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 (→‎Solution 3)
(→‎Solution)
Line 24: Line 24:
  
 
Thus, the answer is: <cmath>\boxed{600}.</cmath>
 
Thus, the answer is: <cmath>\boxed{600}.</cmath>
 +
 +
==Solution==
 +
 +
Similar to the previous problem, we plot the triangle and cut it in half. Then, we find the number of boundary points, which is 100+51+51-3, or just 200. Using Pick's theorem, we know that the area of the half-triangle, which is 2500, is just I+100-1. That means that there are 2401 interior points, plus 200 boundary points, which is 2601. However, (0,0) does not work, so the answer is <cmath>\boxed{600}.</cmath>
  
 
==Solution 2 (less complicated)==
 
==Solution 2 (less complicated)==

Revision as of 12:12, 10 March 2019

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 non-negative

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: \[\boxed{600}.\]

Solution

Similar to the previous problem, we plot the triangle and cut it in half. Then, we find the number of boundary points, which is 100+51+51-3, or just 200. Using Pick's theorem, we know that the area of the half-triangle, which is 2500, is just I+100-1. That means that there are 2401 interior points, plus 200 boundary points, which is 2601. However, (0,0) does not work, so the answer is \[\boxed{600}.\]

Solution 2 (less complicated)

Notice that for $x^2+ax+b$ to be true, for every $a$, $b$ will always be the product of the possibilities of how to add two integers to $a$. For example, if $a=3$, $b$ will be the product of $(3,0)$ and $(2,1)$, as those two sets are the only possibilities of adding two integers to $a$. Note that order does not matter. If we just do some simple casework, we find out that:

if $a$ is odd, there will always be $\left\lceil\frac{a}{2}\right\rceil$ $\left(\text{which is also }\frac{a+1}{2}\right)$ possibilities of adding two integers to $a$.

if $a$ is even, there will always be $\frac{a}{2}+1$ possibilities of adding two integers to $a$.

Using the casework, we have $1+2+2+3+3+...50+50+51$ possibilities. This will mean that the answer is \[\frac{(1+51)\cdot100}{2}\Rightarrow52\cdot50=2600\] possibilities.

Thus, our solution is $2600\bmod {1000}\equiv\boxed{600}$.

Solution by IronicNinja~

Solution 3

Let's write the linear factors as $(x+n)(x+m)$.

Then we can write them as: $a=n+m, b=nm$.

$n$ or $m$ has to be a positive integer as a cannot be 0.

$n+m$ has to be between $1$ and $100$, as a cannot be over $100$.

Excluding $a=1$, we can see there is always a pair of $2$ a-values for a certain amount of b-values.

For instance, $a=2$ and $a=3$ both have $2$ b-values. $a=4$ and $a=5$ both have $3$ b-values.

We notice the pattern of the number of b-values in relation to the a-values: $1, 2, 2, 3, 3, 4, 4…$

The following link is the URL to the graph I drew showing the relationship between a-values and b-values http://artofproblemsolving.com/wiki/index.php?title=File:Screen_Shot_2018-04-30_at_8.15.00_PM.png#file

The pattern continues until $a=100$, and in total, there are $49$ pairs of a-value with the same amount of b-values. The two lone a-values without a pair are, the ($a=1$, amount of b-values=1) in the beginning, and ($a=100$, amount of b-values=51) in the end.

Then, we add numbers from the opposite ends of the spectrum, and quickly notice that there are $50$ pairs each with a sum of $52$. $52\cdot50$ gives $2600$ ordered pairs:

$1+51, 2+50, 2+50, 3+49, 3+49, 4+48, 4+48…$

When divided by $1000$, it gives the remainder $\boxed{600}$, the answer.

Solution provided by- Yonglao

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_1&oldid=104216"