Specify, in ascending order, one space apart, all radixes in which the decimal number 3306 ends in 3.

You need to find all integers N greater than 3 (digit 3 is present in numeration systems only with such a radix), such that the remainder of the division of 3306 and the number N is 3, or (which is the same), 3306 = k * N + 3, where k is a non-negative integer (0, 1, 2, and so on);
From the formula 3306 = k * N + 3 we get that k * N = 3303, so the problem is reduced to finding all divisors of 3303 greater than 3
Answer: 9, 367,1101,3303. Only four divisors (less than three were not considered).