Explanation:
5 digit number
5abcd
a = 2 or a = 3. but a = 3 means the upper limit of 53,000. and the sum of digits is 8 and not 21.
so, a = 2.
d = 3n
c = b
5 + a + b + c + d = 21
5 + a + 2b + 3n = 21
a + 2b + 3n = 16
2 + 2b + 3n = 16
2b + 3n = 14
so, n must be an even number to allow an even result (14). therefore, n = 0 or n = 2 (any larger even number n like 4 would not create a single digit result)
if n = 0, then 2b = 14, b = 7
if n = 2, then 2b + 6 = 14, 2b = 8, b = 4
so, we have the options
52770
52446
I suspect that 0 as 0×3 is not considered a multiple of 3 by your teacher, so only n = 2 (or d = 6) is a desired solution.
therefore, I would use
52446
as answer.