   SEARCH HOME Math Central Quandaries & Queries  Question from Pete, a student: I am a student preparing for a competition and this was one of the prep problems: The product of the digits of a four digit number is 6x5x4x3x2x1. how many such numbers are there with this property? Hi Pete.

Let ABCD be the four digit number x. That is, x = 1000A + 100B + 10C + D.

Then (A)(B)(C)(D) = 6 x 4 x 3 x 2.

This is 144, but in prime factors it is 32 x 24.

The fundamental theorem of arithmetic (look that up using our Quick Search if you like!) says that any positive integer greater than 1 is the product of a unique set of prime factors. That means that each of
A, B, C and D must either be 1s or have 3 and/or 2 as a factor.

This means 1, 2, 3, 4, 6, 8, 9 are the only possibilities for digits.

At this point, you can start counting, ensuring each (factor) is not higher than 9:

32 x 24
= (3x3)(2x2x2)(2)(1)
= (3x3)(2x2)(2x2)(1)
= (3x3)(2x2)(2)(2)

= (3)(3x2)(2x2x2)(1)
= (3)(3x2)(2x2)(2x2)

= (3)(3)(2x2x2)(2)
= (3)(3)(2x2)(2x2)

and that's all 7 combinations. Now you just need to permute each of these. For example, the first combination yields the digits 9, 8, 2, 1 and so that could be 9821 or 2918 or 1892, etc.

Watch out for the doubles as you do the permutations!

Write back if you need more help.

Cheers,
Steve La Rocque.     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.