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January 31, 2015

January 31, 2015

Posted by **Joseph** on Thursday, February 19, 2009 at 3:26pm.

- Math -
**Reiny**, Thursday, February 19, 2009 at 4:44pmsince 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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