Difference between revisions of "2010 AMC 12A Problems/Problem 20"
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Hence <math>n=8</math> is the largest possible <math>n</math>. (There is no need to check <math>n=2</math> anymore.) | Hence <math>n=8</math> is the largest possible <math>n</math>. (There is no need to check <math>n=2</math> anymore.) | ||
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== See also == | == See also == |
Revision as of 20:51, 19 November 2017
Problem
Arithmetic sequences and have integer terms with and for some . What is the largest possible value of ?
Solution
Solution 1
Since and have integer terms with , we can write the terms of each sequence as
where and () are the common differences of each, respectively.
Since
it is easy to see that
.
Hence, we have to find the largest such that and are both integers.
The prime factorization of is . We list out all the possible pairs that have a product of
and soon find that the largest value is for the pair , and so the largest value is .
Solution 2
As above, let and for some .
Now we get , hence . Therefore divides . And as the second term is greater than the first one, we only have to consider the options .
For we easily see that for the right side is less than and for any other it is way too large.
For we are looking for such that . Note that must be divisible by . We can start looking for the solution by trying the possible values for , and we easily discover that for we get , which has a suitable solution .
Hence is the largest possible . (There is no need to check anymore.)
See also
2010 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 |
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