Difference between revisions of "2010 AMC 10A Problems/Problem 10"
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| − | == Problem | + | == Problem 10 == |
| − | + | Marvin had a birthday on Tuesday, May 27 in the leap year <math>2008</math>. In what year will his birthday next fall on a Saturday? | |
<math> | <math> | ||
| − | \mathrm{(A)}\ | + | \mathrm{(A)}\ 2011 |
\qquad | \qquad | ||
| − | \mathrm{(B)}\ | + | \mathrm{(B)}\ 2012 |
\qquad | \qquad | ||
| − | \mathrm{(C)}\ | + | \mathrm{(C)}\ 2013 |
\qquad | \qquad | ||
| − | \mathrm{(D)}\ | + | \mathrm{(D)}\ 2015 |
\qquad | \qquad | ||
| − | \mathrm{(E)}\ | + | \mathrm{(E)}\ 2017 |
</math> | </math> | ||
| Line 49: | Line 49: | ||
== See also == | == See also == | ||
| − | {{AMC10 box|year=2010|ab=A|num-b= | + | {{AMC10 box|year=2010|ab=A|num-b=9|num-a=11}} |
Revision as of 19:09, 1 January 2012
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 is congruent to 
), 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 count forward to find that the first time this date falls on a Saturday is in :
Mon
Tue
Wed
Fri
Sat
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 | ||
