Difference between revisions of "1985 AIME Problems/Problem 13"
m (latex) |
m (→Solution 4) |
||
Line 21: | Line 21: | ||
==Solution 4== | ==Solution 4== | ||
− | We can just plug in Euclidean algorithm, to go from <math>gcd(n^2 + 100, n^2 + 2n + 101)</math> to <math>gcd(n^2 + 100, 2n + 1)</math> to <math>gcd(n^2 + 100 - 100(2n + 1), 2n + 1)</math> to get <math>gcd(n^2 - 200n, 2n + 1)</math>. Now we know that no matter what, <math>n</math> is relatively prime to <math>2n + 1</math>. Therefore the equation can be simplified to: <math>gcd(n - 200, 2n + 1)</math>. Subtracting <math>2n - 400</math> from <math>2n + 1</math> results in <math>gcd(n - 200,401)</math>. The greatest possible value of this is <math>\boxed{401}</math>, an happens when <math>n \equiv 200 \ | + | We can just plug in Euclidean algorithm, to go from <math>\gcd(n^2 + 100, n^2 + 2n + 101)</math> to <math>\gcd(n^2 + 100, 2n + 1)</math> to <math>\gcd(n^2 + 100 - 100(2n + 1), 2n + 1)</math> to get <math>\gcd(n^2 - 200n, 2n + 1)</math>. Now we know that no matter what, <math>n</math> is relatively prime to <math>2n + 1</math>. Therefore the equation can be simplified to: <math>\gcd(n - 200, 2n + 1)</math>. Subtracting <math>2n - 400</math> from <math>2n + 1</math> results in <math>\gcd(n - 200,401)</math>. The greatest possible value of this is <math>\boxed{401}</math>, an happens when <math>n \equiv 200 \pmod{401}</math>. |
+ | |||
== See also == | == See also == | ||
{{AIME box|year=1985|num-b=12|num-a=14}} | {{AIME box|year=1985|num-b=12|num-a=14}} |
Latest revision as of 18:19, 24 October 2020
Problem
The numbers in the sequence , , , , are of the form , where For each , let be the greatest common divisor of and . Find the maximum value of as ranges through the positive integers.
Solution 1
If denotes the greatest common divisor of and , then we have . Now assuming that divides , it must divide if it is going to divide the entire expression .
Thus the equation turns into . Now note that since is odd for integral , we can multiply the left integer, , by a power of two without affecting the greatest common divisor. Since the term is quite restrictive, let's multiply by so that we can get a in there.
So . It simplified the way we wanted it to! Now using similar techniques we can write . Thus must divide for every single . This means the largest possible value for is , and we see that it can be achieved when .
Solution 2
We know that and . Since we want to find the GCD of and , we can use the Euclidean algorithm:
Now, the question is to find the GCD of and . We subtract 100 times from . This leaves us with . We want this to equal 0, so solving for gives us . The last remainder is 0, thus is our GCD.
Solution 3
If Solution 2 is not entirely obvious, our answer is the max possible range of . Using the Euclidean Algorithm on and yields that they are relatively prime. Thus, the only way the GCD will not be 1 is if the term share factors with the . Using the Euclidean Algorithm, . Thus, the max GCD is 401.
Solution 4
We can just plug in Euclidean algorithm, to go from to to to get . Now we know that no matter what, is relatively prime to . Therefore the equation can be simplified to: . Subtracting from results in . The greatest possible value of this is , an happens when .
See also
1985 AIME (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 | ||
All AIME Problems and Solutions |