#math trivia for #February26: #57 = 3*19 — a product using each odd digit once. Are there any other products of this form?

— Burt Kaliski Jr. (@modulomathy) February 26, 2012

Let’s think about what a solution might look like.

The general form needs to be

*ab* = *c***de*

where *a*, *b*, *c*, *d* and *e* are different odd numbers. (The form can’t have four digits on the right, because their product would be too long for the one digit on the left, and it can’t have just two, because their product would be too short.)

Now, consider the possible values of the single-digit multiplier *c.*

It can’t be 1, because that would mean *ab* = *de*, repeating the same digits twice.

It also can’t be 5, because the product of an odd number and 5 always ends in 5, meaning that *b* would be 5, another repeat.

The value also can’t be 9, because the smallest *de* with two different odd digits is 13, and 9*13 = 117, which is too long.

And it can’t be 7, because the only *de* giving a two-digit product would be 13, but the product would be 91, another repeat.

So c must be 3.

Now consider the value of *d*. It can’t be 3, obviously, and 5, 7 and 9 would give too long of a product. So *d* must be 1.

This leaves only three possibilities for *de*: 15, 17 and 19. The corresponding values of *ab *are 45, 51 and 57. The first two involve repeats, so we have only the solution in the example:

57 = 3 * 19 .

I like these kinds of problems because they involve not only properties of numbers but a fair amount of logic.

I just noticed that you offer solutions to your problems. That’s nice! I am glad to learn that I got the answer to this question right :).

Lisa