The mint of England

[Exploited]

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?

[Joka X]

Math again? :'( :'( :'(

[motomaniacs]

i thinks so

[Exploited]

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