Difference between revisions of "2024 AMC 10A Problems/Problem 25"

Latest revision as of 14:53, 20 November 2024

The following problem is from both the 2024 AMC 10A #25 and 2024 AMC 12A #22, so both problems redirect to this page.

Problem

The figure below shows a dotted grid $8$ cells wide and $3$ cells tall consisting of $1''\times1''$ squares. Carl places $1$ -inch toothpicks along some of the sides of the squares to create a closed loop that does not intersect itself. The numbers in the cells indicate the number of sides of that square that are to be covered by toothpicks, and any number of toothpicks are allowed if no number is written. In how many ways can Carl place the toothpicks? $[asy] size(6cm); for (int i=0; i<9; ++i) { draw((i,0)--(i,3),dotted); } for (int i=0; i<4; ++i){ draw((0,i)--(8,i),dotted); } for (int i=0; i<8; ++i) { for (int j=0; j<3; ++j) { if (j==1) { label("1",(i+0.5,1.5)); }}} [/asy]$ $\textbf{(A) }130\qquad\textbf{(B) }144\qquad\textbf{(C) }146\qquad\textbf{(D) }162\qquad\textbf{(E) }196$

(Best) Solution 1

Observations:

1. You can not have a vertical line in any place other than the first and second columns and the last and second-to-last columns.

2. You can place a box around the top row or along the bottom row, otherwise all the solutions have a vertical line at the first and second columns and the last and second-to-last columns.

Thus, Using casework, we can split this problem into 4 cases. However, we can focus on only the first one for right now.

For case 1, we assume that the green lines shown below are given (always have toothpicks on them)

And the only toothpicks we can place that will connect to the red lines are to go horizontally inward:

Now, concentrate on the first row of squares. A toothpick can be placed on either the bottom or top and connected to a continuous squiggle by adding vertical toothpicks, for example,

How many squiggles are possible?

We can summarize this by giving a high squiggle position a 1 and a low position a 0, thus we have a 6-digit binary sequence. Thus, we can have $2^6=64$ ways to make this squiggle.

Case 2: We can also, as the observations state, pull in one of the sides, thus we can have a squiggle with 5 binary digits. only utilizing the first 7 columns:

Here, we only have 5 binary digits to work with, so there are $2^5=32$ ways to make this squiggle.

Case 3: Similarly, we can utilize the last 7 columns.

Again, we only have 5 binary digits to work with, so there are $2^5=32$ ways to make this squiggle.

Case 4: We can use an even smaller section. Using only the middle 6 columns gives us a 4-wide squiggle:

Thus, there are $2^4=16$ ways to make this squiggle.

Adding up all our cases: $64+32+32+16=144$

However, there are two more ways to draw a qualifying shape:

We can draw a rectangle like that in the first row or third row. Thus, we have a grand total of $144+2=\boxed{\textbf{(C) }146}$ ways.

A note to (potential) editors: This answer was not made to be concise or especially professional. It was made to explicitly explain this problem in a way so that it is easy to understand and follow.

~hermanboxcar5

Notes: Remember these are the ONLY possible cases. It is impossible to cross through the rows of boxes of ones to connect the loop around the bottom since then the loop would intersect itself if you are to put only one toothpick on each box with a one (a more clear definition of Observation 1). You are not undercounting by only counting the binary digits on the top, because all of the digits on one side will have corresponding opposite digits on the other.

~juwushu

Solution 2 (Cheese)

Notice that for any case where the closed loop does not connect from the top side of the ones and bottom side of the ones, there are two of these cases. A cheese solution can be found from this; noting that B and C are the only two options to each other, and, being two apart, with people likely to forget this case, $\boxed{\textbf{(C) }146}$ is likely to be the correct answer. Cheese solution done by juwushu.

Solution 3 (Observation)

We have $2$ cases where the loop does not go through the middle.

If the loop goes through the middle, we must have a full column on $(0,8),(0,7),(1,8),(1,7).$ Then we have $6,5,5,4$ empty middle squares. For each one we can have one on top or one on bottom, so $2^6+2^5+2^5+2^4=144.$ Notice that for each case of fixed toothpicks, there is only one way to form the loop. Then we just add $2+144=\boxed{\textbf{(C) } 146.}$

