2010 AMC 12A Problems/Problem 20
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; equivalently, we want to maximize .
The prime factorization of is . We list out all the possible pairs that have a product of , noting that these are the possible values of and we need :
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.)
Alternative Thinking
Since
and
,
it follows that
.
But and are also integers, so must be a factor of smaller than . Notice that . Therefore or and or ; respectively.
Notice that the term is equivalent to the first term plus times the common difference for that particular arithmetic sequence. Let the common difference of be and the common difference of be (not ). Then
(the th term, not the sequence itself)
and
Subtracting one from all the possible values listed above for and , we get
and
In order to maximize , we must maximize . Therefore and are coprime and is the GCF of any corresponding pair. Inspecting all of the pairs, we see that the GCF is always except for the pair which has a GCF of . Therefore the maximum value of is .
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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