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

(Solution 6(simple))
(Solution 6(simple))
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==Solution 6(simple)==
 
==Solution 6(simple)==
By Vietas, the sum of the roots is <math>-a</math> and the product is <math>b</math>. Therefore, both roots are nonpositive. For each value of <math>a</math> from <math>1</math> to <math>100</math>, the number of <math>b</math> values is the number 4 of ways to sum two numbers between <math>0</math> and <math>a-1</math> inclusive to <math>a</math>. This is just <math>1 + 2 + 2 + 3 + 3 +... 50 + 50 + 51 = 2600</math>. Thus, the answer is <math>\boxed{600}</math>
+
By Vietas, the sum of the roots is <math>-a</math> and the product is <math>b</math>. Therefore, both roots are nonpositive. For each value of <math>a</math> from <math>1</math> to <math>100</math>, the number of <math>b</math> values is the number of ways to sum two numbers between <math>0</math> and <math>a-1</math> inclusive to <math>a</math>. This is just <math>1 + 2 + 2 + 3 + 3 +... 50 + 50 + 51 = 2600</math>. Thus, the answer is <math>\boxed{600}</math>
  
 
-bron jiang
 
-bron jiang

Revision as of 09:39, 29 October 2020

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 $101+51+51-3$, 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 nonpositive. 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 $1 + 2 + 2 + 3 + 3 +... 50 + 50 + 51 = 2600$. Thus, the answer is $\boxed{600}$

-bron jiang

Solution 7 (less room for error)

Similar to solution 1 we plot the triangle and half it. From dividing the triangle in half we are removing the other half of answers that are just flipped coordinates. We notice that we can measure the length of the longest side of the half triangle which is just from $0$ to $100$, so the number of points on that line is is $101$. The next row has length $99$, the one after that has length $97$, and so on. We simply add this arithmetic series of odd integers $101+99+97+...+1$. This is \[\frac{(101+1)(51)}{2} = 2601\] Or you can notice that this is the sum of the first $51$ odd terms, which is just $51^2 = 2601$. However, $(0,0)$ is the singular coordinate that does not work, so the answer is $(2601-1)\bmod {1000}\equiv\boxed{600}$

Solution by Damian Kim~

Video Solution

https://www.youtube.com/watch?v=WVtbD8x9fCM ~Shreyas S

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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