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 "2004 AIME I Problems/Problem 7"

m (→‎Solution 2)
(→‎Solutions)
 
(12 intermediate revisions by 8 users not shown)
Line 3: Line 3:
  
 
__TOC__
 
__TOC__
−
== Solution ==
+
== Solutions ==
 
=== Solution 1 ===
 
=== Solution 1 ===
 
Let our [[polynomial]] be <math>P(x)</math>.
 
Let our [[polynomial]] be <math>P(x)</math>.
Line 21: Line 21:
 
<center><math>\sum_{i=1}^{15} S_i = S_{15} + \left(\sum_{i=1}^{7} S_{2i-1} + S_{2i}\right) = -15 + 7 = -8</math></center>
 
<center><math>\sum_{i=1}^{15} S_i = S_{15} + \left(\sum_{i=1}^{7} S_{2i-1} + S_{2i}\right) = -15 + 7 = -8</math></center>
 
and the right-hand sum comes from the formula for the sum of the first <math>n</math> perfect squares. Therefore, <math>|C| = \left|\frac{64-1240}{2}\right| = \boxed{588}</math>.
 
and the right-hand sum comes from the formula for the sum of the first <math>n</math> perfect squares. Therefore, <math>|C| = \left|\frac{64-1240}{2}\right| = \boxed{588}</math>.
 +
 +
=== Solution 3 (Bash)===
 +
 +
Consider the set <math>[-1, 2,-3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15]</math>. Denote by <math>S</math> all size 2 subsets of this set. Replace each element of <math>S</math> by the product of the elements. Now, the quantity we seek is the sum of each element. Since consecutive elements add to <math>1</math> or <math>-1</math>, we can simplify this to <math>588</math>
 +
 +
=== Solution 4 (Easy) ===
 +
We know that this polynomial has roots <math>1, -\frac{1}{2}, \frac{1}{3},\ldots</math> and the coefficient of <math>x^2</math> will be the sum of the product taken by <math>13</math>. However, since this is closer to the constant side, we can create a new polynomial with the reciprocal roots which will make <math>C</math> the coefficient of <math>x^{13}</math> and thus the sum of the reciprocal roots taken by 2. We can calculate this with <math>1\cdot(8-1)-2\cdot(8-(-2))+\ldots+15(8-15)</math>. This will give us <math>2C = -1176</math> giving us a final answer of <math>|C| = \fbox{588}</math>.
 +
 +
~ Vedoral
 +
 +
==Solution 4==
 +
Let set <math>N</math> be <math>\{-1, -3, \ldots -15\}</math> and set <math>P</math> be <math>\{2, 4, \ldots 14\}</math>. The sum of the negative <math>x^2</math> coefficients is the sum of the products of the elements in all two element sets such that one element is from <math>N</math> and the other is from <math>P</math>. Each summand is a term in the expansion of
 +
<cmath>(-1 - 3 - \ldots - 15)(2 + 4 + \ldots + 14)</cmath>
 +
which equals <math>-56 * 64 = -(60^2 - 4^2) = -3584</math>. The sum of the positive <math>x^2</math> coefficients is the sum of the products of all two element sets such that the two elements are either both in <math>N</math> or both in <math>P</math>. By counting, the sum is <math>2992</math>, so the sum of all <math>x^2</math> coefficients is <math>-588</math>. Thus, the answer is <math>\boxed{588}</math>.
 +
 +
 +
== Solution 5==
 +
 +
We can find out the coefficient of <math>x^2</math> by multiplying every pair of two coefficients for <math>x</math>. This means that we multiply <math>-1</math> by <math>2,-3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15</math> and <math>2</math> by <math>3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15</math>. and etc. This sum can be easily simplified and is equal to <math>(-1)(-7)+(-3)(-6)+(-5)(-5)+(-7)(-4)+(-9)(-3)+(-11)(-2)+(-13)(-1)+2(-9)+4(-10)+6(-11)+8(-12)+10(-13)+12(-14)+14(-15)</math> or <math>588</math>.
 +
 +
-David Camacho
 +
 +
==Solution 6==
 +
This is just another way of summing the subsets of 2 from <math>[-1, 2, -3, 4, -5, 6, -7, 8, -9, 10, -11, 12, -13, 14, -15]</math>. Start from the right and multiply -15 to everything on its left. Use the distributive property and add all the 14 integers together to get 7. This gives us <math>-15 * 7</math>. Doing this for 14 gives us <math>14 * -7</math>, and for -13 we get <math>-13 * 6</math>. This pattern repeats where every two integers will multiple 7, 6,... to 0. Combining and simplifying the pattern give us this: <math>-(29 * 7 + 25 * 6 + 21 * 5 + 17 * 4 + 13 * 3 + 9 * 2 + 5*1)</math>. The expression gives us -588, or <math>C = \boxed{588}</math>. This is a good solution because it guarantees we never add a product twice, and the pattern is simple to add by hand.
 +
 +
-jackshi2006
 +
 +
==Solution 7==
 +
We expand and obtain <math>\left(x-1\right)\left(2x-1\right)\left(3x-1\right)\cdots\left(15x-1\right) =  1307674368000x^{15} - 948550176000x^{14} - 689324826240x^{13} + 2733483288464x^{12} + 82808260416x^{11} - 23038684088x^{10} - 3811851848x^9 + 828730833x^8 + 81228128x^7 - 14661124x^6 - 853104x^5 + 132902x^4 + 4256x^3 - \boxed{588}x^2 - 8x + 1.</math>
 +
 +
Do not do this in an actual competition.
 +
 +
~Sliced_Bread
  
