Question 895527: Jon can put his gold coins into either two groups, or three groups, or four groups, with a different primenumber of coins in each group and no coins left over in each case. Which one or more of the following could NOT be the number of gold coins Jon has?
A = 90
B = 75
C = 60
D = 45
E = 30
Answer by Edwin McCravy(20056)   (Show Source):
You can put this solution on YOUR website!
We check each:
A 30 = 13+17 = 2+11+17 = ?+?+?+?
Let's show why we cannot have 4 groups of different primes adding up to 30.
One of the 4 groups cannot be 2, because 2 is the only even prime, and three
more odd primes added to 2 would given an odd number, not 30.
The smallest 4 odd primes have sum 3+5+7+11=26
If we use 13 instead of 11, we have sum 3+5+7+13=28
If we use 17 instead of 11, we have sum 3+5+7+17=32, too much.
We must have a group of 3, because if we left it out, the four smallest
odd primes without it would have sum 5+7+11+13=36, too much.
We must have a group of 5, because if we left it out, the four smallest
odd primes without it would have sum 3+7+11+13=34, too much.
We must have a group of 7, because if we left it out, the four smallest
odd primes without ie would have sum 3+5+11+13=32, too much.
We must have a group of 11, because if we left it out, the four smallest
odd primes would have sum 3+5+7+13=28, too much.
So Jon cannot have exactly 30 coins.
B 45 = 2+43 = 5+17+23 = 2+11+13+19, so Jon could have 45 coins.
C 60 = 29+31 = 2+17+41 = 11+13+17+19, so Jon could have 60 coins.
D 75 = 2+73 = 13+19+43 = 2+19+23+31, so Jon could have 75 coins.
E 90 = 43+47 = 2+41+47 = 17+19+23+31, so Jon could have 90 coins.
Of the choices listed, the only number of coins which Jon could not possibly
have is 30 coins.
Edwin
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