Difference between revisions of "2002 AMC 12P Problems/Problem 13"
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+ | {{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 | |
+ | |||
+ | <cmath>k^2_1 + k^2_2 + ... + k^2_n = 2002?</cmath> | ||
− | <math> \ | + | <math> |
+ | \text{(A) }14 | ||
+ | \qquad | ||
+ | \text{(B) }15 | ||
+ | \qquad | ||
+ | \text{(C) }16 | ||
+ | \qquad | ||
+ | \text{(D) }17 | ||
+ | \qquad | ||
+ | \text{(E) }18 | ||
+ | </math> | ||
== Solution == | == Solution == | ||
− | + | 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 = 18</math>, <math>\frac{n(n+1)(2n+1)}{6} = 1785 + 18^2 = 2109 > 2002</math>. | ||
+ | |||
+ | Therefore, we know <math>n \leq 17</math>. | ||
+ | |||
+ | Now we must show that <math>n = 17</math> works. We replace some integer <math>b</math> within the set <math>\{1, 2, ... 17\}</math> with an integer <math>a > 17</math> to account for the amount under <math>2002</math>, which is <math>2002-1785 = 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 \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> | ||
+ | |||
+ | Solving this system of equations gives <math>a = 109</math> and <math>b = 108</math>. However, <math>108 > 17</math>, so this case does not yield a solution. | ||
+ | |||
+ | Case 2: <math>a+b = 31</math> and <math>a-b = 7</math> | ||
+ | |||
+ | 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>. Since <math>n \geq 18</math> is impossible, the answer is <math>\boxed {\textbf{(D) }17}</math>. | ||
== See also == | == See also == | ||
− | {{AMC12 box|year= | + | {{AMC10 box|year=2002|ab=P|num-b=23|num-a=25}} |
+ | {{AMC12 box|year=2002|ab=P|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 18:20, 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 for which there is a set of distinct positive integers for which
Solution
Note that
When , .
When , .
Therefore, we know .
Now we must show that works. We replace some integer within the set with an integer to account for the amount under , which is .
Essentially, this boils down to writing as a difference of squares. Assume there exist positive integers and where and such that .
We can rewrite this as . Since , either and or and . We analyze each case separately.
Case 1: and
Solving this system of equations gives and . However, , so this case does not yield a solution.
Case 2: and
Solving this system of equations gives and . This satisfies all the requirements of the problem.
The list has terms whose sum of squares equals . Since is impossible, the answer is .
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.