From Mathematics For Pleasure Paperback by Oswald Jacoby and William H. Benson (1965)
A certain article costs an amount which requires a minimum of four standard United States coins to pay for it. To purchase two of these articles would require a minimum of six coins. However, three of the article can be purchased for two coins. What is the price of the article?
The solution simply states that:
A little quick figuring will show that the article costs seventeen cents.
17 cents: 2 cents + 1 nickel + 1 dime = 4 coins
34 cents: 4 cents + 1 nickel + 1 quarter = 6 coins
51 cents: 1 cent + 1 half dollar = 2 coins
My question:
Except trial and error, what would be a methodical way to approach a problem such as this?
Some trial and error is probably inescapable, but perhaps it can be minimized.
The coins are 1, 5, 10, 25, and 50. Considering the 3rd condition, we need two of these to add up to a multiple of three. The only possibilities are $1+5=3\times2$, $1+50=3\times17$, $5+10=3\times5$, $5+25=3\times10$, $10+50=3\times20$, and $25+50=3\times25$. So you just have 6 cases to look at. Now go to the 2nd condition. Do you need six coins for any of 4, 34, 10, 20, 40, 50? Yes, but only for 34, so that's the only possible answer.