OK, this is a quiz based on Ben Law’s puzzle of 17 cows .
1) Assume that time the 2 elder sons still maintain their share as 1/2 and 1/3, the youngest son has the share of 1/x (x could not be 2 or 3), so my question is, in order to let YS has the same solution (please refer back to the post, if you not sure what’s the solution), what is
a) the minium value of the x, and hence, how many cows actually their father left behind?
b) the maximum value of the x, and hence, how many cows actually their father left behind?
2) The other way round, the elder 2 maintain as 1/2 and 1/3, and the youngest son has the share of 1/x (x could not be 2 or 3), so if YS solution is to remove 1 cow first, what will be the
a) the minimum value of x, and hence, how many cows actually their father left behind?
b) the maximum value of the x, and hence, how many cows actually their father left behind?
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OK, this is a quiz based on Ben Law’s puzzle of 17 cows .
1) Assume that time the 2 elder sons still maintain their share as 1/2 and 1/3, the youngest son has the share of 1/x (x could not be 2 or 3), so my question is, in order to let YS has the same solution (please refer back to the post, if you not sure what’s the solution), what is
a) the minium value of the x, and hence, how many cows actually their father left behind?
b) the maximum value of the x, and hence, how many cows actually their father left behind?
2) The other way round, the elder 2 maintain as 1/2 and 1/3, and the youngest son has the share of 1/x (x could not be 2 or 3), so if YS solution is to remove 1 cow first, what will be the
a) the minimum value of x, and hence, how many cows actually their father left behind?
b) the maximum value of the x, and hence, how many cows actually their father left behind?
Should be easy.
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