Difference between revisions of "2002 AMC 12P Problems/Problem 13"
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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 = | + | 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> |
Revision as of 02:05, 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
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
. Therefore, the answer is
.
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.