~nevergonnagiveup

Video Solution by Power Solve

https://www.youtube.com/watch?v=FRNbJ5wIGRo

Video Solution By SpreadTheMathLove

https://www.youtube.com/watch?v=huMQ9J7rIj0&t=77s

@@ Line 1: / Line 1: @@
-<math>\sqrt{\text{sigma}}</math>
+{{duplicate|[[2024 AMC 10A Problems/Problem 25|2024 AMC 10A #25]] and [[2024 AMC 12A Problems/Problem 22|2024 AMC 12A #22]]}}
-<math>\textbf{(A) } alpha \qquad\textbf{(B) } \text{rizzler} \qquad\textbf{(C) } \text{skibidi toilet} \qquad\textbf{(D) }\text{fanum tax}
+==Problem==
- \qquad\textbf{(E) }\text{Mew} \qquad\textbf{(F) }\text{looksmaxxing}</math>
+The figure below shows a dotted grid <math>8</math> cells wide and <math>3</math> cells tall consisting of <math>1''\times1''</math> squares. Carl places <math>1</math>-inch toothpicks along some of the sides of the squares to create a closed loop that does not intersect itself. The numbers in the cells indicate the number of sides of that square that are to be covered by toothpicks, and any number of toothpicks are allowed if no number is written. In how many ways can Carl place the toothpicks?
+<asy>
+size(6cm);
+for (int i=0; i<9; ++i) {
+  draw((i,0)--(i,3),dotted);
+}
+for (int i=0; i<4; ++i){
+  draw((0,i)--(8,i),dotted);
+}
+for (int i=0; i<8; ++i) {
+  for (int j=0; j<3; ++j) {
+    if (j==1) {
+      label("1",(i+0.5,1.5));
+}}}
+</asy>
+<math>\textbf{(A) }130\qquad\textbf{(B) }144\qquad\textbf{(C) }146\qquad\textbf{(D) }162\qquad\textbf{(E) }196</math>
-==Solution 1==
+==(Best) Solution 1==
-YO STOP THE BRAiNROT
-==Solution 2==
+Observations:
-We the People of the United States, in Order to form a more perfect Union, establish Justice, insure domestic Tranquility, provide for the common defence, promote the general Welfare, and secure the Blessings of Liberty to ourselves and our Posterity, do ordain and establish this Constitution for the United States of America.
-Article. I.
+. You can not have a vertical line in any place other than the first and second columns and the last and second-to-last columns.
-Section. 1.
-All legislative Powers herein granted shall be vested in a Congress of the United States, which shall consist of a Senate and House of Representatives.
-Section. 2.
+. You can place a box around the top row or along the bottom row, otherwise all the solutions have a vertical line at the first and second columns and the last and second-to-last columns.
-The House of Representatives shall be composed of Members chosen every second Year by the People of the several States, and the Electors in each State shall have the Qualifications requisite for Electors of the most numerous Branch of the State Legislature.
-No Person shall be a Representative who shall not have attained to the Age of twenty five Years, and been seven Years a Citizen of the United States, and who shall not, when elected, be an Inhabitant of that State in which he shall be chosen.
+Thus, Using casework, we can split this problem into 4 cases. However, we can focus on only the first one for right now.
-Representatives and direct Taxes shall be apportioned among the several States which may be included within this Union, according to their respective Numbers, which shall be determined by adding to the whole Number of free Persons, including those bound to Service for a Term of Years, and excluding Indians not taxed, three fifths of all other Persons. The actual Enumeration shall be made within three Years after the first Meeting of the Congress of the United States, and within every subsequent Term of ten Years, in such Manner as they shall by Law direct. The Number of Representatives shall not exceed one for every thirty Thousand, but each State shall have at Least one Representative; and until such enumeration shall be made, the State of New Hampshire shall be entitled to chuse three, Massachusetts eight, Rhode-Island and Providence Plantations one, Connecticut five, New-York six, New Jersey four, Pennsylvania eight, Delaware one, Maryland six, Virginia ten, North Carolina five, South Carolina five, and Georgia three.
+For case 1, we assume that the green lines shown below are given (always have toothpicks on them)
-When vacancies happen in the Representation from any State, the Executive Authority thereof shall issue Writs of Election to fill such Vacancies.
+[[Image:Scrasdfasd.png|600px]]
-The House of Representatives shall chuse their Speaker and other Officers; and shall have the sole Power of Impeachment.
+[[Image:Screenshot 2024-11-08 192200.png|600px]]
-Section. 3.
+And the only toothpicks we can place that will connect to the red lines are to go horizontally inward:
-The Senate of the United States shall be composed of two Senators from each State, chosen by the Legislature thereof, for six Years; and each Senator shall have one Vote.
-Immediately after they shall be assembled in Consequence of the first Election, they shall be divided as equally as may be into three Classes. The Seats of the Senators of the first Class shall be vacated at the Expiration of the second Year, of the second Class at the Expiration of the fourth Year, and of the third Class at the Expiration of the sixth Year, so that one third may be chosen every second Year; and if Vacancies happen by Resignation, or otherwise, during the Recess of the Legislature of any State, the Executive thereof may make temporary Appointments until the next Meeting of the Legislature, which shall then fill such Vacancies.
+[[Image:Screenshot 2024-11-08 192516.png|600px]]
-No Person shall be a Senator who shall not have attained to the Age of thirty Years, and been nine Years a Citizen of the United States, and who shall not, when elected, be an Inhabitant of that State for which he shall be chosen.
+Now, concentrate on the first row of squares. A toothpick can be placed on either the bottom or top and connected to a continuous squiggle by adding vertical toothpicks, for example,
