Difference between revisions of "2008 AIME II Problems/Problem 15"
m (→Solution 2) |
|||
(7 intermediate revisions by 5 users not shown) | |||
Line 1: | Line 1: | ||
+ | __TOC__ | ||
+ | |||
== Problem == | == Problem == | ||
Find the largest integer <math>n</math> satisfying the following conditions: | Find the largest integer <math>n</math> satisfying the following conditions: | ||
Line 4: | Line 6: | ||
:(ii) <math>2n + 79</math> is a perfect square. | :(ii) <math>2n + 79</math> is a perfect square. | ||
− | |||
== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
Line 22: | Line 23: | ||
(2 + \sqrt {3})^{2k + 1} = a_k + \sqrt {3}b_k | (2 + \sqrt {3})^{2k + 1} = a_k + \sqrt {3}b_k | ||
</cmath> | </cmath> | ||
− | where <math>k = 0,1,2,..</math>. So we can (with very little effort) obtain the following: <math>(k,a_k = 2n) = (0,2),(1,26),(2,362),(3,5042)</math>. It is an AIME problem so it is implicit that <math>n < 1000</math>, so <math>2n < 2000</math>. It is easy to see that <math>a_n</math> is strictly increasing by induction. Checking <math>2n = 362\implies n =boxed{181}</math> in the second condition works (we know <math>b_k</math> is odd so we don't need to find <math>m</math>). So we're done. | + | where <math>k = 0,1,2,..</math>. So we can (with very little effort) obtain the following: <math>(k,a_k = 2n) = (0,2),(1,26),(2,362),(3,5042)</math>. It is an AIME problem so it is implicit that <math>n < 1000</math>, so <math>2n < 2000</math>. It is easy to see that <math>a_n</math> is strictly increasing by induction. Checking <math>2n = 362\implies n =\boxed{181}</math> in the second condition works (we know <math>b_k</math> is odd so we don't need to find <math>m</math>). So we're done. |
+ | |||
+ | === Solution 3 (Big Bash) === | ||
+ | |||
+ | Let us generate numbers <math>1</math> to <math>1000</math> for the second condition, for squares. We know for <math>N</math> to be integer, the squares must be odd. So we generate <math>N = 1, 21, 45, 73, 105, 141, 181, 381, 441, 721, 801</math>. <math>N</math> cannot exceed <math>1000</math> since it is AIME problem. Now take the first criterion, let <math>a</math> be the smaller consecutive cube. We then get: | ||
+ | |||
+ | <math> N^2 = (A + 1)^3 - A^3</math> | ||
+ | |||
+ | <math> N^2 - 1 = 3A^2 + 3A</math> | ||
+ | |||
+ | <math> (N + 1)(N - 1) = 3A(A + 1)</math> | ||
+ | |||
+ | Now we know either <math>N + 1</math> or <math>N - 1</math> must be factor of <math>3</math>, hence <math>N = 1 \pmod 3 </math> or<math> N = 2 \pmod 3</math>. Only <math>1, 73, 181, 721</math> satisfy this criterion. Testing each of the numbers in the condition yields <math>181</math> as the largest that fits both, thus answer <math>= \boxed{181}</math>. | ||
== See also == | == See also == |
Latest revision as of 15:23, 2 January 2024
Problem
Find the largest integer satisfying the following conditions:
- (i) can be expressed as the difference of two consecutive cubes;
- (ii) is a perfect square.
Solution
Solution 1
Write , or equivalently, .
Since and are both odd and their difference is , they are relatively prime. But since their product is three times a square, one of them must be a square and the other three times a square. We cannot have be three times a square, for then would be a square congruent to modulo , which is impossible.
Thus is a square, say . But is also a square, say . Then . Since and have the same parity and their product is even, they are both even. To maximize , it suffices to maximize and check that this yields an integral value for . This occurs when and , that is, when and . This yields and , so the answer is .
Solution 2
Suppose that the consecutive cubes are and . We can use completing the square and the first condition to get: where and are non-negative integers. Now this is a Pell equation, with solutions in the form . However, is even and is odd. It is easy to see that the parity of and switch each time (by induction). Hence all solutions to the first condition are in the form: where . So we can (with very little effort) obtain the following: . It is an AIME problem so it is implicit that , so . It is easy to see that is strictly increasing by induction. Checking in the second condition works (we know is odd so we don't need to find ). So we're done.
Solution 3 (Big Bash)
Let us generate numbers to for the second condition, for squares. We know for to be integer, the squares must be odd. So we generate . cannot exceed since it is AIME problem. Now take the first criterion, let be the smaller consecutive cube. We then get:
Now we know either or must be factor of , hence or. Only satisfy this criterion. Testing each of the numbers in the condition yields as the largest that fits both, thus answer .
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.