2004 AMC 10B Problem 10

P1-10~2 minPrint

Sequences & Series Number Properties Apply an identity: SFFT, sum of squares, difference of cubes, Vieta

A grocer makes a display of cans in which the top row has one can and each lower row has two more cans than the row above it. If the display contains $100$ cans, how many rows does it contain?

Show solution without answering

Key insight

The rows hold 1, 3, 5, 7, ... cans, and adding up the first few odd numbers always gives a perfect square. So 100 cans means 10 rows.

Solution

Write out the rows: $1, 3, 5, 7, \ldots$ cans. Each row has $2$ more than the one above, so these are the odd numbers.
Add up the first few rows and look for a pattern:

Rows	Total cans
$1$	$1$
$2$	$1 + 3 = 4$
$3$	$4 + 5 = 9$
$4$	$9 + 7 = 16$
$5$	$16 + 9 = 25$

The totals are $1, 4, 9, 16, 25$ , the perfect squares. With $n$ rows the total is $n \times n$ .
We need $n \times n = 100$ , so $n = 10$ .
Check another way: $10$ rows go from $1$ to $19$ . Pair the first and last, $1 + 19 = 20$ , then $3 + 17 = 20$ , and so on. That is $5$ pairs of $20$ , which is $100$ .

The answer is $\boxed{\textbf{(D)}\ 10}$ .

Why this works

Adding the first $n$ odd numbers always gives $n^2$ . A picture shows why: a $3 \times 3$ square of dots grows into a $4 \times 4$ square by adding an L-shaped strip of $7$ dots, and each new strip is the next odd number. When a total looks like a perfect square, check whether odd numbers are hiding in the problem.

Alternative approach

Test the nearby choices. $9$ rows give $9 \times 9 = 81$ cans and $11$ rows give $121$ cans. Only $10$ rows give exactly $100$ .

Concise solution

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

The rows contain $1, 3, 5, 7, \dots$ cans: the odd numbers in order. With $n$ rows, the last row has $2n - 1$ cans, and the total is an arithmetic series with $n$ terms, first term $1$ and last term $2n - 1$ :
$1 + 3 + \dots + (2n - 1) = n \cdot \frac{1 + (2n - 1)}{2} = n \cdot n = n^2.$

Setting $n^2 = 100$ gives $n = 10$ .

Quick check: ten rows end with $19$ cans, and $10 \cdot \frac{1 + 19}{2} = 100$ .

The answer is $\boxed{\textbf{(D)}\ 10}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Algebra, Chapter 11: Special Factorizations
Intermediate Algebra, Chapter 17: More Sequences and Series (§17.1, §17.2)