Difference between revisions of "2000 AMC 12 Problems/Problem 18"
LadyKn1ght (talk | contribs) |
(→Solution: correction) |
||
Line 7: | Line 7: | ||
== Solution == | == Solution == | ||
− | There are either <math>65 + 200 = 265</math> or <math>66 + 200 = 266</math> 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>, then it is possible for both dates to be Tuesday; hence year <math>N | + | There are either <math>65 + 200 = 265</math> or <math>66 + 200 = 266</math> 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>, then it is possible for both dates to be Tuesday; hence year <math>N</math> is a leap year and <math>N-1</math> is not a leap year. There are <math>265 + 300 = 565</math> days between the date in years <math>N,N-1</math>, which 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>\mathrm {Thursday} \text{ (A)}</math>. |
== See also == | == See also == |
Revision as of 08:45, 25 February 2016
- 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 or days between the first two dates depending upon whether or not year is a leap year. Since divides into , then it is possible for both dates to be Tuesday; hence year is a leap year and is not a leap year. There are days between the date in years , which 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.