1)Take any three digit number with different digits. Reverse the order of the digits. And find the difference between these two numbers. Consider this result as a three digit number, if you get three digits good, if not pretend there are zeros in front. Reverse the order of this number and add to this number.

Highlight below to see what you have,

1089.

Explain why it works.

2)Prove that for any $\displaystyle n,m>1$ the inequality:

$\displaystyle [(n+m-1)!]^2\leq (2n-1)!(2m-1)!$

Is always satisfied.

And only equality when $\displaystyle n=m$.