1989 AIME Problems/Problem 3
Contents
Problem
Suppose is a positive integer and is a single digit in base 10. Find if
Solution 1
We can express as . We set up the given equation and isolate \begin{align*} \frac{100d+25}{999} &= \frac{n}{810}, \\ \frac{100d+25}{111} &= \frac{n}{90}, \\ 9000d + 2250 &= 111n. \end{align*} We then set up the following modular congruence to solve for \begin{align*}9000d + 2250 &\equiv 0 \pmod {111}, \\ 9d + 30 &\equiv 0 \pmod {111}, \\ 3d + 10 &\equiv 0 \pmod {37}, \\ 36d + 120 &\equiv 0 \pmod {37}, \\ -d &\equiv -9 \pmod {37}, \\ d &\equiv 9 \pmod {37}.\end{align*} Since is a digit, it must be equal to based on our above constraint. When
~vaisri
Solution 2
Repeating decimals represent rational numbers. To figure out which rational number, we sum an infinite geometric series, . Thus so . Since 750 and 37 are relatively prime, must be divisible by 37, and the only digit for which this is possible is . Thus and .
(Note: Any repeating sequence of digits that looks like can be written as , where represents an digit number.)
Solution 3
To get rid of repeating decimals, we multiply the equation by 1000. We get We subtract the original equation from the second to get We simplify to Since is an integer, because is relatively prime to , and d25 is divisible by but not . The only odd number that yields a single digit and 25 at the end of the three digit number is , so the answer is .
Solution 4
Similar to Solution 2, we start off by writing that .Then we subtract this from the original equation to get:
Since n is an integer, we have that .
Since is prime, we can apply Euclid's Lemma to realize that , since . Then we can expand as . Since , by Euclid, we can arrive at . From this we know that . (This is true because )
~qwertysri987
Solution 5
Write out these equations:
Thus divides 25 and 30. The only solution for this under 1000 is .
-jackshi2006
Note: We know since so it suffices to check for numbers under not
Video Solution by OmegaLearn
https://youtu.be/1-iWPCWPsLw?t=600
~ pi_is_3.14
See also
1989 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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