USSR OLYMPIAD PROBLEM
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MATHPRITOM
- Posts:190
- Joined:Sat Apr 23, 2011 8:55 am
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find all x,y,z such that 1!+2!+3!+...+x!=$y^x$
Re: USSR OLYMPIAD PROBLEM
Let $x\ge6$.
Then, $1!+2!+3!+4!+5!+.....+x!\equiv0\pmod9$
So,$9|y^x$
Now since $x>5,27|y^x$.But $1!+2!+3!+4!+5!+.....+x!\equiv9\pmod{27}$, so not divisible by $27$.
Then, $1!+2!+3!+4!+5!+.....+x!\equiv0\pmod9$
So,$9|y^x$
Now since $x>5,27|y^x$.But $1!+2!+3!+4!+5!+.....+x!\equiv9\pmod{27}$, so not divisible by $27$.
One one thing is neutral in the universe, that is $0$.