Since n!, a!, b!, and c! are all greater than 0, it follows that n! is greater than each of a!, b!, and c!. Since a!+b!+c! must necessarily be greater than 3 (since each is greater than or equal to 1), then we also have that that n!>2, hence n>2, or equivalently, n¡Ý3.
Without loss of generality, let a=max{a,b,c}. Then n!=a!+b!+c! ¡Ü3(a!). Hence, n!/a!¡Ü3, thus n(n-1)...(a+1)¡Ü3.
This, together with the fact that n¡Ý3 and 3 is prime forces n to be equal to 3 and thus n!=6. Therefore the only choices to be made for a, b, and c are that a=b=c=2.
Solution. The only solution is n=3, a=b=c=2.
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Posted by Mike C
on 2007-04-25 10:13:54 |
Possible solution?
