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 "2002 AMC 12P Problems/Problem 13"

m (→‎Solution)
(→‎Solution)
 
(6 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2002 AMC 12P Problems|2002 AMC 12P #13]] and [[2002 AMC 10P Problems|2002 AMC 10P #24]]}}
 +
 
== Problem ==
 
== Problem ==
 
What is the maximum value of <math>n</math> for which there is a set of distinct positive integers <math>k_1, k_2, ... k_n</math> for which
 
What is the maximum value of <math>n</math> for which there is a set of distinct positive integers <math>k_1, k_2, ... k_n</math> for which
Line 17: Line 19:
  
 
== Solution ==
 
== Solution ==
−
Note that <math>k^2_1 + k^2_2 + ... + k^2_n = 2002 \leq \frac{n(n+1)(2n+1)}{6}</math>
+
Note that <math>k^2_1 + k^2_2 + ... + k^2_n = 2002 \geq \frac{n(n+1)(2n+1)}{6}</math>
  
 
When <math>n = 17</math>, <math>\frac{n(n+1)(2n+1)}{6} = \frac{(17)(18)(35)}{6} = 1785 < 2002</math>.
 
When <math>n = 17</math>, <math>\frac{n(n+1)(2n+1)}{6} = \frac{(17)(18)(35)}{6} = 1785 < 2002</math>.
Line 29: Line 31:
 
Essentially, this boils down to writing <math>217</math> as a difference of squares. Assume there exist positive integers <math>a</math> and <math>b</math> where <math>a > 17</math> and <math>b \leq 17</math> such that <math>a^2 - b^2 = 217</math>.
 
Essentially, this boils down to writing <math>217</math> as a difference of squares. Assume there exist positive integers <math>a</math> and <math>b</math> where <math>a > 17</math> and <math>b \leq 17</math> such that <math>a^2 - b^2 = 217</math>.
  
−
We can rewrite this as <math>(a+b)(a-b) = 217</math>. Since <math>217 = (7)(31)</math>, either <math>a+b = 217</math> and <math>a-b = 1</math> or <math>a+b = 31</math> and <math>a-b = 7</math>. We analyze each case separately.
+
We can rewrite this as <math>(a+b)(a-b) = 217</math>. Since <math>217 = 7 \cdot 31</math>, either <math>a+b = 217</math> and <math>a-b = 1</math> or <math>a+b = 31</math> and <math>a-b = 7</math>. We analyze each case separately.
  
 
Case 1: <math>a+b = 217</math> and <math>a-b = 1</math>
 
Case 1: <math>a+b = 217</math> and <math>a-b = 1</math>
Line 39: Line 41:
 
Solving this system of equations gives <math>a = 19</math> and <math>b = 12</math>. This satisfies all the requirements of the problem.
 
Solving this system of equations gives <math>a = 19</math> and <math>b = 12</math>. This satisfies all the requirements of the problem.
  
−
The list <math>1, 2 ... 11, 13, 14 ... 17, 19</math> has <math>17</math> terms whose sum of squares equals <math>2002</math>. Therefore, the answer is <math>\boxed {\text{(D) }17}</math>.
+
The list <math>1, 2 ... 11, 13, 14 ... 17, 19</math> has <math>17</math> terms whose sum of squares equals <math>2002</math>. Since <math>n \geq 18</math> is impossible, the answer is <math>\boxed {\textbf{(D) }17}</math>.
  
 
== See also ==
 
== See also ==
 +
{{AMC10 box|year=2002|ab=P|num-b=23|num-a=25}}
 
{{AMC12 box|year=2002|ab=P|num-b=12|num-a=14}}
 
{{AMC12 box|year=2002|ab=P|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:20, 12 October 2024

The following problem is from both the 2002 AMC 12P #13 and 2002 AMC 10P #24, so both problems redirect to this page.

Problem

What is the maximum value of $n$ for which there is a set of distinct positive integers $k_1, k_2, ... k_n$ for which

\[k^2_1 + k^2_2 + ... + k^2_n = 2002?\]

$\text{(A) }14 \qquad \text{(B) }15 \qquad \text{(C) }16 \qquad \text{(D) }17 \qquad \text{(E) }18$

Solution

Note that $k^2_1 + k^2_2 + ... + k^2_n = 2002 \geq \frac{n(n+1)(2n+1)}{6}$

When $n = 17$, $\frac{n(n+1)(2n+1)}{6} = \frac{(17)(18)(35)}{6} = 1785 < 2002$.

When $n = 18$, $\frac{n(n+1)(2n+1)}{6} = 1785 + 18^2 = 2109 > 2002$.

Therefore, we know $n \leq 17$.

Now we must show that $n = 17$ works. We replace some integer $b$ within the set $\{1, 2, ... 17\}$ with an integer $a > 17$ to account for the amount under $2002$, which is $2002-1785 = 217$.

Essentially, this boils down to writing $217$ as a difference of squares. Assume there exist positive integers $a$ and $b$ where $a > 17$ and $b \leq 17$ such that $a^2 - b^2 = 217$.

We can rewrite this as $(a+b)(a-b) = 217$. Since $217 = 7 \cdot 31$, either $a+b = 217$ and $a-b = 1$ or $a+b = 31$ and $a-b = 7$. We analyze each case separately.

Case 1: $a+b = 217$ and $a-b = 1$

Solving this system of equations gives $a = 109$ and $b = 108$. However, $108 > 17$, so this case does not yield a solution.

Case 2: $a+b = 31$ and $a-b = 7$

Solving this system of equations gives $a = 19$ and $b = 12$. This satisfies all the requirements of the problem.

The list $1, 2 ... 11, 13, 14 ... 17, 19$ has $17$ terms whose sum of squares equals $2002$. Since $n \geq 18$ is impossible, the answer is $\boxed {\textbf{(D) }17}$.

See also

2002 AMC 10P (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 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=2002_AMC_12P_Problems/Problem_13&oldid=229685"