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 1digit 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.

Posted by Charlie
on 20050228 16:15:59 