2022 AIME I Problems/Problem 13
Contents
Problem
Let be the set of all rational numbers that can be expressed as a repeating decimal in the form where at least one of the digits or is nonzero. Let be the number of distinct numerators obtained when numbers in are written as fractions in lowest terms. For example, both and are counted among the distinct numerators for numbers in because and Find the remainder when is divided by
Solution 1
, .
Then we need to find the number of positive integers that (with one of more such that ) can meet the requirement .
Make cases by factors of . (A venn diagram of cases would be nice here.)
Case :
and and , aka .
Euler's totient function counts these: values (but it's enough to note that it's a multiple of 1000 and thus does not contribute to the final answer)
Note: You don't need to know this formula. The remaining cases essentially re-derive the same computation for other factors of . This case isn't actually different.
The remaining cases have (or ), , and/or as factors of , which cancel out part of . Note: Take care about when to use vs .
Case : , but and .
Then to leave 3 uncancelled, and , so , giving:
,
,
,
for a subtotal of values.
Case : , but and .
Much like previous case, is , so ,
giving values.
Case : and (so ), but .
Here, is , so ,
giving values.
Case : .
Here, is , so ,
giving values, so we don't need to account for multiples of and .
To sum up, the answer is
Clarification
In this context, when the solution says, "Then to leave 3 uncancelled, and ," it is a bit vague. The best way to clarify this is by this exact example - what is really meant is we need to divide by 9 first to achieve 1111, which has no multiple of 3; thus, given that the fraction x/y is the simplest form, x can be a multiple of 3.
Similar explanations can be said when the solution divides 9999 by 11, 101, and uses that divided result in the PIE calculation rather than 9999.
mathboy282
Solution 2 (similar(?) to solution 1 but reworded (not exactly for clarity))
Video Solution
https://MathProblemSolvingSkills.com
See Also
2022 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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