Re: Two Olympiad Problems
2) There are several methods, but maybe I am too old to
think of what is considered the simplest proof.
1) Too easy:
<font color="white">
For n>3, the last digit of the sum ends in 3, so obviously
can't be a square.
</font>
Better problem IMHO is to show that 1! + 2! +...+ n! is not
a kth power for n>3, k>=2.
Hint for this:
<font color="white">
note the sum is divisible by 9 for n>=5;
for n>=8, look at (mod 27)
and simply check for n<8
</font>
|