How many perfect cubes below 10^6 can be expressed as the sum of two or more distinct integer factorials ?
Rem: only 1! will be considered to represent number 1, 0! is not needed, since 0!+1! can be replaced by 2!, and no factorial can be repeated in the representation of any cube.
The answer is 3.
The factorials and cubes are so widely spaced that a visual search is straightforward:
2^3 = 8 = 3!+2!
3^3 = 27 = 4!+2!+1!
9^3 = 729 = 6!+3!+2!+1!
5^3 - 5! = 5 but this gap is unachievable
18^3 - 7! = 792 which is too wide a gap for the smaller factorials
36^3 - 8! = 6336 which is also too wide
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Posted by Jer
on 2018-09-02 11:37:26 |
Solution
