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Difference between revisions of "2011 AMC 10A Problems/Problem 23"

m (→‎Solution: Corrected a typo, 81* 3 = 243, not 423)
m (→‎Solution)
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== Solution ==
 
== Solution ==
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First look at the numbers Alice says. 1, 3, 4, 6, 7, 9 ... skipping every number that is congruent to 2 mod 3. Thus, Barbara says those numbers EXCEPT every second - being <math>2 + 3^1 = 5</math> mod <math>3^2=9</math>. So Barbara skips every number congruent to 5 mod 9. We continue on and see: (see this for yourself)
+
First look at the numbers Alice says. <math>1, 3, 4, 6, 7, 9 \cdots</math> skipping every number that is congruent to <math>2 \pmod 3</math>. Thus, Barbara says those numbers EXCEPT every second - being <math>2 + 3^1 \equiv 5 \pmod{3^2=9}</math>. So Barbara skips every number congruent to <math>5 \pmod 9</math>. We continue and see:  
  
−
Alice skips 2 mod 3, Barbara skips 5 mod 9, Candice skips 14 mod 27, Debbie skips 41 mod 81, Eliza skips 122 mod 243, and Fatima skips 365 mod 729.
+
Alice skips <math>2 \pmod 3</math>, Barbara skips <math>5 \pmod 9</math>, Candice skips <math>14 \pmod {27}</math>, Debbie skips <math>41 \pmod {81}</math>, Eliza skips <math>122 \pmod {243}</math>, and Fatima skips <math>365 \pmod {729}</math>.
  
−
Since the only number congruent to 365 mod 729 and less than 1,000 is 365, the correct answer is '''(C)'''.
+
Since the only number congruent to <math>365 \pmod {729}</math> and less than <math>1,000</math> is <math>365</math>, the correct answer is <math>\boxed{\text{C}}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 16:28, 4 February 2012

Problem

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?

$\textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998$

Solution

First look at the numbers Alice says. $1, 3, 4, 6, 7, 9 \cdots$ skipping every number that is congruent to $2 \pmod 3$. Thus, Barbara says those numbers EXCEPT every second - being $2 + 3^1 \equiv 5 \pmod{3^2=9}$. So Barbara skips every number congruent to $5 \pmod 9$. We continue and see:

Alice skips $2 \pmod 3$, Barbara skips $5 \pmod 9$, Candice skips $14 \pmod {27}$, Debbie skips $41 \pmod {81}$, Eliza skips $122 \pmod {243}$, and Fatima skips $365 \pmod {729}$.

Since the only number congruent to $365 \pmod {729}$ and less than $1,000$ is $365$, the correct answer is $\boxed{\text{C}}$.

See Also

2011 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
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