2006 AMC 10A Problem 9

P1-10~3 minPrint

Number Properties Sequences & Series Organized listing / direct enumeration

How many sets of two or more consecutive positive integers have a sum of $15$ ?

Show solution without answering

Key insight

Sum = (how many numbers) times (their average). Try runs of 2, 3, 4, 5, 6 numbers and see which give 15 using only positive numbers.

Solution

For consecutive numbers, the sum is (how many numbers) $\times$ (the average), and the average is the middle of the run. So for a run of $n$ numbers, the average must be $15 \div n$ . Try each length in turn.

Length $n$	Average $15 \div n$	Run	Works?
$2$	$7.5$	$7, 8$	yes
$3$	$5$	$4, 5, 6$	yes
$4$	$3.75$	none	no
$5$	$3$	$1, 2, 3, 4, 5$	yes
$6$	$2.5$	$0, 1, 2, 3, 4, 5$	no, $0$ is not positive

Why length $4$ fails: four consecutive numbers have a middle halfway between two whole numbers, like $2.5$ or $3.5$ . The average $3.75$ is not of that kind.

Why we can stop: a run of $7$ or more positive numbers adds up to at least $1 + 2 + \cdots + 7 = 28$ , which is more than $15$ .

So there are exactly $3$ sets.

The answer is $\boxed{\textbf{(C)}\ 3}$ .

Why this works

Consecutive numbers are evenly spaced, so their sum is the count times the middle value. That turns "find all runs" into "try each possible length," and there are only a few lengths before the sum gets too big. Keep an eye on the word "positive": it is what rules out the run that starts at $0$ .

Alternative approach

Just search by the first number. Starting at $1$ : $1 + 2 + 3 + 4 + 5 = 15$ . Starting at $2$ : $2 + 3 + 4 = 9$ , $+5 = 14$ , $+6 = 20$ , skips $15$ . Starting at $3$ : $3 + 4 = 7$ , $+5 = 12$ , $+6 = 18$ , skips. Starting at $4$ : $4 + 5 + 6 = 15$ . Starting at $5$ : $5 + 6 = 11$ , $+7 = 18$ , skips. Starting at $6$ : $6 + 7 = 13$ , $+8 = 21$ , skips. Starting at $7$ : $7 + 8 = 15$ . From $8$ on, two numbers already add to more than $15$ . Three sets.

Concise solution

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

A run of $n$ consecutive integers has sum $n$ times its average, and that average is the middle term when $n$ is odd, or a number ending in $.5$ when $n$ is even. So we need $15/n$ to have the right form, and the run built around it must stay positive.

$n = 2$ : average $7.5$ , run $7, 8$ . Works.
$n = 3$ : average $5$ , run $4, 5, 6$ . Works.
$n = 4$ : average $3.75$ , not a half-integer. Fails.
$n = 5$ : average $3$ , run $1, 2, 3, 4, 5$ . Works.
$n = 6$ : average $2.5$ , run $0, 1, \dots, 5$ , but $0$ is not positive. Fails.

Any longer run has average below $2.5$ and would start at or below $0$ , so the search is complete: exactly $3$ sets.

The answer is $\boxed{\textbf{(C)}\ 3}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)
Introduction to Algebra, Chapter 20: Special Functions (§20.3)