Q: An imaginary country mints coins in three denominations. Each denomination has an integral number, 1, 2, 3, etc. The amounts, 20, 23 and 29, can each be made with exactly three coins. What are the three denominations minted?Since the deadline has passed, I can tell you my answer to the puzzle. I arrived at this answer with a little bit of experimentation and algebra. If you don't want the answer, stop reading.
Edit: 5, 9, 10 was given as another acceptable answer (using 2a + c = 20, 2b + a = 23, 2c + b = 29), but I personally don't like it. I read the question as if each combination could only be made with 3 coins, not more nor less. 20 could be made with two "dimes" or four "nickels" rather than 5+5+10. However, I guess it technically answers the puzzle question which is why they must have decided to allow it. I think 6, 7, 11 must have been the intended answer.A: Assume a, b and c are integers.
Let's try the following equations:
1) a + b + b = 20
2) a + a + c = 23
3) b + c + c = 29
Combining #1 and #2 we have:
2a + c = 23
2a + 4b = 40
4) 4b - c = 17
Next combining #3 and #4 we have:
4b - c = 17
4b + 8c = 116
9c = 99
5) c = 11
Combining #3 and #5 we have:
b + 22 = 29
6) b = 7
Combining #1 and #6 we have:
a + 14 = 20
7) a = 6
Thus the coins are 6, 7 and 11
6 + 7 + 7 = 20
6 + 6 + 11 = 23
7 + 11 + 11 = 29
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