Posted by Kayla on .
What are the positive integers between 1 and 100 that are only divisible by three numbers: one, the number itself and a prime number?

Math 
drwls,
All numbers are evenly divisible by 1 and the number itself. So you are asking for the number of integers divisible by prime numbers.
Start with all the even numbers (50). Add the numbers divisible by three that are not even: 17. Add the numbers that are divisible by 5 but not 2 or 3 (5,25,35,55,65,85,95) That number is 7. You already have 50 + 17 + 7 = 74
There are a few others: 7, 49, 77, 91, 11, 13, 17, 19, 23
You finish the list 
Math 
Count Iblis,
Suppose that such an integer, n, is not itself a prime number (if it is a prime number, then it can only be divided by two different numbers). Then n will be divisible by some prime number p and we can then factor n/p into further prime numbers. But if these additional factors cannot contain any other prime factors than p, so n must be a power of p.
If n = p^k, then besides 1 and n, the numbers p^r with 1<=r<k divide n, so k must be 2..
So, the integers are squares of primes, i.e. 2^2, 3^2, 5^2, and 7^2. 
Math 
Damon,
The question is not clear to me. I interpret it as allowing two or three different primes to be factors, for example 2*3*5 = 30
There are two ways I can see to do that, hard one first,easy after :)
There are 25 prime numbers less than 100. You can find them with a quick search on google. They qualify for your list.
In addition products of those primes that are less than 100 also qualify and you will have to find those.
For example 2 3 5 and 7 start the list of primes
2*3 = 6 is less than 100 and only divisible by 1, itself, and two primes (I am assuming that the wording of the question allows division by any number of different primes, not just one prime, itself)
2*2*2*2*2*2 = 2^6 = 64 NO because 32 not prime
2*5*7 = 70  that will do
3*5*7 = will not do, greater than 100
I suppose a way to do this is to list all the nonprime numbers between 3 and 100 and see which nonprimes are divisible only by primes
Like
4 yes 2*2
6 yes 2*3
8 no 2*2*2 bcause4 not prime
9 yes 3*3 Yes
10 yes 2*5
12 no 2*2*3 n, because 4 not prime
14 yes
15 yes
16 no, 4 and 8 not prime
18 no, 9 not prime
20 no, 4 not prime
21 yes
22 yes
24 no
25 yes
26 yes
27 yes
28 no
30 yes 3*2*5
32 no
33 yes
34 yes
35 yes
36 no
38 yes
39 yes
40 no
42
another way  better
take each prime and find all the products with other primes that are less than 100
2*3
2*5
2*7
2*11
etc up to 2*47
then
2*3*5
2*3*7
2*3*11
etc up to
2*3*13
then
2*3*5*7 NO TOO BIG
So we go on to starting with 3
3*5
3*7
3*11
3*13
3*17
3*19
3*23
3*29
3*31
Then try 3*5*7 no too big, no need for triples any more
5*7
5*11
5*13
5*17
5*19 no more there
7*11
7*13 no more there
the next, 11*13 is too big, so we are finished 
Math 
drwls,
I interpreted the question as numbers divisible by only 1, itself and ANY (and usually more than one) prime number. That was probably not what you wanted.