-The Vice President of the United States shall be President of the Senate, but shall have no Vote, unless they be equally divided.
+[[Image:Screenshot 2024-11-08 193401.png|600px]]
-The Senate shall chuse their other Officers, and also a President pro tempore, in the Absence of the Vice President, or when he shall exercise the Office of President of the United States.
+How many squiggles are possible?
-The Senate shall have the sole Power to try all Impeachments. When sitting for that Purpose, they shall be on Oath or Affirmation. When the President of the United States is tried, the Chief Justice shall preside: And no Person shall be convicted without the Concurrence of two thirds of the Members present.
+[[Image:Screenshot 2024-11-08 194804.png|600px]]
-Judgment in Cases of Impeachment shall not extend further than to removal from Office, and disqualification to hold and enjoy any Office of honor, Trust or Profit under the United States: but the Party convicted shall nevertheless be liable and subject to Indictment, Trial, Judgment and Punishment, according to Law.
+We can summarize this by giving a high squiggle position a 1 and a low position a 0, thus we have a 6-digit binary sequence. Thus, we can have <math>2^6=64</math> ways to make this squiggle.
-Section. 4.
+Case 2: We can also, as the observations state, pull in one of the sides, thus we can have a squiggle with 5 binary digits. only utilizing the first 7 columns:
-The Times, Places and Manner of holding Elections for Senators and Representatives, shall be prescribed in each State by the Legislature thereof; but the Congress may at any time by Law make or alter such Regulations, except as to the Places of chusing Senators.
-The Congress shall assemble at least once in every Year, and such Meeting shall be on the first Monday in December, unless they shall by Law appoint a different Day.
+[[Image:Screenshot 2024-11-08 195135.png|600px]]
-Section. 5.
+Here, we only have 5 binary digits to work with, so there are <math>2^5=32</math> ways to make this squiggle.
-Each House shall be the Judge of the Elections, Returns and Qualifications of its own Members, and a Majority of each shall constitute a Quorum to do Business; but a smaller Number may adjourn from day to day, and may be authorized to compel the Attendance of absent Members, in such Manner, and under such Penalties as each House may provide.
-Each House may determine the Rules of its Proceedings, punish its Members for disorderly Behaviour, and, with the Concurrence of two thirds, expel a Member.
+Case 3: Similarly, we can utilize the last 7 columns.
-Each House shall keep a Journal of its Proceedings, and from time to time publish the same, excepting such Parts as may in their Judgment require Secrecy; and the Yeas and Nays of the Members of either House on any question shall, at the Desire of one fifth of those Present, be entered on the Journal.
+[[Image:Screenshot 2024-11-08 195553.png|600px]]
-Neither House, during the Session of Congress, shall, without the Consent of the other, adjourn for more than three days, nor to any other Place than that in which the two Houses shall be sitting.
+Again, we only have 5 binary digits to work with, so there are <math>2^5=32</math> ways to make this squiggle.
-Section. 6.
+Case 4: We can use an even smaller section. Using only the middle 6 columns gives us a 4-wide squiggle:
+[[Image:Screenshot 2024-11-08 195945.png|600px]]
+Thus, there are <math>2^4=16</math> ways to make this squiggle.
+Adding up all our cases: <math>64+32+32+16=144</math>
+However, there are two more ways to draw a qualifying shape:
+[[Image:Screenshot 2024-11-08 200416.png|600px]]
+We can draw a rectangle like that in the first row or third row. Thus, we have a grand total of
+<math>144+2=\boxed{\textbf{(C) }146}</math> ways.
+A note to (potential) editors:
+This answer was not made to be concise or especially professional. It was made to explicitly explain this problem in a way so that it is easy to understand and follow.
+~hermanboxcar5
+Notes:
+Remember these are the ONLY possible cases. It is impossible to cross through the rows of boxes of ones to connect the loop around the bottom since then the loop would intersect itself if you are to put only one toothpick on each box with a one (a more clear definition of Observation 1).
+You are not undercounting by only counting the binary digits on the top, because all of the digits on one side will have corresponding opposite digits on the other.
+~juwushu
+==Solution 2 (Cheese)==
+Notice that for any case where the closed loop does not connect from the top side of the ones and bottom side of the ones, there are two of these cases. A cheese solution can be found from this; noting that B and C are the only two options to each other, and, being two apart, with people likely to forget this case, <math>\boxed{\textbf{(C) }146}</math> is likely to be the correct answer.
+Cheese solution done by [[User:Juwushu|juwushu]].
+==Solution 3 (Observation)==
+We have <math>2</math> cases where the loop does not go through the middle.
+If the loop goes through the middle, we must have a full column on <math>(0,8),(0,7),(1,8),(1,7).</math> Then we have <math>6,5,5,4</math> empty middle squares. For each one we can have one on top or one on bottom, so <math>2^6+2^5+2^5+2^4=144.</math> Notice that for each case of fixed toothpicks, there is only one way to form the loop. Then we just add <math>2+144=\boxed{\textbf{(C) } 146.}</math>
+~nevergonnagiveup
+== Video Solution by Power Solve ==
+https://www.youtube.com/watch?v=FRNbJ5wIGRo
+==Video Solution By SpreadTheMathLove==
+https://www.youtube.com/watch?v=huMQ9J7rIj0&t=77s
+==See also==
+{{AMC10 box|year=2024|ab=A|num-b=24|after=Last Problem}}
+{{AMC12 box|year=2024|ab=A|num-b=21|num-a=23}}
+{{MAA Notice}}

2024 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

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