2018 AIME I Problems/Problem 1

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 2

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 3 (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 4

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

Solution 5

Let's say that the quadratic $x^2 + ax + b$ can be factored into $(x+c)(x+d)$ where $c$ and $d$ are non-negative numbers. We can't have both of them zero because $a$ would not be within bounds. Also, $c+d \leq 100$ . Assume that $c < d$ . $d$ can be written as $c + x$ where $x \geq 0$ . Therefore, $c + d = 2c + x \leq 100$ . To find the amount of ordered pairs, we must consider how many values of $x$ are possible for each value of $c$ . The amount of possible values of $x$ is given by $100 - 2c + 1$ . The $+1$ is the case where $c = d$ . We don't include the case where $c = d = 0$ , so we must subtract a case from our total. The amount of ordered pairs of $(a,b)$ is: $\left(\sum_{c=0}^{50} (100 - 2c + 1)\right) - 1$ This is an arithmetic progression. $\frac{(101 + 1)(51)}{2} - 1 = 2601 - 1 = 2600$ When divided by $1000$ , it gives the remainder $\boxed{600}$

~Zeric Hang

Solution 6(simple)

By vietas, the sum of the roots is -a and the product is b. Therefore both roots are negative. For each value of a from 1 to 100, the number of b values is the number of ways to sum two numbers between 0 and a-1 inclusive to a. This is just $1 + 2 + 2 + 3 + 3 +... 50 + 50 + 51 = 2600$ . So the answer is $\boxed{600}$

-bron jiang

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

Page

Toolbox

Search