2011
Problem - 1457
Seven students count from $1$ to $1000$ as follows:
- Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says $1$, $3$, $4$, $6$, $7$, $9$, . . ., $997$, $999$, $1000$.
- Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.
- Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.
- Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.
- Finally, George says the only number that no one else says.
What number does George say?
Answer
C
Such a puzzle is clearly related to modular arithmetics. Let's try to find the pattern.
- Alice skips all the numbers $n$ which satisfies $n\equiv 2\pmod{3}$. Her group contains $3$ numbers. Each group produces one number for Barbara to process.
- Barbara's group contains $3$ numbers provided by Alice which is equivalent to $3^2=9$ original numbers. It is easy to find out that all the number Barbara skips are $n\equiv (3+2)\pmod{3^2}$ or $n\equiv 5\pmod{9}$
- Similarly, Candice's group will contain $3^3=27$ original numbers and skip those $n\equiv (9+5)\pmod{3^3}$ or $n\equiv 14\pmod{27}$
- Debbie: $n\equiv 27+41\pmod{3^4}$ or $n\equiv 41\pmod{81}$
- Eliza: $n\equiv 81+41\pmod{3^5}$ or $n\equiv 122\pmod{243}$
- Fatima: $n\equiv 243+122\pmod{3^6}$ or $n\equiv 365\pmod{729}$
There is only number $n$ not exceeding $1000$ which satifies $n\equiv 365\pmod{729}$ which is $\boxed{365}$.