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Math Help - a difficult problem

  1. #1
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    a difficult problem

    A number has 24 factors.The multiplication of its prime factors is 30(taking 1 at a time).What is the ratio of the square roots of the highest and the smallest such number possible?
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by jashansinghal View Post
    A number has 24 factors.The multiplication of its prime factors is 30(taking 1 at a time).What is the ratio of the square roots of the highest and the smallest such number possible?

    The hard part is the first sentence. You know the primes that divide such a number (as these are the primes that multiply to get 30), and the powers of the primes sum to 24 and are all non-zero (as otherwise the product would not be 24). So, what is the largest number possible? What is the smallest?

    Big hint: let x=p_1^{a_1}p_2^{a_2} \ldots p_n^{a_n} be such a number, with p_1<p_2 < \ldots < p_n. Then the largest is the one with a_1=a_2= \ldots = a_{n-1} = 1 and a_n = 24-(n-1).

    (Unless I'm missing something here?)
    Last edited by Swlabr; June 23rd 2009 at 08:36 AM.
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  3. #3
    Senior Member TheAbstractionist's Avatar
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    I took jashansinghals first line to mean 24 factors, not 24 prime factors. But then Im unable to find any integer with exactly 24 factors. Maybe Im missing something as well.
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by TheAbstractionist View Post
    I took jashansinghals first line to mean 24 factors, not 24 prime factors. But then Im unable to find any integer with exactly 24 factors. Maybe Im missing something as well.
    Well, I wasn't sure what "24 factors" meant otherwise. 24 ways of factorising it, perhaps? 5 has one way, 6 has two ways (6 and 2.3) etc, but I thought that a bit far fetched...
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  5. #5
    Moo
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    Hello,

    For example, 6 has 4 factors : 1,2,3,6

    If I'm not mistaking (hasn't seen this for a while) : the number of factors of a number n=p_1^{\alpha_1}\cdot p_2^{\alpha_2} \cdots p_m^{\alpha_m} is \prod_{k=1}^m \{\alpha_k+1\}

    So here, we have \prod_{k=1}^m \{\alpha_k+1\}=24

    And the product of its prime factors is 30 : \prod_{k=1}^m p_k=30

    Then, we have to be able to find the smallest and greatest possible numbers responding to these conditions... So I'm pretty sure we have to study the prime decomposition of 30
    But now... I have to take a shower
    Last edited by Moo; June 23rd 2009 at 11:40 AM.
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  6. #6
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    Each of these has 24 factors: 2^3\cdot 3^2\cdot 5=360~\&~2\cdot 3^2\cdot 5^3=2250.
    If I understand the bit about the product being 30 then they are smallest and largest.
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  7. #7
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by Plato View Post
    Each of these has 24 factors: 2^3\cdot 3^2\cdot 5=360~\&~2\cdot 3^2\cdot 5^3=2250.
    So they do! Now how in the world did I miss those?
    Last edited by TheAbstractionist; June 23rd 2009 at 12:15 PM.
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    Moo
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    2\cdot 3\cdot 5^5>2\cdot 3^2 \cdot 5^3 and satisfies all the conditions...
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    Quote Originally Posted by Moo View Post
    2\cdot 3\cdot 5^5>2\cdot 3^2 \cdot 5^3 and satisfies all the conditions...
    I am willing to bet that who ever wrote that question also missed that solution.
    Because it almost makes this question pointless: What is the ratio of the square roots of the highest and the smallest such number possible?
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  10. #10
    Moo
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    Quote Originally Posted by Plato View Post
    I am willing to bet that who ever wrote that question also missed that solution.
    Because it almost makes this question pointless: What is the ratio of the square roots of the highest and the smallest such number possible?
    Completely agree.
    Yours is aesthetically more beautiful lol
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  11. #11
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    Quote Originally Posted by Plato View Post
    Each of these has 24 factors: 2^3\cdot 3^2\cdot 5=360~\&~2\cdot 3^2\cdot 5^3=2250.
    If I understand the bit about the product being 30 then they are smallest and largest.
    how can you get the solution without doing hit and trial method
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  12. #12
    Senior Member TheAbstractionist's Avatar
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    All right guys, I just remembered this: the tau function \tau(n) which gives the number of factors of each integer n > 1. \tau has the following property: if n=p_1^{k_1}\cdots p_r^{k_r} where the p_i are distinct primes, then

    \tau(n)\ =\ \prod_{i\,=\,1}^r(1+k_i)

    So for this particular problem, the distinct primes are 2, 3, and 5 and the number is of the form 2^a3^b5^c where (1+a)(1+b)(1+c)=24. The possibilities are thus

    (i) a=1,\,b=1,\,c=5
    (ii) a=1,\,b=5,\,c=1
    (iii) a=5,\,b=1,\,c=1
    (iv) a=1,\,b=2,\,c=3
    (v) a=1,\,b=3,\,c=2
    (vi) a=2,\,b=1,\,c=3
    (vii) a=2,\,b=3,\,c=1
    (viii) a=3,\,b=1,\,c=2
    (ix) a=3,\,b=2,\,c=1

    The largest number is therefore 2^13^15^5 and the smallest is 2^33^25^1. Hence the answer to the question is \sqrt{\frac{2^13^15^5}{2^33^25^1}}\,=\,\frac{25}{2  \sqrt3}.
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  13. #13
    Moo
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    Quote Originally Posted by TheAbstractionist View Post
    All right guys, I just remembered this: the tau function \tau(n) which gives the number of factors of each integer n > 1. \tau has the following property: if n=p_1^{k_1}\cdots p_r^{k_r} where the p_i are distinct primes, then

    \tau(n)\ =\ \prod_{i\,=\,1}^r(1+k_i)

    So for this particular problem, the distinct primes are 2, 3, and 5 and the number is of the form 2^a3^b5^c where (1+a)(1+b)(1+c)=24. The possibilities are thus

    (i) a=1,\,b=1,\,c=5
    (ii) a=1,\,b=5,\,c=1
    (iii) a=5,\,b=1,\,c=1
    (iv) a=1,\,b=2,\,c=3
    (v) a=1,\,b=3,\,c=2
    (vi) a=2,\,b=1,\,c=3
    (vii) a=2,\,b=3,\,c=1
    (viii) a=3,\,b=1,\,c=2
    (ix) a=3,\,b=2,\,c=1

    The largest number is therefore 2^13^15^5 and the smallest is 2^33^25^1. Hence the answer to the question is \sqrt{\frac{2^13^15^5}{2^33^25^1}}\,=\,\frac{25}{2  \sqrt3}.
    Well, basically, you're just rewording and explaining the steps of what has been said above
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  14. #14
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by Moo View Post
    Well, basically, you're just rewording and explaining the steps of what has been said above
    Yes, and I should have done it much sooner.
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