Posted by Joseph on Thursday, February 19, 2009 at 3:26pm.
since 15 is a common factor the number of cards each has must be a multiple of 15
let the number of cards that Joe has be 15x, let Anna's number of cards be 15y
then 15x + 15y = 75
x + y = 5
also x > y , and of course both x and y must be whole numbers.
there are only 2 possibilities for x and y
(4,1), (3,2)
(4,1) : Joe has 60, Anna has 15
(3,2) : Joe has 45, Anna has 30
both cases satisfy your 3 conditions.
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