Find out all natural numbers with 6 different divisors and no more than 100

Find out all natural numbers with 6 different divisors and no more than 100

Since the number of divisors is equal to the product of the exponents of each factor plus one, let x = a ^ 2 * B;
The results are as follows
When a = 2, B = 3, 5, 7, 11, 13, 17, 19, 23
X=12、20、28、44、52、68、76、92；
When a = 3, B = 2, 5, 7, 11
X=18、45、63、99；
When a = 5, B = 2, 3
X=50、75；
When a = 7, B = 2
X=98

Finding all natural numbers with 6 divisors less than 100

6 = 5 + 1 = (1 + 1) * (2 + 1) so the numbers are as follows: 2 ^ 5 = 322 * 3 ^ 2 = 182 * 5 ^ 2 = 502 * 7 ^ 2 = 983 * 2 ^ 2 = 123 * 5 ^ 2 = 755 * 2 ^ 2 = 205 * 3 ^ 2 = 457 * 2 ^ 2 = 287 * 3 ^ 2 = 6311 * 2 ^ 2 = 4411 * 3 ^ 2 = 9913 * 2 ^ 2 = 5217 * 2 ^ 2 = 6819 * 2 ^ 2 = 7623 * 2 ^ 2 = 92

There are only 12 divisors of those natural numbers within 100

From 12 = 2 * 6 = 3 * 4 = 2 * 2 * 3, we can see that natural numbers with 12 divisors can be decomposed into the following forms: P ^ 11, p * q ^ 5, P ^ 2 * q ^ 3, p * Q * R ^ 2, where p, q, R are prime numbers. (1) P ^ 11 form does not exist within 100. (2) P * q ^ 5 form does not exist within 100

What are the natural numbers with only 3 divisors between 100 and 200? There is a process. Thank you

The divisor of a number must contain 1 and itself [except 1], so there can only be one other divisor except these two divisors. Therefore, the number is the square of the divisor. Moreover, the divisor cannot have its own divisor. Otherwise, the divisor of the divisor will also be the divisor of the original number. So the unique divisor must also be a prime number