1987 AIME Problems/Problem 8
Problem
What is the largest positive integer for which there is a unique integer such that ?
Solution 1
Multiplying out all of the denominators, we get:
Since , . Also, , so . Thus, . is unique if it is within a maximum range of , so .
Solution 2
Flip all of the fractions for
Continue as in Solution 1.
Solution 3
Flip the fractions and subtract one from all sides to yield Multiply both sides by to get This is equivalent to find the largest value of such that there is only one multiple of 56 within the open interval between and . If then and is the unique value. For there is at least possible numbers for and there is one every 56 numbers. Hence, there must be at least two values of that work. So, the largest value of is .
Solution 4
Notice that in order for to be unique, and must be true. Solving these inequalities for yields .
Thus, we want to find such that . Solving this inequality yields , and plugging this into in the original equation yields so the answer is .
See also
1987 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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