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

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Notice that for <math>x^2+ax+b</math> to be true, for every <math>a</math>, <math>b</math> will always be the product of the possibilities of how to add two integers to <math>a</math>. For example, if <math>a=3</math>, <math>b</math> will be the product of <math>(3,0)</math> and <math>(2,1)</math>, as those two sets are the only possibilities of adding two integers to <math>a</math>. Note that order does not matter. If we just do some simple casework, we find out that:  
 
Notice that for <math>x^2+ax+b</math> to be true, for every <math>a</math>, <math>b</math> will always be the product of the possibilities of how to add two integers to <math>a</math>. For example, if <math>a=3</math>, <math>b</math> will be the product of <math>(3,0)</math> and <math>(2,1)</math>, as those two sets are the only possibilities of adding two integers to <math>a</math>. Note that order does not matter. If we just do some simple casework, we find out that:  
  
if <math>a</math> is odd, there will always be <math>\lceil\frac{a}{2}\rceil</math> <math>(</math>which is also <math>\frac{a+1}{2}</math><math>)</math> possibilities of adding two integers to <math>a</math>.
+
if <math>a</math> is odd, there will always be <math>\left\lceil\frac{a}{2}\right\rceil</math> <math>\left(\text{which is also }\frac{a+1}{2}\right)</math> possibilities of adding two integers to <math>a</math>.
  
 
if <math>a</math> is even, there will always be <math>\frac{a}{2}+1</math> possibilities of adding two integers to <math>a</math>.  
 
if <math>a</math> is even, there will always be <math>\frac{a}{2}+1</math> possibilities of adding two integers to <math>a</math>.  

Revision as of 23:36, 25 June 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: \[\boxed{600}.\]

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. 52x50 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 600, the answer.

Solution provided by- Yonglao

See Also

2018 AIME I (ProblemsAnswer KeyResources)
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

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