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

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Thus, the answer is: <cmath>\boxed{600}.</cmath>
 
Thus, the answer is: <cmath>\boxed{600}.</cmath>
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 +
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Solution 2:
 +
 +
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 amount of b-values in relation to the a values:
 +
1, 2, 2, 3, 3, 4, 4…
 +
 +
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.
 +
 +
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
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2018|n=I|before=First Problem|num-a=2}}
 
{{AIME box|year=2018|n=I|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 01:21, 1 May 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:

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 amount of b-values in relation to the a values: 1, 2, 2, 3, 3, 4, 4…

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.

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

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

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