Find all possible values of a five-digit base ten positive integer N such that N is equal to 45 times the product of its digits.

Prove that there are no others.

Extra Challenge: A non computer program aided solution.

we seek a solution to 45abcde = 10000a+1000b+100c+10d+e
clearly none of the digits can be zero
this means all of the digits are odd or the LHS will end in zero.
The LHS will end in 5 so e=5
The equation becomes
225abcd = 10000a+1000b+100c+10d+5
multiples of 225 end in 25, 50 or 75 so d must be a 7.
1575abc = 10000a+1000b+100c+75
multiples of 1575 only end in 75 when they end in 175, 575, or 875 so c could be either 1 or 5.

If c=1
1575ab = 10000a+100b+175
there is only one five digit multiple of 1575 that ends in 175 and has all odd digits:  77175 is the solution.
[I'd admit a touch of brute force here.]

If c=5
7875ab = 10000a+1000b+575
there are no 5 digit multiples of 7875 that end in 575 so there are no other solutions.

Posted by Jer on 2013-10-28 15:05:15