2009 AMC 10A Problem 2

P1-10~2 minPrint

Number Properties Modular Arithmetic Use the answer choices (mod checks, size, form) to eliminate or select Bound the quantity above/below or estimate to pin it down

Four coins are picked out of a piggy bank that contains a collection of pennies, nickels, dimes and quarters. Which of the following could not be the total value of the four coins, in cents?

Show solution without answering

Key insight

Every choice is a multiple of 5, so the pennies must total a multiple of 5, which means none; four coins of at least 5 cents make at least 20.

Solution

1. Look at the choices: $15, 25, 35, 45, 55$ . Each one is a multiple of $5$ .
2. Nickels ( $5$ ), dimes ( $10$ ) and quarters ( $25$ ) are all worth a multiple of $5$ cents. So if the total is a multiple of $5$ , the pennies by themselves must also add up to a multiple of $5$ .
3. There are only four coins, so there are at most $4$ pennies, worth at most $4$ cents. The only multiple of $5$ from $0$ to $4$ is $0$ . So there are no pennies at all.
4. That means every coin is worth at least $5$ cents, and four coins are worth at least $4 \times 5 = 20$ cents. A total of $15$ cents cannot happen.
5. Check that each other total really can happen:
- $25 = 5 + 5 + 5 + 10$
- $35 = 5 + 10 + 10 + 10$
- $45 = 5 + 5 + 10 + 25$
- $55 = 10 + 10 + 10 + 25$

The answer is $\boxed{\textbf{(A)}\ 15}$ .

Why this works

For a "which one is impossible" question, look for a reason that rules out one choice, then show the other choices by giving an example of each. Here the reason is divisibility by $5$ : it forces the pennies out, and then the smallest possible total is easy to see.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Every answer choice is a multiple of $5$ . Nickels, dimes and quarters are all worth multiples of $5$ cents, so the pennies alone must contribute a multiple of $5$ cents. With only four coins in hand, the number of pennies is at most $4$ , so it must be $0$ .

Therefore all four coins are worth at least $5$ cents each, and the total is at least $4 \cdot 5 = 20$ cents. A total of $15$ cents is impossible.

The other totals are all achievable:
- $25 = 5 + 5 + 5 + 10$
- $35 = 5 + 10 + 10 + 10$
- $45 = 5 + 5 + 10 + 25$
- $55 = 10 + 10 + 10 + 25$

The answer is $\boxed{\textbf{(A)}\ 15}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Number Theory, Chapter 12: Introduction to Modular Arithmetic
Introduction to Algebra, Chapter 20: Special Functions (§20.3)
Introduction to Number Theory, Chapter 10: Units Digits