Difference between revisions of "2022 AIME I Problems/Problem 13"
m (→Solution 1) |
|||
Line 23: | Line 23: | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2022|n=I|num-b=12|num-a=14}} | {{AIME box|year=2022|n=I|num-b=12|num-a=14}} | ||
+ | |||
+ | [[Category:Intermediate Number Theory Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 15:44, 28 March 2022
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 less than 10000 can meet the requirement.Suppose the number is x.
Case 1: (9999, x)=1. Clearly x satisfies.
Case 2: 3|x but x is not a multiple of 11 or 101. Then the least value of abcd is 9x, so that , 334 values from 3 to 1110.
Case 3: 11|x but x is not a multiple of 3 or 101. Then the least value of abcd is 11x, so that , 55 values from 11 to 902.
Case 4: 101|x. None.
Case 5: 3, 11|x. Then the least value of abcd is 11x, 3 values from 33 to 99.
To sum up, the answer is .
See Also
2022 AIME I (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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.