1!+2!+3!=9=32.
Prove that k=3 is the only case of the sum 1!+2!+3!+...k!
resulting in an integer power of an integer number.
k=4 gives 33
Every factorial above 4 ends in 0 so the sum will end in 3.
No perfect square ends in 3, so that rules out any even power.
Of course, odd powers can end in 3.
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Posted by Jer
on 2015-02-24 12:13:27 |