2020 AMC 12A Problems/Problem 13
Problem
There are integers and each greater than such that
for all . What is ?
Solution
can be simplified to
The equation is then which implies that
has to be since . is the result when and are and
being will make the fraction which is close to .
Finally, with being , the fraction becomes . In this case and work, which means that must equal ~lopkiloinm
Solution 2
As above, notice that you get
Now, combine the fractions to get .
WLOG, let and .
From the first equation we get . Note also that from the second equation, and must both be factors of 36.
After some trial and error we find that and works, with . So our answer is
~Silverdragon
Edits by ~Snore
Solution 3
Collapsed, . Comparing this to , observe that and . The first can be rewritten as . Then, has to factor into 24 while 1 less than that also must factor into 36. The prime factorizations are as follows and . Then, , as only 4 and 3 factor into 36 and 24 while being 1 apart.
~~BJHHar
See Also
2020 AMC 12A (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 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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