Consider if the problem is rewritten as:
n! - c! = a! + b!
The sum of two factorials must equal the difference of two others. Fr the first six factorials, the five differences between consecutive factorials is {1, 4, 18, 96, 600}. The factorials of the first six are {1, 2, 6, 24, 120, 720}. It is plain to see that the higher one goes, no possibility exists that the sum of any two factorials can equal the ever-increasing gaps. For example, to fill the gap of 600 (6! - 5!) would require 5! + 5! + 5! + 5! + 5!, a sum of 5 factorials as opposed to the 2 defined.
From that, it is easier to narrow down to the area where the integer approaches zero. Comparing the two sets above, (4 = 2 + 2) or:
3! - 2! = 2! + 2!
Therefore, the only solution is:
2! + 2! + 2! = 3!
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Posted by hoodat
on 2007-04-26 08:57:19 |
Solution
