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]]}} | ||
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== 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 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>. Since <math>n | + | 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 17:45, 15 July 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.