+321 Each of the digits 1,2,3,4,5,6 is used exactly once in forming the 3-digit integers \(X\) and \(Y\). How many possible values of \(X+Y\) are there if \(|X-Y| = 111\)
+600 $X=abc=100a+10b+c$
$Y=def=100d+10e+f$
So $|100(a-d)+10(b-e)+(c-f)|=111$ and then $|a-d|=1, |b-e|=1, |c-f|=1$.
Now do casework on the possible values and then sum.
