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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)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: Done. -spoiler | Comment 3 of 7 |
(In reply to Done. -spoiler by Ady TZIDON)

But

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       B+ B+B=2+2+2+2+2+2

has 5 1's, 5 3's and 6 2's, rather than an equal number of each denomination.

You want brother 3 to get A + B + B = 1 + 2 + 2 + 2 +2 + 3.

Edited on February 6, 2013, 5:09 pm
 Posted by Charlie on 2013-02-06 17:06:41

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