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Difference between revisions of "2002 AMC 12B Problems/Problem 8"
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Let <math>M</math> be a 31-day month and <math>W</math> a day of the week. We can easily see that <math>W</math> occurs five times in <math>M</math> if and only if one of the first three days of <math>M</math> falls on a <math>W</math>. This is because the 5th occurrence of <math>W</math> is 28 days after the first one, so the only possibilities for their dates are <math>(1,29)</math>, <math>(2,30)</math>, and <math>(3,31)</math>.
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{{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #8]] and [[2002 AMC 10B Problems|2002 AMC 10B #8]]}}
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== Problem ==
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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.)
  
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We now know that one of July 1, July 2 and July 3 was a Monday.
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<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>
  
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In these three cases, August 1 is a Thursday, a Wednesday, and a Tuesday.
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== Solution 1 ==
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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 answer is a day of the week that is guaranteed to fall on one of August 1, August 2, and August 3. We can easily see that the only such day of the week is <math>\boxed{\mathrm{Thursday}}</math> (August 1 in case 1, August 2 in case 2, and August 3 in case 3) <math>\Longrightarrow \mathrm{(D)}.</math>
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In the first case August starts on a Thursday, and there are five Thursdays, Fridays, and Saturdays in August.
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In the second case August starts on a Wednesday, and there are five Wednesdays, Thursdays, and Fridays in August.
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In the third case August starts on a Tuesday, and there are five Tuesdays, Wednesdays, and Thursdays in August.
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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>.
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== Solution 2 (visualization) ==
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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.
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Then, the rest of the steps in Solution 1 can be followed.
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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>.
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~bearjere
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== See Also ==
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{{AMC10 box|year=2002|ab=B|num-b=7|num-a=9}}
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{{AMC12 box|year=2002|ab=B|num-b=7|num-a=9}}
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[[Category:Introductory Algebra Problems]]
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{{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 $N$ has five Mondays. Which of the following must occur five times in the August of year $N$? (Note: Both months have $31$ days.)

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

Solution 1

If there are five Mondays, there are only three possibilities for their dates: $(1,8,15,22,29)$, $(2,9,16,23,30)$, and $(3,10,17,24,31)$.

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 $\boxed{\textrm{(D)}\ \text{Thursday}}$.

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 $\boxed{\textrm{(D)}\ \text{Thursday}}$.

~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

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