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    When cashing a cheque, a bank clerk makes an error and swaps the pounds and pence figures. If you were to spend 1.18 pounds of the money you recieved, you would have exactly double the value of the cheque. What was the amount on the cheque?

    Saw this one today in the newspaper, it has a very interesting solution.
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    £26.54
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    (Original post by AntiMagicMan)
    £26.54
    Well I knew the answer, but want a proper solution. SO would you like to post your method?
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    Solution : 26.54 was the value of the check.
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    (Original post by IntegralNeo)
    Well I knew the answer, but want a proper solution. SO would you like to post your method?
    My method was brute force computation.
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    (Original post by AntiMagicMan)
    My method was brute force computation.
    But there is a mathematical method of doing it,as I did it.
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    My method was like this:
    First, the equation that we want to solve is:
    x + 100y - 118 = 200x +2y

    Where x and y are integers, y < 100 .
    Rearranging the equation we get 199x = 98y - 118
    Let x = 2k,
    199k = 49y - 59
    7*28k + 3k = 7*7y - 7*8 - 3

    Rearranging, we get
    7(7y - 8 - 28k) = 3k + 3
    7(7y - 8 - 28k) = 3(k+1)

    Considering the factors on both sides, let k+1 = 7n
    Then,
    7y - 8 - 28(7n - 1) = 3n
    199n = 7y + 20

    Now, through trial and error, try n = 1, 2, ....
    n = 2 gives a solution, where y = 54
    Hence, k = 13, -> x=26, when y = 54
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    Well that technically is brute force computation. So I assume Integral has a better solution.
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    (Original post by sephonline)
    My method was like this:
    First, the equation that we want to solve is:
    x + 100y - 118 = 200x +2y
    hey r u waiting for alevel results too?
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    (Original post by AntiMagicMan)
    Well that technically is brute force computation. So I assume Integral has a better solution.
    (100D + L) - 118 = 2(100L + D)

    This is:

    98D = 199L + 118

    D = (2L+1) + [(3L+20)/98]

    Now we know that [(3L+20)/98] must be whole and L <100

    so,

    3L + 20 = 98, and L =26, D=54. Cheque was worth 26.54 pounds.
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    hey r u waiting for alevel results too?
    Yes.
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    (Original post by sephonline)
    Yes.
    what alevels did u do?

    why dont u fill in the official UKL Who Are You Questionnaire?

    Click here

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    (Original post by IntegralNeo)
    (100D + L) - 118 = 2(100L + D)

    This is:

    98D = 199L + 118

    D = (2L+1) + [(3L+20)/98]

    Now we know that [(3L+20)/98] must be whole and L <100

    so,

    3L + 20 = 98, and L =26, D=54. Cheque was worth 26.54 pounds.
    Very nice solution
    Just one comment, though.
    Basically, what you get down to is,

    [(3L+20)/98] must be a whole number, but this gives you,

    3L + 20 = 98k, k = 1,2,3, ...

    this time, there was a aolution for k=1, but in other circumstances, the value of k could have been anything.

    Problems like this where you have one equation with two unknowns but the solutions are integers, are known as linear diophantine equations (part of Number Theory), and can be solved mathematically.
 
 
 
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