Difference between revisions of "2008 AIME II Problems/Problem 15"
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<math> (N + 1)(N - 1) = 3A(A + 1)</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 | + | 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 == |
Revision as of 21:29, 23 December 2020
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.
Contents
[hide]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 BAYUS)
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 |
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