Difference between revisions of "2002 AMC 12P Problems/Problem 13"
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Therefore, we know <math>n \leq 17</math>. | Therefore, we know <math>n \leq 17</math>. | ||
− | Now we must show that <math>n = 17</math> works. We replace | + | Now we must show that <math>n = 17</math> works. We replace some integer <math>b</math> within the set <math>\{1, 2, ... 17}\$ 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< | + | 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>, | + | 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. |
− | Case 1: <math>a+b = 217< | + | Case 1: </math>a+b = 217<math> and </math>a-b = 1<math> |
− | Solving this system of equations gives <math>a = 109< | + | 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< | + | Case 2: </math>a+b = 31<math> and </math>a-b = 7<math> |
− | Solving this system of equations gives <math>a = 19< | + | 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< | + | 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}$. |
== See also == | == See also == | ||
{{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}} |
Revision as of 03:03, 2 July 2024
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 $\{1, 2, ... 17}$ with an integer$ (Error compiling LaTeX. Unknown error_msg)a > 17
2002
2002-1785 = 217$.
Essentially, this boils down to writing$ (Error compiling LaTeX. Unknown error_msg)217a
b
a > 17
b \leq 17
a^2 - b^2 = 217$.
We can rewrite this as$ (Error compiling LaTeX. Unknown error_msg)(a+b)(a-b) = 217217 = (7)(31)
a+b = 217
a-b = 1
a+b = 31
a-b = 7$. We analyze each case separately.
Case 1:$ (Error compiling LaTeX. Unknown error_msg)a+b = 217a-b = 1
a = 109
b = 108
108 > 17$, so this case does not yield a solution.
Case 2:$ (Error compiling LaTeX. Unknown error_msg)a+b = 31a-b = 7
a = 19
b = 12$. This satisfies all the requirements of the problem.
The list$ (Error compiling LaTeX. Unknown error_msg)1, 2 ... 11, 13, 14 ... 17, 1917
2002
\boxed {\text{(D) }17}$.
See also
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.