Results 1 to 9 of 9

Math Help - A prime factor problem.

  1. #1
    Newbie
    Joined
    Apr 2011
    Posts
    2

    Question A prime factor problem.

    A certain integer has only two distinct prime factors. The number of its divisors is 6, and the sum of its divisors is 28. The integer is?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7
    Quote Originally Posted by Incognitoi View Post
    A certain integer has only two distinct prime factors. The number of its divisors is 6, and the sum of its divisors is 28. The integer is?
    By trial and error, the divisors are 5, 5, 5, 5, 5, 3. So, the integer is 5x5x5x5x5x3=9375.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2011
    Posts
    2
    Quote Originally Posted by alexmahone View Post
    By trial and error, the divisors are 5, 5, 5, 5, 5, 3. So, the integer is 5x5x5x5x5x3=9375.
    Thanks for the reply. To be frank, i was already aware of the trial & error method. A much concrete method(if it exists) would be much appreciated.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,845
    Thanks
    320
    Awards
    1
    Quote Originally Posted by Incognitoi View Post
    A certain integer has only two distinct prime factors. The number of its divisors is 6, and the sum of its divisors is 28. The integer is?
    Ugh. I just had this whole thing typed in and lost it. Ah well.

    I have a systematic way of solving the problem, but I admit that it's long and trial and error is much shorter. Here it is, briefly. You can easily fill in the steps.

    We want to find a z such that z = (x^a)(y^b) where x and y are distinct primes, a and b are positive integers such that a + b = 6, and ax + by = 28.

    We can say that b = 6 - a, so the second condition becomes ax + (6 - a)y = 28. We need to solve this equation.

    First note that, according to the problem statement x and y are interchangeable. Let's start then, by letting x = 2. When we do this we run into a contradiction. So x is not 2 and since x and y are interchangeable, neither is y.

    So x and y are both (distinct) odd primes.

    Now, since ax + (6 - a)y = 28 either ax and (6 - a)y are both even or they are both odd.

    Assume ax and (6 - a)y are both even. Then a must be even since x and y are odd. Let a = 2n. Then n = 1 or n = 2. Both cases create a contradiction, so ax and (6 - a)y both cannot be even.

    So ax and (6 - a)y must both be odd. Thus a is odd. Let a = 2n - 1. Then n = 1, 2, or 3. Since x and y are interchangeable n = 1 and n = 2 both provide the same result. n = 3 gives a contradiction. Thus let n = 1. So we (finally) have that x + 5y = 28.

    Now, x and y are distinct primes, not equal to 2. Thus x = 3, y = 5; x = 5, y = 7; x = 3, y = 7, etc. Of these only x = 3, y = 5 can be a solution since all other possibilities give x + 5y > 28.

    Thus the only solution is x = 3, y = 5 and since n = 1 we know that a = 1 and b = 5, which gives z = (3^1)(5^5) = 9375.

    One would hope for a shorter solution method....

    -Dan
    Last edited by topsquark; April 27th 2011 at 08:32 AM. Reason: Copied wrong.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2011
    Posts
    71
    But 9375 has 12 divisors. We should be looking for an integer of the form p^2q where p and q are distinct primes. That's how I would interpret it at least. Since the factors add up to a small number, it's not hard to find p and q.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,845
    Thanks
    320
    Awards
    1
    Quote Originally Posted by LoblawsLawBlog View Post
    But 9375 has 12 divisors. We should be looking for an integer of the form p^2q where p and q are distinct primes. That's how I would interpret it at least. Since the factors add up to a small number, it's not hard to find p and q.
    Oooh! Good catch.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7
    Quote Originally Posted by LoblawsLawBlog View Post
    But 9375 has 12 divisors. We should be looking for an integer of the form p^2q where p and q are distinct primes. That's how I would interpret it at least. Since the factors add up to a small number, it's not hard to find p and q.
    1+p+q+pq+p^2+p^2q=28

    1+p+p^2+q(1+p+p^2)=28

    (1+p+p^2)(1+q)=28

    By inspection p=2 and q=3

    n=2^2 \times 3=12

    PS: This means that posts #2 and #4 are incorrect.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member abhishekkgp's Avatar
    Joined
    Jan 2011
    From
    India
    Posts
    495
    Thanks
    1
    Quote Originally Posted by topsquark View Post
    Ugh. I just had this whole thing typed in and lost it. Ah well.

    I have a systematic way of solving the problem, but I admit that it's long and trial and error is much shorter. Here it is, briefly. You can easily fill in the steps.

    We want to find a z such that z = (x^a)(y^b) where x and y are distinct primes, a and b are positive integers such that a + b = 6, and ax + by = 28.

    why . shouldn't it be . also why is the expression for the sum of divisors ax+by. i guess it should be product of two geometric series?? .

    We can say that b = 6 - a, so the second condition becomes ax + (6 - a)y = 28. We need to solve this equation.

    First note that, according to the problem statement x and y are interchangeable. Let's start then, by letting x = 2. When we do this we run into a contradiction. So x is not 2 and since x and y are interchangeable, neither is y.

    So x and y are both (distinct) odd primes.

    Now, since ax + (6 - a)y = 28 either ax and (6 - a)y are both even or they are both odd.

    Assume ax and (6 - a)y are both even. Then a must be even since x and y are odd. Let a = 2n. Then n = 1 or n = 2. Both cases create a contradiction, so ax and (6 - a)y both cannot be even.

    So ax and (6 - a)y must both be odd. Thus a is odd. Let a = 2n - 1. Then n = 1, 2, or 3. Since x and y are interchangeable n = 1 and n = 2 both provide the same result. n = 3 gives a contradiction. Thus let n = 1. So we (finally) have that x + 5y = 28.

    Now, x and y are distinct primes, not equal to 2. Thus x = 3, y = 5; x = 5, y = 7; x = 3, y = 7, etc. Of these only x = 3, y = 5 can be a solution since all other possibilities give x + 5y > 28.

    Thus the only solution is x = 3, y = 5 and since n = 1 we know that a = 1 and b = 5, which gives z = (3^1)(5^5) = 9375.

    One would hope for a shorter solution method....

    -Dan
    am i mistaken?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,845
    Thanks
    320
    Awards
    1
    why . shouldn't it be . also why is the expression for the sum of divisors ax+by. i guess it should be product of two geometric series?? .
    I was under the mistaken impression (as was the OP it seems) that the number of factors of z was the sum of each factor of the form p*p*p*p*q*q, say. Which is 6. As LoblawsLawBlog pointed out, this is a mistake.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. How do you factor an un-prime quadratic?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: March 29th 2011, 08:30 PM
  2. Replies: 1
    Last Post: December 14th 2009, 05:13 PM
  3. Factor or Prime?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: May 4th 2008, 02:08 PM
  4. Prime factor problem
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 29th 2008, 11:12 AM
  5. Find a prime factor of 99!-1
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: June 16th 2005, 05:19 AM

Search Tags


/mathhelpforum @mathhelpforum