 
== See also ==
 
== See also ==

Latest revision as of 10:37, 27 May 2024

Problem

Let $C$ be the coefficient of $x^2$ in the expansion of the product $(1 - x)(1 + 2x)(1 - 3x)\cdots(1 + 14x)(1 - 15x).$ Find $|C|.$

Solutions

Solution 1

Let our polynomial be $P(x)$.

It is clear that the coefficient of $x$ in $P(x)$ is $-1 + 2 - 3 + \ldots + 14 - 15 = -8$, so $P(x) = 1 -8x + Cx^2 + Q(x)$, where $Q(x)$ is some polynomial divisible by $x^3$.

Then $P(-x) = 1 + 8x + Cx^2 + Q(-x)$ and so $P(x)\cdot P(-x) = 1 + (2C - 64)x^2 + R(x)$, where $R(x)$ is some polynomial divisible by $x^3$.

However, we also know $P(x)\cdot P(-x) = (1 - x)(1 + x)(1 +2x)(1 - 2x) \cdots (1 - 15x)(1 + 15x)$ $= (1 - x^2)(1 - 4x^2)\cdots(1 - 225x^2)$ $= 1 - (1 + 4 + \ldots + 225)x^2 + R(x)$.

Equating coefficients, we have $2C - 64 = -(1 + 4 + \ldots + 225) = -1240$, so $-2C = 1176$ and $|C| = \boxed{588}$.

Solution 2

Let $S$ be the set of integers $\{-1,2,-3,\ldots,14,-15\}$. The coefficient of $x^2$ in the expansion is equal to the sum of the product of each pair of distinct terms, or $C = \sum_{1 \le i \neq j}^{15} S_iS_j$. Also, we know that \begin{align*}\left(\sum_{i=1}^{n} S_i\right)^2 &= \left(\sum_{i=1}^{n} S_i^2\right) + 2\left(\sum_{1 \le i \neq j}^{15} S_iS_j\right)\\ (-8)^2 &= \frac{15(15+1)(2\cdot 15+1)}{6} + 2C\end{align*} where the left-hand sum can be computed from:

$\sum_{i=1}^{15} S_i = S_{15} + \left(\sum_{i=1}^{7} S_{2i-1} + S_{2i}\right) = -15 + 7 = -8$

and the right-hand sum comes from the formula for the sum of the first $n$ perfect squares. Therefore, $|C| = \left|\frac{64-1240}{2}\right| = \boxed{588}$.

