2011 AMC 10A Problem 23

Question

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,

\ldots,

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?

37

242

365

728

998

Answer: C

Concepts:arithmetic sequence process simulation pattern recognition

Difficulty rating: 2110

Small hint:

Track the first remaining term and the common difference

Big hint:

Each student triples the common difference and shifts to the middle term

Solution:

We can walk through all the iterations to find what is left.

Alice does not say the numbers

2, 5, 8, 11, 14, 17, \ldots, 998.

After Barbara says her numbers, the remaining ones are

5, 14, 23, 32, 41, \ldots, 995.

Note that both of these are arithmetic sequences where the common difference is increased by a multiple of

3.

This pattern continues as the numbers remaining after Candice says hers are

14, 41, 68, 95, \ldots, 986.

Then after Debbie, they are

41, 122, 203, \ldots, 932

and after Eliza, they are

122, 365, 608, 851.

Finally, the only number left after Fatima goes is

365,

which is the number that George will have to say.

Thus, C is the correct answer.