Monday

January 23, 2017
Posted by **Kayla** on Tuesday, January 22, 2008 at 7:12pm.

- Math -
**drwls**, Tuesday, January 22, 2008 at 7:41pmAll 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**, Tuesday, January 22, 2008 at 7:45pmSuppose 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**, Tuesday, January 22, 2008 at 8:05pmThe 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 non-prime numbers between 3 and 100 and see which non-primes 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**, Wednesday, January 23, 2008 at 2:09amI 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.