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 Commemorations (Posted on 2013-02-06)
3 jealous brothers are to divvy up a set of commemorative coins.

There are an equal number of coins of each of the values: \$1, \$2, \$3.

They discover that it is possible to do this in such a way that each brother gets a different assortment of coins, yet each gets the same number of coins and the same total value of coins.

What's the smallest possible number of coins in the set?

 No Solution Yet Submitted by Jer Rating: 3.0000 (1 votes)

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 Done. -spoiler | Comment 1 of 7

Sol:

36 BUCKS IN 18 COINS............ (1+2+3)*6

Let's  agree that 1+3=2+2, thus expressing 4 by two coins
in 2 different ways.

if A=1+3   AND  B=2+2

then

Brother 1 gets         A+A+A=1+1+1+3+3+3

Brother 2 gets          A+A+B=1+1+2+2+3+3

Brother 3    gets       A+ B+B=1+2+2+2+2+3

End

Edited on February 6, 2013, 8:42 pm
 Posted by Ady TZIDON on 2013-02-06 14:45:57

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