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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 \(|X-Y| = 111\)

 Apr 7, 2021
edited by itachiuchihakujo  Apr 7, 2021
 #1
avatar+556 
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$X=abc=100a+10b+c$

$Y=def=100d+10e+f$

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.

 Apr 7, 2021

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