.

Then:

$\sqbrk {abcd} = c^d$

Apart from the above powers which have $4$ digits, we also have:

 $\ds 3^7$ $=$ $\ds 2187$ $\ds 4^5$ $=$ $\ds 1024$ $\ds 6^4$ $=$ $\ds 1296$

Hence, by inspection, it is seen that none of these fit the pattern $\sqbrk {1bcd}$ or $\sqbrk {a0cd}$.

Similarly, suppose $c = 1$ or $d = 0$.

Then:

$\sqbrk {abcd} = a^b$

Again, by inspection, it is seen that none of the above $4$-digit powers fit the pattern $\sqbrk {ab1d}$ or $\sqbrk {abc0}$.

Suppose either $a^b$ or $c^d$ has $4$ digits.

Then the other is less than $10$, giving:

 $\ds n^1$ $=$ $\ds n$ $\ds 2^2$ $=$ $\ds 4$ $\ds 2^3$ $=$ $\ds 8$ $\ds 3^2$ $=$ $\ds 9$

We try multiplying these by all the above $4$-digit powers such that their product is $4$ digits (there are not many).

First note that $1^1$ can be ruled out as none of these $4$-digit powers either begins or ends with $11$.

 $\ds 3^7 \times 2^1$ $=$ $\ds 2187 \times 2$ $\ds$ $=$ $\ds 4374$ $\ds 3^7 \times 3^1$ $=$ $\ds 2187 \times 3$ $\ds$ $=$ $\ds 6561$ $\ds 3^7 \times 4^1$ $=$ $\ds 3^7 \times 2^2$ $\ds$ $=$ $\ds 2187 \times 4$ $\ds$ $=$ $\ds 8748$

 $\ds 4^5 \times 2^1$ $=$ $\ds 1024 \times 2$ $\ds$ $=$ $\ds 2048$ $\ds 4^5 \times 3^1$ $=$ $\ds 1024 \times 3$ $\ds$ $=$ $\ds 3072$ $\ds 4^5 \times 4^1$ $=$ $\ds 4^5 \times 2^2$ $\ds$ $=$ $\ds 1024 \times 4$ $\ds$ $=$ $\ds 4096$ $\ds 4^5 \times 5^1$ $=$ $\ds 1024 \times 5$ $\ds$ $=$ $\ds 5120$ $\ds 4^5 \times 6^1$ $=$ $\ds 1024 \times 6$ $\ds$ $=$ $\ds 6144$ a near miss: $6^1 \times 4^5 = 6144$ $\ds 4^5 \times 7^1$ $=$ $\ds 1024 \times 7$ $\ds$ $=$ $\ds 7168$ $\ds 4^5 \times 8^1$ $=$ $\ds 4^5 \times 2^3$ $\ds$ $=$ $\ds 1024 \times 8$ $\ds$ $=$ $\ds 8192$ $\ds 4^5 \times 9^1$ $=$ $\ds 4^5 \times 3^2$ $\ds$ $=$ $\ds 1024 \times 9$ $\ds$ $=$ $\ds 9216$

 $\ds 4^6 \times 2^1$ $=$ $\ds 4096 \times 2$ $\ds$ $=$ $\ds 8192$

 $\ds 5^5 \times 2^1$ $=$ $\ds 3125 \times 2$ $\ds$ $=$ $\ds 6250$ $\ds 5^5 \times 3^1$ $=$ $\ds 3125 \times 3$ $\ds$ $=$ $\ds 9375$

 $\ds 6^4 \times 2^1$ $=$ $\ds 1296 \times 2$ $\ds$ $=$ $\ds 2592$ $\ds 6^4 \times 3^1$ $=$ $\ds 1296 \times 3$ $\ds$ $=$ $\ds 3888$ $\ds 6^4 \times 4^1$ $=$ $\ds 6^4 \times 2^2$ $\ds$ $=$ $\ds 1296 \times 4$ $\ds$ $=$ $\ds 5184$ $\ds 6^4 \times 5^1$ $=$ $\ds 1296 \times 5$ $\ds$ $=$ $\ds 6480$ $\ds 6^4 \times 6^1$ $=$ $\ds 1296 \times 6$ $\ds$ $=$ $\ds 7776$ $\ds 6^4 \times 7^1$ $=$ $\ds 1296 \times 7$ $\ds$ $=$ $\ds 9072$

 $\ds 7^4 \times 2^1$ $=$ $\ds 2401 \times 2$ $\ds$ $=$ $\ds 4802$ $\ds 7^4 \times 3^1$ $=$ $\ds 2401 \times 3$ $\ds$ $=$ $\ds 7203$ $\ds 7^4 \times 4^1$ $=$ $\ds 7^4 \times 2^2$ $\ds$ $=$ $\ds 2401 \times 4$ $\ds$ $=$ $\ds 9604$

 $\ds 8^4 \times 2^1$ $=$ $\ds 4096 \times 2$ $\ds$ $=$ $\ds 8192$

So none of the above $4$-digit powers multiplied by any of these single-digit powers satisfies the condition.

We have that $\sqbrk {abcd}$ cannot end in $0$, as that can also be ruled out by inspection of the $4$-digit powers.

Similarly, because $a \ne 1$, $\sqbrk {abcd} > 2000$.

Also, it cannot be the case that both $a^b$ and $c^d$ are smaller than $44$, as $44^2 < 2000$.

## Source of Name

This entry was named for Henry Ernest Dudeney.

## Historical Note

This result is generally attributed to Henry Ernest Dudeney, who stated it in his $1917$ book Amusements in Mathematics.

It continues to crop up occasionally in puzzle pages of journals.

The name Dudeney's Property of 2592 has hence been coined by $\mathsf{Pr} \infty \mathsf{fWiki}$, as a convenient shorthand for what would otherwise be tediously unwieldy.