Difference between revisions of "2010 AMC 10A Problems/Problem 10"
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+ | == 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> | ||
+ | \mathrm{(A)}\ 2011 | ||
+ | \qquad | ||
+ | \mathrm{(B)}\ 2012 | ||
+ | \qquad | ||
+ | \mathrm{(C)}\ 2013 | ||
+ | \qquad | ||
+ | \mathrm{(D)}\ 2015 | ||
+ | \qquad | ||
+ | \mathrm{(E)}\ 2017 | ||
+ | </math> | ||
+ | |||
+ | |||
+ | |||
==Solution== | ==Solution== | ||
Revision as of 10:32, 9 April 2019
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
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
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?
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 |
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