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my question is- if u want too build a pile of 100coins(1and 2dinars) such that, number of coins between any two '2' dinar coins is not equal to five, then what are the maximum number of 2 dinar coins that can be included in the pile?
we have a pile of 100 coins which has both 1dinar and 2 dinar coins. the questions asks us to tell the maximum number of 2 dinar coins in the pile if the number of coins between any '2' 2dinar coins is nt five.
wait r u taking the value of total coins to be 100 dinars??
:) v don't want the value to be 100 we want total number of coins to be 100 :)
So total 100 coins, with both kinds. Since question does not specify that there \(has\) to be 1 dinar coins between the two-dinars, then Minimum number of 1 dinar coins = 1 what is the maximum number of two-dinars?
we have 1 more condition @mathmate that is - there are no '2' 2dinar coins with a gap of 5 coins between then
i m getting the answer 50
in this case there will be no '2' 2dinar coins with a gap of 5coins between them so in this case we can see that the number of 1dinar nd 2dinar coins is equal so there are 50 2dinar coins
but i am not sure that 50 is the MAXIMUM value of 2dinar coins...this answer is obtained frm the case which i made i donno there can be another case in which the answer can exceed 50.
My bad, I didn't read the question properly. As a start, we can try with 6x2, then 6x1, then 6... 222222111111222222111111....2222221111112222 for a total of 52 2-dinars. We'll investigate to see if there is a better arrangement.
the answer by my method is nt 50 its less than that
@imqwerty I find 8 patterns of 222222111111 giving 48 2's out of 96 coins. The remaining 4 coins can be 2's, giving a total of 48+4=52 coins of 2 dinars. Would you like to show your calculations?