Difference between revisions of "2000 AMC 12 Problems/Problem 18"
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In addition, there are either <cmath>65 + 200 = 265</cmath> or <cmath>66 + 200 = 266</cmath> days between the first two dates depending upon whether or not year <math>N</math> is a leap year. Since <math>7</math> divides into <math>266</math> but not <math>265</math>, for both days to be a Tuesday, year <math>N</math> must be a leap year. | In addition, there are either <cmath>65 + 200 = 265</cmath> or <cmath>66 + 200 = 266</cmath> days between the first two dates depending upon whether or not year <math>N</math> is a leap year. Since <math>7</math> divides into <math>266</math> but not <math>265</math>, for both days to be a Tuesday, year <math>N</math> must be a leap year. | ||
| − | Hence, year <math>N-1</math> is not a leap year, and so since there are <cmath>265 + 300 = 565</cmath> days between the date in years <math>N,\text{ }N-1</math>, this leaves a remainder of <math>5</math> upon division by <math>7</math>. Since we are subtracting days, we count 5 days before Tuesday, which gives us <math>\ | + | Hence, year <math>N-1</math> is not a leap year, and so since there are <cmath>265 + 300 = 565</cmath> days between the date in years <math>N,\text{ }N-1</math>, this leaves a remainder of <math>5</math> upon division by <math>7</math>. Since we are subtracting days, we count 5 days before Tuesday, which gives us <math>\boxed{\mathbf{(A)} \ \text{Thursday}.}</math> |
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== See also == | == See also == | ||
Revision as of 11:40, 11 December 2018
- The following problem is from both the 2000 AMC 12 #18 and 2000 AMC 10 #25, so both problems redirect to this page.
Problem
In year 
, the 
day of the year is a Tuesday. In year 
, the 
day is also a Tuesday. On what day of the week did the 
th day of year 
occur?
Solution
There are either ![\[265 + 300 = 565\]](http://latex.artofproblemsolving.com/9/b/b/9bb28446bd7a88948b838d007a8a00e9e3f96965.png)
or ![\[266 + 300 = 566\]](http://latex.artofproblemsolving.com/f/9/f/f9f073c4e44651cdc2dee1f710087a9f3f63e448.png)
days between year and year , depending on whether year is a leap year.
In addition, there are either ![\[65 + 200 = 265\]](http://latex.artofproblemsolving.com/6/2/9/62997404eccda827729e180344fb549322e5596b.png)
or ![\[66 + 200 = 266\]](http://latex.artofproblemsolving.com/4/1/0/410b22f951be4fd0f5d55cc96280621336ed4ed4.png)
days between the first two dates depending upon whether or not year is a leap year. Since 
divides into 
but not 
, for both days to be a Tuesday, year must be a leap year.
Hence, year is not a leap year, and so since there are days between the date in years 
, this leaves a remainder of 
upon division by . Since we are subtracting days, we count 5 days before Tuesday, which gives us 
See also
| 2000 AMC 12 (Problems • Answer Key • Resources) | |
| Preceded by Problem 17 |
Followed by Problem 19 |
| 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 12 Problems and Solutions | |
| 2000 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last Problem | |
| 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. 
