2009 AMC 10B Problem 7

P1-10~3 minPrint

Basic Counting Organized listing / direct enumeration

By inserting parentheses, it is possible to give the expression $2\times3 + 4\times5$ several values. How many different values can be obtained?

Show solution without answering

Key insight

Only the single plus sign can be moved earlier. It can join the 3 with the left product, the right product, both, or neither, giving 26, 46, 50 and 70.

Solution

Parentheses around a string of multiplications change nothing, because $2 \times 3 \times 4$ is the same in any order. So the only thing parentheses can really change is when the addition happens.

1. Without any parentheses, the multiplications go first: $2 \times 3 + 4 \times 5 = 6 + 20 = 26$ .
2. Make the addition happen before the multiplication on its right. The $3$ and the $4 \times 5$ get added first:
$2 \times (3 + 4 \times 5) = 2 \times 23 = 46$ .
3. Make the addition happen before the multiplication on its left. The $2 \times 3$ and the $4$ get added first:
$(2 \times 3 + 4) \times 5 = 10 \times 5 = 50$ .
4. Make the addition happen before both multiplications:
$2 \times (3 + 4) \times 5 = 2 \times 7 \times 5 = 70$ .
5. Any other way of placing parentheses does one of these four things, so it gives one of these four numbers.

The four values $26, 46, 50, 70$ are all different, so there are $4$ values.

The answer is $\boxed{\textbf{(C)}\ 4}$ .

Why this works

Parentheses only matter where they change the order of different operations. Focus on the one operation that is different (the addition) and ask which multiplications it can come before: the left one or not, the right one or not. That gives $2 \times 2 = 4$ cases. The question asks for different values, so compute each case and check none repeat.

Concise solution

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

Multiplication is associative, so regrouping a product alone changes nothing. The only thing parentheses can do is decide whether the addition happens before one or both of the neighboring multiplications. There are four possibilities:

Neither: $2\times3 + 4\times5 = 26$ (the usual value).
Addition before the right product: $2\times(3 + 4\times5) = 2 \times 23 = 46$ .
Addition before the left product: $(2\times3 + 4)\times5 = 10 \times 5 = 50$ .
Addition first, before both: $2\times(3+4)\times5 = 70$ .

Any other placement of parentheses reproduces one of these, so there are $4$ distinct values.

The answer is $\boxed{\textbf{(C)}\ 4}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques
Introduction to Counting & Probability, Chapter 4: Committees and Combinations