Find all possible values of y for which the number 53y1 is divisible by 3. Also, find each such number.
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.