Difference between revisions of "2000 AMC 10 Problems/Problem 25"
5849206328x (talk | contribs) m (→See Also) |
|||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
| + | |||
| + | In year <math>N</math>, the <math>300^\text{th}</math> day of the year is a Tuesday. In year <math>N+1</math>, the <math>200^\text{th}</math> day is also a Tuesday. On what day of the week did the <math>100^\text{th}</math> day of year <math>N-1</math> occur? | ||
| + | |||
| + | <math>\mathrm{(A)}\ \text{Thursday} \qquad\mathrm{(B)}\ \text{Friday} \qquad\mathrm{(C)}\ \text{Saturday} \qquad\mathrm{(D)}\ \text{Sunday} \qquad\mathrm{(E)}\ \text{Monday}</math> | ||
==Solution== | ==Solution== | ||
| + | Clearly, identifying what of these years may/must/may not be a leap year will be key in solving the problem. | ||
| + | |||
| + | Let <math>A</math> be the <math>300^\text{th}</math> day of year <math>N</math>, <math>B</math> the <math>200^\text{th}</math> day of year <math>N+1</math> and <math>C</math> the <math>100^\text{th}</math> day of year <math>N-1</math>. | ||
| + | |||
| + | If year <math>N</math> is not a leap year, the day <math>B</math> will be <math>(365-300) + 200 = 265</math> days after <math>A</math>. As <math>265 \bmod 7 = 6</math>, that would be a Monday. | ||
| + | |||
| + | Therefore year <math>N</math> must be a leap year. (Then <math>B</math> is <math>266</math> days after <math>A</math>.) | ||
| + | |||
| + | As there can not be two leap years after each other, <math>N-1</math> is not a leap year. Therefore day <math>A</math> is <math>265 + 300 = 565</math> days after <math>C</math>. We have <math>565\bmod 7 = 5</math>. Therefore <math>C</math> is <math>5</math> weekdays before <math>A</math>, i.e., <math>C</math> is a <math>\boxed{\text{Thursday}}</math>. | ||
| + | |||
| + | (Note that the situation described by the problem statement indeed occurs in our calendar. For example, for <math>N=2004</math> we have <math>A</math>=Tuesday, October 26th 2004, <math>B</math>=Tuesday, July 19th, 2005 and <math>C</math>=Thursday, April 10th 2003.) | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2000|num-b=24|after=Last Question}} | {{AMC10 box|year=2000|num-b=24|after=Last Question}} | ||
Revision as of 11:41, 11 January 2009
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 
day of year 
occur?
Solution
Clearly, identifying what of these years may/must/may not be a leap year will be key in solving the problem.
Let 
be the day of year , 
the day of year and 
the day of year .
If year is not a leap year, the day will be 
days after . As 
, that would be a Monday.
Therefore year must be a leap year. (Then is 
days after .)
As there can not be two leap years after each other, is not a leap year. Therefore day is 
days after . We have 
. Therefore is 
weekdays before , i.e., is a 
.
(Note that the situation described by the problem statement indeed occurs in our calendar. For example, for 
we have =Tuesday, October 26th 2004, =Tuesday, July 19th, 2005 and =Thursday, April 10th 2003.)
See Also
| 2000 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last Question | |
| 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 | ||
