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MAA Problem 9/22/10

  1. Sep 24, 2010 #1
    Not homework, but a really neat problem I stumbled across. MAA has a Facebook thing-- page? I don't know, I only use Facebook to share pictures with my family and read about the MAA stuff. They post problems often, some of which are very cool. Such as this one.

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

    A number of three digits in base 7 has the same three digits in reverse order when expressed in base 9. What is the number?

    2. Relevant equations
    I re-phrased the question as follows:

    Let n be an integer in base 10 such that n = abc7 and n = cba9. What is n?

    3. The attempt at a solution

    I've got two approaches; a brute force approach that I used and another that a friend and I came up with. I'd love something a bit more generalized. Someone on Facebook said they had three distinct solutions. So...

    i. I started by looking for some boundaries. By definition, the only integers we can use for base 7 are 0-6. So, 6667 = 34210 is our upper bound. But, 6009 = 48610. So, c cannot be 6. By the same argument c cannot be 5.

    From the problem stating that both abc7 and cba9 are three digits, I don't think that a and c can be 0.

    These are the restrictions I came up with:
    1 < a < 6
    0 < b < 6
    1 < c < 4

    I flailed around a bit, and decided to try this: Three number lines

    Base 10 --------------------------------------
    Base 7 ---------------------------------------
    Base 9 ---------------------------------------

    I had an upper and lower boundary, so I checked the base 10 value of 1007, 2007, 3007, 4007, 5007, and 6007. Then I checked 1009, 2009, 3009, and 4009. I was looking for base 10 values that appeared to converge. 5007 and 3009 were very close; 245 and 243 in base 10, respectively. Since the base 7 and base 9 numbers have to be inverted, I tried 5037 and 3059 and was successful: 24810.

    I kept the same restrictions in mind from approach (i).
    1 < a < 6
    0 < b < 6
    1 < c < 4

    Let n = abc7 and n = cba9. Then,
    n = a72 + b71 + c70 = 49a + 7b + c and
    n = c92 +b91 + c90 = 81c + 9b + a.

    So, 49a +7b + c = 81c + 9b + a. Or,
    48a -2b -80c = 0.

    Then, b = 24a -40c = 8(3a-5c)
    Then, b/8 = 3a-5c.

    But, a, b, and c are all integers. So, 3a is an integer, 5c is an integer, and 3a-5c is an integer. So, b/8 must be an integer, and an integer divisible by 8. The only integer in the boundaries of b that meets that criteria is 0. Therefore, b = 0.

    3a-5c = 0

    So, a = 5 and c=3.

    Comments: (i) is really ugly. It felt kind of like... well, like trying to get into a room and accomplishing that by kicking the door in. Sure, it works, but you feel a bit silly when someone walks up behind you with the key.
    (ii) is the result of trying to find a solution that wasn't quite so ugly, and one that, perhaps, yielded a more generalized result. Am I correct in thinking that this solution is the only nontrivial answer to the problem? Someone claimed to have found three solutions, but I can't for the life of me see how.

    Finally, is there a still better way of approaching this problem?
  2. jcsd
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