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Another number theory problem

  1. Mar 15, 2007 #1
    this is question found as part of a 25question 1hour long maths problem quiz i took today. It was just about the only one i couldn't do in the time, and even with another look after I can't do it :S I think it may involve maths I havnt come across or at least methodology i havn't (I 16 and only just started looking beyond class studies)

    1. The problem statement, all variables and given/known data

    Let N be the smallest integer such that 10 x N is a perfect square and 6 x N is a perfect cube.
    How many positive factors does N have?


    A 30 B 40 C 54 D 72 E 96

    2. Relevant equations

    3. The attempt at a solution

    right, Ill talk you through my attempt so far...


    I first of all tried a few values for N (the basics as tests like 0,1,2,3,5,10, etc..) and came to the logical conclusion that it wasn't going to be as simple as that, and probably would be a very large number (probably shoulda concluded that from the options but o well

    o yeh, I found out that 0 kinda works, but I'm guessing from the options that 0 doesn't count because 0 isn't a cube or a square? am I right?

    Then I tried using modular arithematic (I still a super noobie with it)

    N = 0 (mod 10) is a perfect square, while N = 0 (mod 6) is a perfect cube.

    From that I started working my way through the first half, trying lots of options for what makes the first bit work. I came up with stuff like N = 10, 40, 90, etc.. Not in a very logical approach but hey, mighta saved me time, but it didnt...

    I can't figure out how to find N?!?!?

    Have I approached the question all wrong though? is finding N entirely necessary or can you derive the factors somehow???!?!?

    I tried the later approach, looking for similarities between the factors of squares and cubes but I dunno if it was me being slow, or wierd, or whatever, (or oblvivious to the obvious) but I couldn't find anything that way...

    any help please :D

    and as per usualy, one step at a time ;)

  2. jcsd
  3. Mar 16, 2007 #2


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    Clearly N has a prime factorization of the form 2^i*3^j*5^k. What constraints do your premises put on i,j,k?
  4. Mar 16, 2007 #3
    :S whats prime factorization?

  5. Mar 16, 2007 #4
    you can play some tricks involving divisibility.

    for instance, you have 10*N is a perfect square. so
    so 2 divides both sides, hence 2 divides a^2, so 2 divides a. (if 2 doesn't divide a, then 2 can't divide a^2 because 2 is a prime)

    so you have:


    cancel out the two

    rinse and repeat for 5. and similarly for 6N=b^3
  6. Mar 16, 2007 #5
    Sorry to be off topic but was that test the AMC?
  7. Mar 16, 2007 #6
    If [tex]10N = a^2[/tex], then [tex]2^{x},5^{y}, k_{1}^{z_{1}}, k_{2}^{z_{2}}... | N[/tex] where x, y can be expressed as 2*q - 1, where q is any positive integer. k refers to any prime, and z an even number.

    For [tex]6N = b^{3}[/tex], we have [tex]2^{a},3^{b}, l_{1}^{c_{1}}... | N[/tex] where a, b can be expressed as 3*q - 1, where q is any positive integer. l refers to any prime and c is a multiple of 3.

    Now we have concluded that 2, 3 and 5 must be factors of N. Also, we know what the factors of the respective exponents must be. Can you go on from there?
    Last edited: Mar 16, 2007
  8. Mar 17, 2007 #7
    it was the second round of the UKMT Maths Challenge (from the UK), and is called something like "the european pink kangaroo round", haha, funny name, lol

    anyways, yeh i reks i didnt through to the olympiad cause i working out i probs only got like 90 outa 135 or whatever it is...

    whats the AMC?
  9. Mar 17, 2007 #8
    o yeh, thnx for the help, ill try and go from there and get back to ya if i get stuck...
  10. Mar 17, 2007 #9


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    If you are going to keep looking at number theory problem it would be good to get to know about prime factorization. It's not even hard.
  11. Mar 20, 2007 #10
    o rite i know what prime factorization is lol

    just never heard it called that before cause at school i just get told it when you split a number into prime factors hehe, stupid me

    right i've decided i have no idea whats going on lol

    unfortunately, i not sure what any of your explainations mean or where to go from there.

    thnx for the help anyways, ill come to this problem once i read some number theory
  12. Mar 20, 2007 #11


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    So then do you agree N=2^i*3^j*5^k for some i,j,k? You can do it now.
  13. Mar 21, 2007 #12


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    In order that 10*N be a square, since 10= 2*5, N must have factors of 2 and 5. In order that 6*N be a cube, since 6= 2*3, N must have factors of 22 and 32. Since N is the smallest such number, it must be 22*32*5= 180. Now, I'll leave the hard part to you: how many distinct factors does 180 have?
  14. Mar 21, 2007 #13


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    Sorry, but the number of factors of 180 won't do him any good, since 10*180 is not a perfect square and 6*180 is not a perfect cube.
  15. Mar 21, 2007 #14
    Halls, something you've let something slip there...

    By the way Trailer Builder, the answer is ugly.
  16. Mar 21, 2007 #15


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    What's ugly about 2^5*3^2*5^3=36000?
  17. Mar 21, 2007 #16


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    Considering the prime factorization, counting factors is really easy as well... Trail Builder gave up too fast.
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