Math

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

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  1. 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

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  2. 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.

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  3. 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 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

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  4. 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.

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