Difference between revisions of "2002 AMC 12B Problems/Problem 8"
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{{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #8]] and [[2002 AMC 10B Problems|2002 AMC 10B #8]]}} | {{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #8]] and [[2002 AMC 10B Problems|2002 AMC 10B #8]]}} | ||
== Problem == | == Problem == | ||
| − | + | Suppose July of year <math>N</math> has five Mondays. Which of the following must occur five times in the August of year <math>N</math>? (Note: Both months have <math>31</math> days.) | |
| − | Suppose July of year <math>N</math> has five Mondays. Which of the following must | ||
<math>\textrm{(A)}\ \text{Monday} \qquad \textrm{(B)}\ \text{Tuesday} \qquad \textrm{(C)}\ \text{Wednesday} \qquad \textrm{(D)}\ \text{Thursday} \qquad \textrm{(E)}\ \text{Friday}</math> | <math>\textrm{(A)}\ \text{Monday} \qquad \textrm{(B)}\ \text{Tuesday} \qquad \textrm{(C)}\ \text{Wednesday} \qquad \textrm{(D)}\ \text{Thursday} \qquad \textrm{(E)}\ \text{Friday}</math> | ||
| − | ==Solution== | + | == Solution 1 == |
If there are five Mondays, there are only three possibilities for their dates: <math>(1,8,15,22,29)</math>, <math>(2,9,16,23,30)</math>, and <math>(3,10,17,24,31)</math>. | If there are five Mondays, there are only three possibilities for their dates: <math>(1,8,15,22,29)</math>, <math>(2,9,16,23,30)</math>, and <math>(3,10,17,24,31)</math>. | ||
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The only day of the week that is guaranteed to appear five times is therefore <math>\boxed{\textrm{(D)}\ \text{Thursday}}</math>. | The only day of the week that is guaranteed to appear five times is therefore <math>\boxed{\textrm{(D)}\ \text{Thursday}}</math>. | ||
| − | ==See Also== | + | == Solution 2 (visualization) == |
| + | A diagram can be constructed to visualize the possibilities for the dates. First, a 4x7 rectangle representing 28 days can be constructed with three extra cells on the bottom to represent the three extra days. | ||
| + | |||
| + | Then, the rest of the steps in Solution 1 can be followed. | ||
| + | |||
| + | For example, if the first of the bottom three cells is a Monday, it is clear that the following two days are Tuesday and Wednesday. This means that the new bottom three cells (as both months have 31 days) of August will be Thursday, Friday, and Saturday. This same process can be repeated so that the second cell is represented as Monday and so forth; then, the only day of the week that appears when all three cases are considered is <math>\boxed{\textrm{(D)}\ \text{Thursday}}</math>. | ||
| + | |||
| + | ~bearjere | ||
| + | |||
| + | == See Also == | ||
{{AMC10 box|year=2002|ab=B|num-b=7|num-a=9}} | {{AMC10 box|year=2002|ab=B|num-b=7|num-a=9}} | ||
{{AMC12 box|year=2002|ab=B|num-b=7|num-a=9}} | {{AMC12 box|year=2002|ab=B|num-b=7|num-a=9}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 14:48, 14 October 2023
- The following problem is from both the 2002 AMC 12B #8 and 2002 AMC 10B #8, so both problems redirect to this page.
Problem
Suppose July of year 
has five Mondays. Which of the following must occur five times in the August of year ? (Note: Both months have 
days.)
Solution 1
If there are five Mondays, there are only three possibilities for their dates: 
, 
, and 
.
In the first case August starts on a Thursday, and there are five Thursdays, Fridays, and Saturdays in August.
In the second case August starts on a Wednesday, and there are five Wednesdays, Thursdays, and Fridays in August.
In the third case August starts on a Tuesday, and there are five Tuesdays, Wednesdays, and Thursdays in August.
The only day of the week that is guaranteed to appear five times is therefore 
.
Solution 2 (visualization)
A diagram can be constructed to visualize the possibilities for the dates. First, a 4x7 rectangle representing 28 days can be constructed with three extra cells on the bottom to represent the three extra days.
Then, the rest of the steps in Solution 1 can be followed.
For example, if the first of the bottom three cells is a Monday, it is clear that the following two days are Tuesday and Wednesday. This means that the new bottom three cells (as both months have 31 days) of August will be Thursday, Friday, and Saturday. This same process can be repeated so that the second cell is represented as Monday and so forth; then, the only day of the week that appears when all three cases are considered is .
~bearjere
See Also
| 2002 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 | ||
| 2002 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 7 |
Followed by Problem 9 |
| 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. 
