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Sandpit

January 16th, 2017

A new sandpit for long side discussions, conspiracy theories, idees fixes and so on.

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  1. Ikonoclast
    January 19th, 2017 at 05:52 | #1

    I have a little maths puzzle which I cannot work out completely even with a spreadsheet. I am such a maths illiterate! I will try to give all the relevant data as clearly as I can and I hope there is enough information to enable a solution. Yes, I am crowd-sourcing here.

    First, there are six sets of numbers which yield ratios. Express the ratio as A/B which is easy of course. I give the example of the first ratio.

    A B

    100 20 Ratio = 5
    100 80
    100 20
    100 80
    100 7

    100 80
    100 160
    100 40
    100 160
    100 15

    100 20
    100 10
    100 10
    100 40
    100 3

    100 80
    100 40
    100 160
    100 160
    100 15

    100 20
    100 10
    100 40
    100 10
    100 3

    100 213
    100 106
    100 426
    100 106
    100 426

    You will notice that some ratios result in long strings of decimals. What is the lowest common multiple of the whole six sets such that all these ratios with decimals can be made whole numbers (integers)? The complicating factor is that some of the numbers in the above six sets have been rounded to integers to themselves: probably all rounded down but maybe some rounded up for .5 and above.

    The final data you may need are these. All these numbers represent trades in a (computer game) market with six commodities. In each set, a commodity is being used to buy the other five. Market prices never change, they are unaffected by trades. You will notice this is a very unrealistic market for both this reason and because trading one way and then back the other would result in excessive losses which no real market (except perhaps a pawn shop) would replicate.

  2. Ikonoclast
    January 19th, 2017 at 05:55 | #2

    Oops, I should have said, “What is the lowest common multiple of the ratios… “

  3. Ikonoclast
    January 19th, 2017 at 08:13 | #3

    And it is only the numbers in the second column that may have been rounded to integers.

  4. J-D
    January 19th, 2017 at 09:38 | #4

    If your question means what I think it means, then the answer is 632184.

  5. Ikonoclast
    January 19th, 2017 at 09:47 | #5

    @J-D

    That seems large but it might well be right. Did you allow for the fact that some unknown sub-set of column B figures could be rounded to integers?

  6. Ikonoclast
    January 19th, 2017 at 09:54 | #6

    @J-D

    What I mean is this. There could be a lower common multiple (much lower maybe) depending on the rounding of numbers to integers in column B.

    The real question, if I am now posing it correctly, is this. What is the lowest possible lower common multiple if any number of numbers in column B have been rounded to integers?

  7. Ikonoclast
    January 19th, 2017 at 09:56 | #7

    Arrhgh typos, “the lowest possible lowest common multiple”.

  8. J-D
    January 19th, 2017 at 11:12 | #8

    If the true values of the divisors are not known, then the true values of the ratios can’t be calculated, and therefore it would be impossible to calculate their lowest common integer multiple.

  9. Ikonoclast
    January 19th, 2017 at 14:40 | #9

    @J-D

    Hang on, the values of the divisors are known within a narrow range. Coming from a computer program, the most likely rounding function used is INT(n). INT(n) would round the decimal down so 9.1 and 9.9 would both become 9. Using that information we should be able to find the lowest common multiple… I think

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