2016 AMC 12B Problems/Problem 22
Contents
[hide]Problem
For a certain positive integer less than , the decimal equivalent of is , a repeating decimal of period of , and the decimal equivalent of is , a repeating decimal of period . In which interval does lie?
Solution
Solution by e_power_pi_times_i
If , must be a factor of . Also, by the same procedure, must be a factor of . Checking through all the factors of and that are less than , we see that is a solution, so the answer is .
Note: and are both solutions, which invalidates this method. However, we need to examine all factors of that are not factors of , , or , or . Additionally, we need to be a factor of but not , , or . Indeed, satisfies these requirements.
We can see that and are not solutions by checking it in the requirements of the problem: , period 1, and , period 3. Thus, is the only answer.
For anyone who wants more information about repeating decimals, visit: https://en.wikipedia.org/wiki/Repeating_decimal
Solution 2 (Faster Approach)
Notice that the repeating fraction can be represented as and thereby, Also, notice that so However, we have to make some restrictions here. For instance, if then could be expressed as which cannot happen. Therefore, from this, we see that the smallest such that is Also, the smallest number such that is by similar reasoning.
Proceeding, we can factorize after which we see that must contain a prime factor of as it cannot divide but must divide However, is prime, so ! Looking at the answer choices, all of the intervals are less than so we know that (the minimum value of) is thereby either or Testing, we see that gives which in fact is a divisor of while not being a divisor of Therefore, the answer is
~ Professor-Mom
Video Solution by CanadaMath (Problem 21-25)
Fast Forward to 11:20 for problem 22 https://www.youtube.com/watch?v=P3jJDLGyF2w&t=1546s
~THEMATHCANADIAN
See Also
2016 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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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