Difference between revisions of "2011 AMC 10A Problems/Problem 23"
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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{365\ \mathbf{(C)}} </math>. | 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{365\ \mathbf{(C)}} </math>. | ||
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| + | Tip: We look at the answer choices and notice that there are 365 days in the year. Therefore, the answer is likely to be <math> \boxed{365} </math>. | ||
== See Also == | == See Also == | ||
Revision as of 12:33, 12 January 2014
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?
Solution
First look at the numbers Alice says. 
skipping every number that is congruent to 
. Thus, Barbara says those numbers EXCEPT every second - being 
. So Barbara skips every number congruent to 
. We continue and see:
Alice skips , Barbara skips , Candice skips 
, Debbie skips 
, Eliza skips 
, and Fatima skips 
.
Since the only number congruent to and less than 
is 
, the correct answer is 
.
Tip: We look at the answer choices and notice that there are 365 days in the year. Therefore, the answer is likely to be 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
