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Alan and Bob’s cash

Posted by Chris on August 29, 2014 – 6:09 pm

Alan and Bob have a whole number of dollars. Alan says to Bob, “If you give me $3, I will have n times as much as you”. Bob says to Alan, “If you give me $n, I will have 3 times as much as you”.

If n is a positive integer, what are its possible values?


This post is under “MathsChallenge” and has 7 respond so far.
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7 Responds so far- Add one»

  1. 1. Sumit Swain Said:

    The only possible value of n turns out to be 7.

  2. 2. Sumit Swain Said:

    The only possible value of n turns out to be 1.

  3. 3. Chris Said:

    Hi Sumit. That’s two of the four possible values.

  4. 4. Sumit Said:

    Hello Chris. According to the condition above of n being a positive integer, I think n = 1. The other three values I cannot think of them. And apologies, the post with n = 7 was my mistake.

  5. 5. Chris Said:

    Hi Sumit. Unfortunately this site has a moderation policy. So neither of your posts appeared until I approved them. Despite your wording, in fact both 1 and 7 are correct. There are two more solutions.

    If we relax n (and the initial values) being positive, then there are a further four solutions.

  6. 6. jan janse Said:

    starting with:
    a+3=n*(b-3)
    b+n=3*(a-n)
    From this i conclude that the following must be satisfied:
    n=(b+9)/(3*b-13)

    since n and b are integers the only solution i can find are:
    a=5, b=11, n=1
    a=5, b=7, n=2
    a=6, b=6, n=3
    a=11, b=5, n=7

  7. 7. Chris Said:

    Hi Jan, nicely done, that’s the solution set I was after.

    Because n ≥ 1, trying n = 1 => b+9 = 3b – 13 => b = 11.

    If try b = 12, we get 21/23 < 1, so b ≤ 11.
    For 3b – 13 > 0, we have b > 13/3 => b ≥ 5.
    It happens that b = 5 => n = 7. For the rest we only need to try b = 6,7,8,9 and 10. So that could be worse.

    It happens that n = (b+9)/(2b -13), after some work, can be rewritten as
    (3n – 1)(3b – 13) = 40 = 1*40 = 2*20 = 4*10 = 8*5

    Hence 3n -1 = 1,2,4,5,8,10,20,40 are candidates. The rest is straightforward. Sadly that methodical approach seems to involve more labour that Jan’s approach.

    If allow any integers, then (n,a,b) =
    (0,-3,-9), (-1,-1,1), (-3,-3,3), (-13,-16,4) also works.

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