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.
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 |