Factorial problem
Number Theory
Level
pending
Find the number of solutions in natural number of the equation a!+b!+c!=3^d where a≥b≥c.
The answer is 3.
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1 solution
By trial and error. 3^0=1 not possible. 3^1=1!+1!+1!; 3^2=3!+2!+1!; 3^3=4!+2!+1!; 3^4=81<5!. Other combinations less than 5 dont work out. Looking for an analytic solution. Anybody?