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"

(→‎Solution)
m (→‎Solution 3)
Line 40: Line 40:
 
==Solution 3==
 
==Solution 3==
  
−
Let's write the linear factors as (x+n)(x+m).
+
Let's write the linear factors as <math>(x+n)(x+m)</math>.
  
−
Then we can write them as: a=n+m, b=nm.
+
Then we can write them as: <math>a=n+m, b=nm</math>.
  
−
n or m has to be a positive integer as a cannot be 0.
+
<math>n</math> or <math>m</math> 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.
+
<math>n+m</math> has to be between <math>1</math> and <math>100</math>, as a cannot be over <math>100</math>.
  
−
Excluding a=1, we can see there is always a pair of 2 a-values for a certain amount of b-values.  
+
Excluding <math>a=1</math>, we can see there is always a pair of <math>2</math> 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.
+
For instance, <math>a=2</math> and <math>a=3</math> both have <math>2</math> b-values. <math>a=4</math> and <math>a=5</math> both have <math>3</math> b-values.
  
 
We notice the pattern of the number of b-values in relation to the a-values:
 
We notice the pattern of the number of b-values in relation to the a-values:
−
1, 2, 2, 3, 3, 4, 4…
+
<math>1, 2, 2, 3, 3, 4, 4…</math>
  
 
The following link is the URL to the graph I drew showing the relationship between a-values and b-values
 
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
 
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.  
+
The pattern continues until <math>a=100</math>, and in total, there are <math>49</math> pairs of a-value with the same amount of b-values. The two lone a-values without a pair are, the (<math>a=1</math>, amount of b-values=1) in the beginning, and (<math>a=100</math>, 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:
+
Then, we add numbers from the opposite ends of the spectrum, and quickly notice that there are <math>50</math> pairs each with a sum of <math>52</math>. <math>52\cdot50</math> gives <math>2600</math> ordered pairs:
  
−
1+51, 2+50, 2+50, 3+49, 3+49, 4+48, 4+48…
+
<math>1+51, 2+50, 2+50, 3+49, 3+49, 4+48, 4+48…</math>
  
−
When divided by 1000, it gives the remainder 600, the answer.
+
When divided by <math>1000</math>, it gives the remainder <math>\boxed{600}</math>, the answer.
  
 
Solution provided by- Yonglao
 
Solution provided by- Yonglao

Revision as of 14:43, 12 February 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 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

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=101920"