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  • Dec 7, 2010, 04:22 PM
    dfsacxzsdefa
    Number Theory
    Here's a toughie :):

    Two six-digit numbers, A and B, have such a property that when cocatenated (like 5 and 34 to make 534), it is divisible by the product AB. Find A and B

    There is only one answer!
  • Dec 8, 2010, 12:48 PM
    ebaines
    A=166667
    B=333334

    166667333334 is divisible by 166667*333334

    :)
  • Dec 8, 2010, 02:50 PM
    galactus

    Interesting problem.

    The concatenation of A and B equals

    And AB must divide into it



    where p is some integer and p>1.

    Rearranging with some algebra, we get:



    From here, with some head-scratching, the numbers ebaines posted can be found. Try letting p=2 and p=3.
  • Dec 8, 2010, 03:22 PM
    ebaines

    Exactly my approach, except that from



    you can rearrange to get



    and you can see that B must be a multiple of A. So let B = qA, and you get:



    Note that if we try p = 1 then 10^6 = q(A-1), which means B=qA is greater than 10^6 and hence not a 6-digit number. Hence as we proceed we can ignore the possibility that p = 1. So p>1.

    In order for A to be a 6 digit number the product of q and p must be less than 10. Since q is a factor of , this means that q must be 1, 2, 4, 5, or 8. But since p >1, we can throw out q=5 and q=8. So let's try q=1, 2,or 4:

    If q=1, then pA = 1,000,001. But the only prime factors of 1,000,001 are 101 and 9901, so the product qp is too big.

    If q = 2, then pA = 500,001. The factors of 500,001 are 3 and 166,667. Hence p=3, A = 166,667, and B = 2* 166,667=333334. So this works!

    if q = 4, then pA = 250,001. The smallest prime factor of this number is 53, and so qp > 10, which is no good.

    Hence the solution I gave earlier is the only solution.

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