There are an infinite number of numbers that are a multiple of 3 & 5 and have [exactly] 8 factors. The lowest is 30.
The next few are: 105, 135, 165, 195, 255, 285, 375, 435, ...
Any number of the form 15p where p is a Prime number (which is 2 or greater than 5) will be a multiple of 3 & 5 and have exactly 8 factors.
p=2 gives 15 x 2 = 30, p=7 gives 15 x 7 = 105, etc.
There are two further numbers (listed above): 135 (= 15 x 9) & 375 (= 15 x 25).
No. 9 is a multiple of 3, an equal or larger number which is the product of factors.
9 isn't prime, but one three-digit multiple of 18 is 108.
24.
If the sum of the digits is a multiple of three, the whole number is a multiple of three.
For a number to be a perfect square, its prime factors must be separable into two identical sets. For a number to be a perfect cube, its prime factors must be separable into three identical sets. For a number to be a perfect 'fifth', its prime factors must be separable into five identical sets. ---- If N x 2 is separable into two identical sets of prime factors, then: * N must have an odd number of 2's in its prime factorisation. * All other prime factors of N must occur an even number of times. If N x 3 is separable into three identical sets of prime factors, then: * The prime factorisation of N must contain a number of 3's that is one less than a multiple of three. * All other prime factors of N must occur a number of times that is already a multiple of three. If N x 5 is separable into five identical sets of prime factors, then: * The prime factorisation of N must contain a number of 5's that is one less than a multiple of five. * All other prime factors of N must occur a number of times that is already a multiple of five. ---- From the above, we know that N's prime factorisation must contain 2's, 3's and 5's. How many times does 2 appear in the prime factorisation of N? - It must be an odd number. - It must be a multiple of three. - It must be a multiple of five. The smallest valid frequency is 15. How many times does 3 appear in the prime factorisation of N? - It must be an even number. - It must be one less than a multiple of three. - It must be a multiple of five. The smallest valid frequency is 20. How many times does 5 appear in the prime factorisation of N? - It must be an even number. - It must be a multiple of three. - It must be one less than a multiple of five. The smallest valid frequency is 24. ---- Therefore N = 215 x 320 x 524 = 6810125783203125000000000000000. That seems to be the smallest possible value.
Eight factors.
It's difficult to tell what number you're asking about. 17 has two factors. 70 has eight factors.
No. 9 is a multiple of 3, an equal or larger number which is the product of factors.
Any three-digit multiple of 60, from 120 to 960, has the first five counting numbers as factors.
9 isn't prime, but one three-digit multiple of 18 is 108.
13 is a prime number. 13, like every other prime number, has only two factors. The factors of 13 are 1 and 13. Any multiple of any number greater than one would have at least three factors: 1, the number, and the number it is a multiple of.
Any multiple of 24.
24
24.
The number 9: factors are 9, 3, and 1.
If the sum of the digits is a multiple of three, the whole number is a multiple of three.
If you're asking about distinct prime factors, there are eight numbers tied with three of them. If not, 64 has six twos.