Here, our given number is 53y1, which is divisible by 3.

We know that if number is divisible by 3 only when sum of digits of given number is divisible by 3.

Hence, if 53y1 is divisible by 3, then,

5 + 3 + y + 1 = 0, 3, 6, 9,12,…

∴ 9 + y = multiple of 3.

Hence, y = (multiple of 3) – 9

Here, possible values for multiples of 3 are 9, 12, 15 and 18

Hence, y = 0, 3, 6 and 9

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

(In given answer, 5301 is not considered)