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Difference between revisions of "2006 AMC 10B Problems/Problem 16"

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*Note how we did not require any complex or time-consuming computation, and thus it will save crucial time, especially during a test like the AMC 10.
 
*Note how we did not require any complex or time-consuming computation, and thus it will save crucial time, especially during a test like the AMC 10.
 +
(psst buddy 5844 mod 7 isnt that hard)
  
 
== See Also ==
 
== See Also ==

Revision as of 17:59, 9 September 2019

Problem

Leap Day, February 29, 2004, occurred on a Sunday. On what day of the week will Leap Day, February 29, 2020, occur?

$\mathrm{(A) \ } \textrm{Tuesday} \qquad \mathrm{(B) \ } \textrm{Wednesday} \qquad \mathrm{(C) \ } \textrm{Thursday} \qquad \mathrm{(D) \ } \textrm{Friday} \qquad \mathrm{(E) \ } \textrm{Saturday}$

Solution

There are $365$ days in a year, plus $1$ extra day if there is a Leap Day, which occurs on years that are multiples of 4 (with a few exceptions that don't affect this problem).

Therefore, the number of days between Leap Day 2004 and Leap Day 2020 is:

$16 \cdot 365 + 4 \cdot 1 = 5844$

Since the days of the week repeat every $7$ days and $5844 \equiv -1 \bmod{7}$, the day of the week Leap Day 2020 occurs is the day of the week the day before Leap Day 2004 occurs, which is $\textrm{Saturday} \Rightarrow E$.

Solution 2 (Feasible Shortcuts)

Since every non-leap year there is one day extra after the $52$ weeks, we can deduce that if we were to travel forward $x$ amount of non-leap years, then the answer would be "Sunday + $x$" days.

However, we also have leap-years in our pool of years traveled forward, and for every leap-year that passes, we have two extra days after the 52 weeks. So we would travel "Sunday + $2x$" days.

Mixing these two together, we get $4$ leap years $(2008, 2012, 2016, 2020)$, and $12$ "normal years". We thus get $12 + 2\cdot 4 = 20$ "extra days" after Sunday. Since seven days are in a week, we can get $20\mod 7 = 6$, and six days after Sunday is $\textrm{Saturday} \Rightarrow E$.

  • Note how we did not require any complex or time-consuming computation, and thus it will save crucial time, especially during a test like the AMC 10.

(psst buddy 5844 mod 7 isnt that hard)

See Also

2006 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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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