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Products of 22 (Posted on 2002-12-23) Difficulty: 2 of 5
The sum of the elements of a set of positive integers is 22.
What is the greatest possible product of the integers in this set if:

A Duplicates are allowed?
B Duplicates are not allowed?

Problem modified from UNL Math Day with help from friedlinguini

See The Solution Submitted by cges    
Rating: 3.2222 (9 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Puzzle solution with explanation | Comment 6 of 7 |
(In reply to answer by K Sengupta)

At the outset, for the given product to be a maximum, each of the numbers must be as close as possible to e.

(i) If repetitions are allowed, then the numbers must consist of repetitions of 2's and 3's. Let  2 be repeated p times while 3 is repeated q times.

Then we have to consider the maximum possible product of
2^p*3^q, subject to the constraint 2p+3q =22
A litle trial and error suggests that (p, q) = (6, 2)  maximises the said product.

This gives 2^2*3^6 = 2916 as the maximal product whenever repetitions are permitted.

(ii) If repetitions are not permitted then, it is evident that one of (1, 2, 3, 4, 5, 7 ) or (2, 3, 4, 6, 7) would give the maximal product.

Clearly, Prod (2, 3, 4, 6, 7) =  1008> Prod (1,2,3,4,5,7) = 840
Thus, the maximal product is 1008 when no repetitions are permissible and this is generated by the set ( 2, 3 , 4, 6, 7)

  Posted by K Sengupta on 2007-05-16 06:10:45
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