2020 AMC 12A Problems/Problem 13
Problem
There are integers and each greater than such that
for all . What is ?
Solution 1
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 .
Let us assume and as this is the most convenient. (EDIT: This used to say WLOG but that is inaccurate)
From the first equation, we get . Note also that from the second equation, and must be factors of 36.
After listing out the factors of 36 and utilising trial and error, we find that and works, with . So our answer is
~Silverdragon
Edits by ~Snore, ~Swaggergotcha
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
Note: 3 and 2 also factor into 24 and 36 while being 1 apart, but we can eliminate by quick guess and check. ~ Edit by Soupnoodle
Solution 4
can be rewritten as . The trick is to focus on the properties of the denominators and numerators. must be either be a factor of 36, because in any other situation the denominator of the final expression is preserved, and is forced to be 36, which is not an answer choice. Next, realize that the numerator of must be , and even more importantly, that the denominator must be large enough so that when is subtracted from it, it will form a fraction less than . The only number large enough to do this will also being a fraction of is 18. This means that is 2. Moving onward, we are left with . Since must be a factor of , just plug in values from the answer choices to find that and .
~jackshi2006
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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