Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2018 AIME I Problems/Problem 1"

m (→‎Question)
(→‎Solution)
Line 24: Line 24:
  
 
Thus, the answer is: <cmath>\boxed{600}.</cmath>
 
Thus, the answer is: <cmath>\boxed{600}.</cmath>
 +
 +
==See Also==
 +
{{AIME box|year=2018|n=I|before=First Problem|num-a=2}}
 +
{{MAA Notice}}

Revision as of 23:18, 7 March 2018

Problem 1

Let $S$ be the number of ordered pairs of integers $(a,b)$ with $1 \leq a \leq 100$ and $b \geq 0$ such that the polynomial $x^2+ax+b$ can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when $S$ is divided by $1000$.

Solution

You let the linear factors be as $(x+c)(x+d)$.

Then, obviously $a=c+d$ and $b=cd$.

We know that $1\le a\le 100$ and $b\ge 0$, so $c$ and $d$ both have to be positive.

However, $a$ cannot be $0$, so at least one of $c$ and $d$ must be greater than $0$, ie positive.

Also, $a$ cannot be greater than $100$, so $c+d$ must be less than or equal to $100$.

Essentially, if we plot the solutions, we get a triangle on the coordinate plane with vertices $(0,0), (0, 100),$ and $(100,0)$. Remember that $(0,0)$ does not work, so there is a square with top right corner $(1,1)$.

Note that $c$ and $d$ are interchangeable, since they end up as $a$ and $b$ in the end anyways. Thus, we simply draw a line from $(1,1)$ to $(50,50)$, designating one of the halves as our solution (since the other side is simply the coordinates flipped).

We note that the pattern from $(1,1)$ to $(50,50)$ is $2+3+4+\dots+51$ solutions and from $(51, 49)$ to $(100,0)$ is $50+49+48+\dots+1$ solutions, since we can decrease the $y$-value by $1$ until $0$ for each coordinate.

Adding up gives \[\dfrac{2+51}{2}\cdot 50+\dfrac{50+1}{2}\cdot 50.\] This gives us $2600$, and $2600\equiv 600 \bmod{1000}.$

Thus, the answer is: \[\boxed{600}.\]

See Also

2018 AIME I (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_1&oldid=92981"