Difference between revisions of "2000 AMC 12 Problems/Problem 18"
(→Solution 2) |
(→Solution 1) |
||
(3 intermediate revisions by 3 users not shown) | |||
Line 7: | Line 7: | ||
== Solution 1 == | == Solution 1 == | ||
− | 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. | + | 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+1</math> is a leap year (since the February 29th of the leap year would come between the 300th day of year <math>N</math> and 200th day of year <math>N + 1</math>). 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>\boxed{\mathbf{(A)} \ \text{Thursday}.}</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> | ||
Line 15: | Line 15: | ||
~dolphin7 | ~dolphin7 | ||
+ | |||
+ | ==Video Solution == | ||
+ | https://youtu.be/aFJtbvTAmm4 | ||
+ | |||
+ | https://www.youtube.com/watch?v=YfiUn8hLlpA ~David | ||
== See also == | == See also == |
Latest revision as of 07:46, 24 August 2024
- 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 1
There are either or
days between the first two dates depending upon whether or not year
is a leap year (since the February 29th of the leap year would come between the 300th day of year
and 200th day of 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
Solution 2
The day of year
and the
day of year
are Tuesdays. If there were the same number of days in year
and year
the
day of year
will be a Tuesday. But year
is a leap year because
so the
day of year
is a Tuesday. It follows that the
day of year
is
~dolphin7
Video Solution
https://www.youtube.com/watch?v=YfiUn8hLlpA ~David
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.