Solution 3 (Bash)

Consider the set $[-1, 2,-3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15]$. Denote by $S$ all size 2 subsets of this set. Replace each element of $S$ by the product of the elements. Now, the quantity we seek is the sum of each element. Since consecutive elements add to $1$ or $-1$, we can simplify this to $588$

Solution 4 (Easy)

We know that this polynomial has roots $1, -\frac{1}{2}, \frac{1}{3},\ldots$ and the coefficient of $x^2$ will be the sum of the product taken by $13$. However, since this is closer to the constant side, we can create a new polynomial with the reciprocal roots which will make $C$ the coefficient of $x^{13}$ and thus the sum of the reciprocal roots taken by 2. We can calculate this with $1\cdot(8-1)-2\cdot(8-(-2))+\ldots+15(8-15)$. This will give us $2C = -1176$ giving us a final answer of $|C| = \fbox{588}$.

~ Vedoral

Solution 4

Let set $N$ be $\{-1, -3, \ldots -15\}$ and set $P$ be $\{2, 4, \ldots 14\}$. The sum of the negative $x^2$ coefficients is the sum of the products of the elements in all two element sets such that one element is from $N$ and the other is from $P$. Each summand is a term in the expansion of \[(-1 - 3 - \ldots - 15)(2 + 4 + \ldots + 14)\] which equals $-56 * 64 = -(60^2 - 4^2) = -3584$. The sum of the positive $x^2$ coefficients is the sum of the products of all two element sets such that the two elements are either both in $N$ or both in $P$. By counting, the sum is $2992$, so the sum of all $x^2$ coefficients is $-588$. Thus, the answer is $\boxed{588}$.


Solution 5

We can find out the coefficient of $x^2$ by multiplying every pair of two coefficients for $x$. This means that we multiply $-1$ by $2,-3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15$ and $2$ by $3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15$. and etc. This sum can be easily simplified and is equal to $(-1)(-7)+(-3)(-6)+(-5)(-5)+(-7)(-4)+(-9)(-3)+(-11)(-2)+(-13)(-1)+2(-9)+4(-10)+6(-11)+8(-12)+10(-13)+12(-14)+14(-15)$ or $588$.

-David Camacho

Solution 6

This is just another way of summing the subsets of 2 from $[-1, 2, -3, 4, -5, 6, -7, 8, -9, 10, -11, 12, -13, 14, -15]$. Start from the right and multiply -15 to everything on its left. Use the distributive property and add all the 14 integers together to get 7. This gives us $-15 * 7$. Doing this for 14 gives us $14 * -7$, and for -13 we get $-13 * 6$. This pattern repeats where every two integers will multiple 7, 6,... to 0. Combining and simplifying the pattern give us this: $-(29 * 7 + 25 * 6 + 21 * 5 + 17 * 4 + 13 * 3 + 9 * 2 + 5*1)$. The expression gives us -588, or $C = \boxed{588}$. This is a good solution because it guarantees we never add a product twice, and the pattern is simple to add by hand.

-jackshi2006

Solution 7

We expand and obtain $\left(x-1\right)\left(2x-1\right)\left(3x-1\right)\cdots\left(15x-1\right) = 1307674368000x^{15} - 948550176000x^{14} - 689324826240x^{13} + 2733483288464x^{12} + 82808260416x^{11} - 23038684088x^{10} - 3811851848x^9 + 828730833x^8 + 81228128x^7 - 14661124x^6 - 853104x^5 + 132902x^4 + 4256x^3 - \boxed{588}x^2 - 8x + 1.$

Do not do this in an actual competition.

~Sliced_Bread

See also

2004 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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=2004_AIME_I_Problems/Problem_7&oldid=219810"