X+Y= 501
Digits we can use are 2,4,5,6,7
X has 3 digits and Y has 2.
The only numbers which can give 1 as the last digit are 5 and 6. Let's keep them in either X or Y.
So, we have 2 cases.
_ _ 5 and _ 6 or _ _ 6 and _ 5
Now to get a 0 in ten's place we need a 9 before 6 and 5 give a carry of 1.
We can get 9 via only 7 and 2.
Again 2 and 7 can go to either X or Y.
Now, let's fix 4 as the remaining hundreds digit. It can only go to X.
We can have a permutation of 6 n 5 and 7 n 2.
Which will give us 2*2= 4 pairs.
We can list them as-
425+76
426+75
475+26
476+25
IMO C
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