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

Revision as of 15:08, 7 March 2018

Question

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: $\boxed{600}.$

@@ Line 1: / Line 1: @@
+==Question==
+Let <math>S</math> be the number of ordered pairs of integers <math>(a,b)</math> with <math>1 \leq a \leq 100</math> and <math>b \geq 0</math> such that the polynomial <math>x^2+ax+b</math> can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when <math>S</math> is divided by <math>1000</math>.
+==Solution==
+You let the linear factors be as <math>(x+c)(x+d)</math>.
+Then, obviously <math>a=c+d</math> and <math>b=cd</math>.
+We know that <math>1\le a\le 100</math> and <math>b\ge 0</math>, so <math>c</math> and <math>d</math> both have to be positive.
+However, <math>a</math> cannot be <math>0</math>, so at least one of <math>c</math> and <math>d</math> must be greater than <math>0</math>, ie positive.
+Also, <math>a</math> cannot be greater than <math>100</math>, so <math>c+d</math> must be less than or equal to <math>100</math>.
+Essentially, if we plot the solutions, we get a triangle on the coordinate plane with vertices <math>(0,0), (0, 100),</math> and <math>(100,0)</math>. Remember that <math>(0,0)</math> does not work, so there is a square with top right corner <math>(1,1)</math>.
+Note that <math>c</math> and <math>d</math> are interchangeable, since they end up as <math>a</math> and <math>b</math> in the end anyways. Thus, we simply draw a line from <math>(1,1)</math> to <math>(50,50)</math>, designating one of the halves as our solution (since the other side is simply the coordinates flipped).
+We note that the pattern from <math>(1,1)</math> to <math>(50,50)</math> is <math>2+3+4+\dots+51</math> solutions and from <math>(51, 49)</math> to <math>(100,0)</math> is <math>50+49+48+\dots+1</math> solutions, since we can decrease the <math>y</math>-value by <math>1</math> until <math>0</math> for each coordinate.
+Adding up gives <cmath>\dfrac{2+51}{2}\cdot 50+\dfrac{50+1}{2}\cdot 50.</cmath>
+This gives us <math>2600</math>, and <math>2600\equiv 600 \bmod{1000}.</math>
+Thus, the answer is: <cmath>\boxed{600}.</cmath>
-$

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

Revision as of 15:08, 7 March 2018

Question

Solution