Find the three least positive integers that cannot be written as a!/b! + c!/d! + e!/f! for positive integers a,b,c,d,e,f less than or equal to 5.
I did this by hand but I think it's correct.
The possible summands are 1, 2, 3, 4, 5, 6, 12, 20, 24, 60, 120 and their reciprocals.
The reciprocals are only useful for making the two smallest numbers:
1 = 1/2 + 1/4 + 1/4
2 = 1 + 1/2 + 1/2
It's also apparent the odd sums will be more difficult since the only odds are 1,3,5. Also 6 is the largest even not divisible by 4.
Everything up to
54 = 24+24+6 is possible.
55 is not possible (it requires a 1,3,5 along with two that sum to 50, 52, or 54.)
56 = 24+20+12.
57 is not possible (like 55 but two can't sum to 52, 54, 56.)
58 is not possible (not a multiple of 4, so requires a 2 or 6 and the other two sum to 52 or 56)
59 also not possible
60 = 20+20+20 = 24+24+12
above this there are many gaps.
The solution is 55, 57, 58.
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Posted by Jer
on 2024-11-26 13:13:44 |
Solution
