Find nine single digit numbers (other than 1, 2, 3, ..., and 9) with a sum of 45 and a product of 9! (362,880).

For instance, 2, 2, 3, 4, 5, 6, 7, 8, 8 add up to 45, but their product is 645,120.

Try finding the answer without the use of a program.

Since we need 9 1-digit numbers, but not the one where each is present once, there must be repeats and omissions allowable.

What immediately comes to mind is the fact that 6 is a perfect number, being equal to 1x2x3 and 1+2+3.  Playing around with this I didn't find any way to  get rid of other numbers to compensate and keep the count of numbers equal to nine.

Factoring 9! we get 2^7 x 3^4 x 5 x 7.  The 5 and the 7 can't be combined with anything, so must remain as is.

In the original 9!, two of the 2's are tied together for 4; three 2's are tied together for 8; two of the 3's for a 9 and a 2 and a 3 for the 6.

A tying together (multiplying) of a 2 and a 3 adds 1 to their total (2+3=5, becomes 6); a 2 tied with a 2 does not affect the total:4; two 3's tied together add 3 (3+3=6 vs 9); a 4 with an additional 2 adds 2 (4+2=6 vs 8).

Now one must break apart (factor) and tie together (multiply) numbers from the original 1x2x...x9, keeping in mind that multiplying never subtracts from the total and factoring never adds.