Difference between revisions of "2000 AMC 12 Problems/Problem 18"
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== Solution == | == Solution == | ||
| − | There are either <math> | + | 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 Thursday <math>\mathrm{(A)}</math>. |
== See also == | == See also == | ||
Revision as of 23:10, 8 May 2011
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 Thursday 
.
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 | |
