raghav4202 wrote:
5,6,10 can go in this order
5,18,10 (1,3,1)
10,12,10(2,2,1
15,6,10(3,1,1)
5,6,20(1,1,2)
20,6,10(4,1,1)
10,6,20(2,1,2)
these itself become 6 , so shouldn't the answer be E ?
correct me if i am wrong
i guess it's right...
i try to solve like that atleast 1 of each is require so 21$ ( 5+6+10) definitely spend then remains (30-21)&(36-21)= 9 to 15$
(10)
(5,5)
(5,6)
(5,10)
(5,5,5)
(6,6)
yeah so more than 5...
i was just hoping if someone knows simpler way.