Difference between revisions of "2010 AMC 10A Problems/Problem 10"
m (→Problem 10) |
Math Kirby (talk | contribs) m |
||
Line 15: | Line 15: | ||
==Solution== | ==Solution== | ||
− | |||
− | |||
There are <math>365</math> days in a non-leap year. There are <math>7</math> days in a week. Since <math>365 = 52 \cdot 7 + 1</math> (or <math>365\equiv 1 \pmod{ 7}</math>), the same date (after February) moves "forward" one day in the subsequent year, if that year is not a leap year. | There are <math>365</math> days in a non-leap year. There are <math>7</math> days in a week. Since <math>365 = 52 \cdot 7 + 1</math> (or <math>365\equiv 1 \pmod{ 7}</math>), the same date (after February) moves "forward" one day in the subsequent year, if that year is not a leap year. | ||
Line 44: | Line 42: | ||
<math>5/27/17</math> Sat | <math>5/27/17</math> Sat | ||
+ | |||
+ | <math>\boxed{(E) 2017}</math> | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 10:45, 30 October 2024
Contents
[hide]Problem 10
Marvin had a birthday on Tuesday, May 27 in the leap year . In what year will his birthday next fall on a Saturday?
Solution
There are days in a non-leap year. There are days in a week. Since (or ), the same date (after February) moves "forward" one day in the subsequent year, if that year is not a leap year.
For example:
Tue
Wed
However, a leap year has days, and . So the same date (after February) moves "forward" two days in the subsequent year, if that year is a leap year.
For example: Fri
Sun
You can keep counting forward to find that the first time this date falls on a Saturday is in :
Mon
Tue
Wed
Fri
Sat
Video Solution
https://youtu.be/P7rGLXp_6es?t=628
~IceMatrix
See also
2010 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 |
2010 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 1 |
Followed by Problem 3 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.