UPSC 2014 CDS II Math Paper

Z is an odd number i.e. 1, 3, 5, 7 or 9.
Z≠1 as XY5×1≠X215
Let Z=3, so 5×3=15 but there exists no value of Y for which Y×3+1 ends in 1. So, Z≠3
Let Z=5,so 5×5=25 bur there exists no value of Y for which Y×5+2 ends in 1.So, Z≠5
Let Z=7,so 5×7=35.Y×7+3 can end in 1 for Y=4.X×7+3 end in 2 if X=7
But, this also not possible as value of X and Z cannot be same.
Let Z=9,so 5×9=45.Y×9+4 can end in 1 for Y=3.X×9+3 can end in 2 if X=1
So, X=1,Y=3 and Z=9
∴X+Y+Z=1+3+9=13