$2^5 \times 9^2 =2592$ [duplicate]

Solution 1:

The 2592 puzzle apparently originated with Henry Ernest Dudeney's 1917 book Amusements in Mathematics where it is given as puzzle 115, "A PRINTER'S ERROR":

In a certain article a printer had to set up the figures $5^4\times2^3,$ which of course meant that the fourth power of $5\ (625)$ is to be multiplied by the cube of $2\ (8),$ the product of which is $5,000.$ But he printed $5^4\times2^3$ as $5423,$ which is not correct. Can you place four digits in the manner shown, so that it will be equally correct if the printer sets it up aright, or makes the same blunder?

[. . . .]

The answer is that $2^5\times9^2$ is the same as $2592,$ and this is the only possible solution to the puzzle.

It was apparently rediscovered fifteen years later and published in the American Mathematical Monthly, vol. 40, December 1933, p. 607, as problem E69, proposed by Raphael Robinson, in the following form:

Instead of a product of powers, $a^bc^d,$ a printer accidentally prints the four digit number, $abcd.$ The value is however the same. Find the number and show that it is unique.

A solution by C. W. Trigg was published in vol. 41, May 1934, p. 332; the problem was also solved by Florence E. Allen, W. E. Buker, E. P. Starke, Simon Vatriquant, and the proposer.

Solution 2:

Asking Mathematica to do this

Do[If[a^b*c^d==a*1000+b*100+c*10+d,Print[{a,b,c,d}];,],{a,1,9},{b,1,9},{c,1,9},{d,1,9}]