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Difference between revisions of "2024 AMC 10A Problems/Problem 25"
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</asy>
 
</asy>
 
<math>\textbf{(A) }130\qquad\textbf{(B) }144\qquad\textbf{(C) }146\qquad\textbf{(D) }162\qquad\textbf{(E) }196</math>
 
<math>\textbf{(A) }130\qquad\textbf{(B) }144\qquad\textbf{(C) }146\qquad\textbf{(D) }162\qquad\textbf{(E) }196</math>
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==Solution==
 
==Solution==
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Call the first row <math>A</math> and the second row <math>B</math>. To cross over from <math>A</math> to <math>B</math>, we must touch two boxes during each cross, and we will cross twice. We can cross at the first column or the second column on the left side, and the eighth column or the ninth column on the right side. For each crossing case, the remaining boxes not touched (<math>6,5,5,4</math> boxes depending on the crossing points) can be touched at either the top or the bottom but not both, so there are <math>2^n</math> cases for <math>n</math> remaining boxes. Thus there are <math>2^6+2^5+2^5+2^4=144</math> cases. However, we have forgotten the case where the loop only resides in <math>A</math> or only in <math>B</math>, so the final count is <math>144+2=\boxed{\textbf{(C) }146}</math>.
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~eevee9406

Revision as of 16:33, 8 November 2024

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] $\textbf{(A) }130\qquad\textbf{(B) }144\qquad\textbf{(C) }146\qquad\textbf{(D) }162\qquad\textbf{(E) }196$

Solution

Call the first row $A$ and the second row $B$. To cross over from $A$ to $B$, we must touch two boxes during each cross, and we will cross twice. We can cross at the first column or the second column on the left side, and the eighth column or the ninth column on the right side. For each crossing case, the remaining boxes not touched ($6,5,5,4$ boxes depending on the crossing points) can be touched at either the top or the bottom but not both, so there are $2^n$ cases for $n$ remaining boxes. Thus there are $2^6+2^5+2^5+2^4=144$ cases. However, we have forgotten the case where the loop only resides in $A$ or only in $B$, so the final count is $144+2=\boxed{\textbf{(C) }146}$.

~eevee9406

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