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16 January, 20:32

A student at Harvard named Kevin

Was counting his stones by 11

He messed up n times

And instead counted 9s

And wound up at 2007.

To be precise, Kevin divides 2007 stones into groups of 9 stones and 11 stones, where n of the groups have 9 stones in them. How many values of could make this limerick true?

(Kevin can mess up every time)

+1
Answers (1)
  1. 16 January, 22:56
    0
    The answer is 21.

    Work:

    Some groups of 11,

    and some groups of 9,

    together make 2007.

    We may make an equation

    11x + 9y = 2007,

    and we need to find how many possibilities there are for y.

    2007/9=223

    So, the highest value for y is 223.

    Let k be a value for y.

    If k is a valid value for y then (2007-9k) must be divisible by 11.

    Let's just brute force this by doing this validity test with every number from 1 to 223. Then we can count how many valid options there are.

    0 is NOT valid

    1 is valid

    2 is NOT valid

    3 is NOT valid

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    11 is NOT valid

    12 is valid

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    210 is valid

    211 is NOT valid

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    213 is NOT valid

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    219 is NOT valid

    220 is NOT valid

    221 is valid

    222 is NOT valid

    223 is NOT valid

    21 valid options for y!
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