Tuesday, September 15, 2009

Nine single-digit numbers

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

2 comments:

Anonymous said...

Any cluez pls???
ld we use 0! or something like dis??

Unknown said...

To solve for 9 digits from 1 to 9 summing to 45 with product of 362880...

Identify prime factors 1,2,3,5,7
no numbers from 1 to 9 share 5 and 7 as factors, thus 5 and 7 are fixed, solving for 2 numbers

product of 10368 for the remainder or 2^7+3^4=10368. with sum of 33.

Rough checks for value range:
-average value between 4 and 5
-2*7+3*4=29 ...so will need to use higher exponents to bring up sum
-using nine for first 2 digits, 9*2=18, 33-18=15 for remaining 5 digits averaging 3... so use 4s then 2s, then 1s
-Yielding:9,9,4,4,4,2,1
Reordered and incorporating 5 and 7 for final solution...

Answer:
1,2,4,4,4,5,7,9,9