2014 AMC 10A Problem 21

P21-25~3 minPrint

Diophantine Equations Coordinate Geometry Linear Equations & Word Problems Organized listing / direct enumeration Set up the equation/formula and compute; no special trick needed

Positive integers $a$ and $b$ are such that the graphs of $y=ax+5$ and $y=3x+b$ intersect the $x$ -axis at the same point. What is the sum of all possible $x$ -coordinates of these points of intersection?

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Key insight

Setting the two x-intercepts equal gives a times b = 15. So a is 1, 3, 5 or 15, and the crossing points are -5, -5/3, -1 and -1/3.

Solution

A graph crosses the $x$ -axis where $y = 0$ , so find that point for each line.

First line. $0 = ax + 5$ gives $x = -\frac{5}{a}$ .
Second line. $0 = 3x + b$ gives $x = -\frac{b}{3}$ .
Same point. We need $\frac{5}{a} = \frac{b}{3}$ . Cross-multiply: $ab = 15$ .
List the options. $a$ and $b$ are positive whole numbers that multiply to $15$ :

$a$	$b$	crossing point $-\frac{5}{a}$
$1$	$15$	$-5$
$3$	$5$	$-\frac53$
$5$	$3$	$-1$
$15$	$1$	$-\frac13$

Add them. Group the fractions: $-\frac53 - \frac13 = -2$ , and $-5 - 1 = -6$ . The total is $-6 - 2 = -8$ .

The answer is $\boxed{\textbf{(E)}\ {-8}}$ .

Why this works

"Same $x$ -intercept" gives one equation linking $a$ and $b$ . Because they must be positive whole numbers, the equation $ab = 15$ has only a few solutions, one for each way to split $15$ into two factors. Listing them and computing the crossing point for each is then routine. Watch for this: a geometry condition that turns into "two whole numbers multiply to a fixed number".

Alternative approach

Use the other formula for the crossing point, $-\frac{b}{3}$ , as a check. The possible $b$ values are $1, 3, 5, 15$ , which add to $24$ . So the crossing points add to $-\frac{24}{3} = -8$ .

Concise solution

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

Set $y = 0$ in each equation. The line $y = ax + 5$ meets the $x$ -axis at $x = -\frac{5}{a}$ , and $y = 3x + b$ meets it at $x = -\frac{b}{3}$ . These are equal exactly when
$\frac{5}{a} = \frac{b}{3} \quad\Longleftrightarrow\quad ab = 15.$

Since $a$ and $b$ are positive integers, $(a, b)$ is one of $(1, 15),\ (3, 5),\ (5, 3),\ (15, 1)$ . The common $x$ -intercept is $-\frac{5}{a}$ , giving
$-5,\quad -\frac{5}{3},\quad -1,\quad -\frac{1}{3}.$

Their sum is
$-\left(5 + 1 + \frac{5}{3} + \frac{1}{3}\right) = -(6 + 2) = -8.$

The answer is $\boxed{\textbf{(E)}\ {-8}}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 7: Algebra With Integers
Introduction to Geometry, Chapter 17: Analytic Geometry
Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)