Difference between revisions of "2010 AMC 10A Problems/Problem 10"
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<math>\boxed{(E)}</math> <math> 2017 </math> | <math>\boxed{(E)}</math> <math> 2017 </math> | ||
− | 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 | + | 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. |
For example: | For example: | ||
Line 35: | Line 35: | ||
<math>5/27/12</math> Sun | <math>5/27/12</math> Sun | ||
− | You can keep | + | You can keep counting forward to find that the first time this date falls on a Saturday is in <math> 2017</math>: |
<math>5/27/13</math> Mon | <math>5/27/13</math> Mon | ||
Line 47: | Line 47: | ||
<math>5/27/17</math> Sat | <math>5/27/17</math> Sat | ||
+ | ==Video Solution== | ||
+ | https://youtu.be/P7rGLXp_6es?t=628 | ||
+ | |||
+ | ~IceMatrix | ||
== See also == | == See also == | ||
{{AMC10 box|year=2010|ab=A|num-b=9|num-a=11}} | {{AMC10 box|year=2010|ab=A|num-b=9|num-a=11}} | ||
+ | {{AMC12 box|year=2010|ab=A|num-b=1|num-a=3}} | ||
+ | {{MAA Notice}} |
Latest revision as of 11:59, 29 March 2021
Contents
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.