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Title: The mint of England Post by: Exploited on October 08, 2007, 11:43:52 am We have a problem in the mint of England (the monetary holder bank).
Suddenly they decieded that they must release new hard cash (monets). The rule was that every cash sum from 1 to 20 must be paid with one or maximum 2 monets without need for change (no returning of odd money)! What is the minimal number of different hard cash (monets) that they can release and still follow the rules? Title: Re: The mint of England Post by: Joka X on October 08, 2007, 10:48:09 pm Math again? :'( :'( :'(
Title: Re: The mint of England Post by: motomaniacs on October 10, 2007, 08:43:41 pm i thinks so ;)
Title: Re: The mint of England Post by: Exploited on October 24, 2007, 09:48:01 am Solution:
With one coin we can pay 2 sums (for example 1=1 and 1+1=2). Just a note here that we must have coin 1 for sure! With two coins we can pay 4 sums With three coins we can pay 9 sums With four coins we can pay 18 sums - still not sufficient, because we have 20 sums => we must have at least 5 coins. We will assume for a moment that we will need 5 coins total - let's name them A, B, C, D and E (and we have A<B<C<D<E) Every sum must be presented on one way only (so we can minimize the number of the coins). A=1 => we do not need a coin with amount 2, because A+A=1+1=2. However we need a coin 3 => B=3. The ready sums are 1+1=2, 1=1, 3=3, 1+3=4 and 3+3=6 => we have a missing sum of 5 => we need a coing C=5, but we have a duplicate here, becase 3+3=6 and 1+5=6 => 5 coins total is not enough... We will try to continue with adding a 6th coin F (F>E) the coins 1,3 and 5 can pay the sums of 1,2,3,4,5,6 - unfortunately they cannot pay the sum 7 => our fourth coin D=7 The coins 1,3,5 and 7 can make 1,2,3,4,5,6,7,8, but they cannot pay 9 => the next coin E=9 1,3,5,7 and 9 can pay almost everything (as we suspected), but not all - they cannot make 20... => our final 6th coin will be F=10 All coins are 1,3,5,7,9 and 10... Case solved |