Find all possible values of y for which the number 53y1 is divisible by 3.

If a number is divisible by 3, then the sum of the digits is also divisible by 3 .

Sum of the digits $=5+3+y+1=9+y$

The sum of the digits is divisible by 3 in the following cases:

$9+y=9$, or $y=0$

Then the number is 5301 .

$9+y=12$, or $y=3$

Then the number is 5331 .

$9+y=15$, or $y=6$

Then the number is 5361 .

$9+y=18$, or $y=9$

Then the number is 5391 .

∴ y = 0, 3, 6 or 9

The possible numbers are 5301, 5331, 5361 and 5391.