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Topic: Hourglasses: generalized (Read 1335 times) |
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JanClaesen
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Posts: 2
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Hourglasses: generalized
« on: May 11th, 2012, 10:50am » |
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You have two hourglasses.
(What about more?)
The greater time period isn't a multiple of the smaller time period, both are a multiple of the same time period.
Prove that you can measure any time period that is a multiple of the greatest common divisor of the two time periods and that includes and lies between the greatest common divisor of the two time periods and the sum of the smaller time period and the greater time period. If this is true, you can measure any time period that is a multiple of the greatest common divisor of the two time periods.
An example with time periods 7 and 4:
(1 ) 7 - 4 = 11
(2 ) 3 - 4 = 7
(3 ) 7 - 1 = 8
(4 ) 6 - 4 = 10
(5 ) 2 - 4 = 6
(6 ) 7 - 2 = 9
(7 ) 5 - 4 = 9
(8 ) 1 - 4 = 5
(9 ) 7 - 3 = 10
(10 ) 4 - 4 = 8
(1 ) 7 - 4 = 11
...
(1 ) 7 - 4
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(2 ) 3 - 4
= =
10 8
is also true but irrelevant to the solution.
(The translation of the symbols and syntax of the example are a bonus riddle.)
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| « Last Edit: May 12th, 2012, 6:50am by JanClaesen » |
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