MathBin	Diophantine Equation - Coins in a Jar	Member

This bin is based on the November 17, 2019 column of Ask Marilyn by Marilyn vos Savant, where Marilyn responds to a question posed by Leda Metro of Darien, Conn.

In a contest, a friend guessed the exact number of coins in a jar: 1,692. She won all of the coins, which totaled $100.54. They consisted of pennies, nickels, dimes and quarters. How many coins of each kind where in the jar? Marilyn asks the reader to start with a guess, and reallocate the coins until they sum to the requested total. More than one answer is possible, and the two Marilyn provides are given below.

Analysis

This problem gives two linear equations in four unknowns, so the variables cannot be found using straight-up algebra. Using the variables q, d, n, and p to represent quarters, dimes, nickels, and pennies, we have the following two equations. Note that I am using cents instead of dollars in order to use integers instead of decimal numbers.

1. q + d + n + p = 1692
2. 25q + 10d + 5n + p = 10054

It's clear that at least 4 pennies are required, but what is the maximum number of pennies? Having all pennies (p = 1692) does not come close to the required amount in equation (1). To maximize p, we need to use pennies and quarters (why?). The equation to solve is 25(1692 - p) + p ≥ 10054. This yields p ≤ 1343.583, so p = 1343 and q = 349. Plugging these values into the right-hand-side of equation (1) gives 25(349) + 1343 = 10068. Since the result is a little too big, all this means is the maximum number of quarters is 348, the maximum number of pennies is still 1343, and the extra coin is a dime. To wit, 25(348) + 10(1) + 1343 = 10053, which is almost a solution!

If I solved for p in the first equation, I could eliminate p in the second equation.

1. p = 1692 - q - d - n
2. 24q + 9d + 4n = 8362

It's fairly obvious there are constraints on the variables.

• q + d + n < 1689, because we need at least 4 pennies.
• q < 349, even if d and n are zero in equation (1). I already proved this.
• d < 930, even if q and n are zero in equation (1).
• Both q and d cannot be zero. Why?
• n < 1609 because 1608 nickels and 80 quarters is still less than 8362 in equation (1).
• p < 1344. I already proved this.