Difference between revisions of "2024 AMC 10A Problems/Problem 19"

Revision as of 21:28, 20 March 2025

Problem

A bee is moving in three-dimensional space. A fair six-sided die with faces labeled $A^{+}, A^{-}, B^{+}, B^{-}, C^{+}$ , and $C^{-}$ is rolled. Suppose the bee occupies the point $(a, b, c)$ . If the die shows $A^{+}$ , then the bee moves to the point $(a + 1, b, c)$ and if the die shows $A^{-}$ , then the bee moves to the point $(a - 1, b, c)$ . Analogous moves are made with the other four outcomes. Suppose the bee starts at the point $(0, 0, 0)$ and the die is rolled four times. What is the probability that the bee traverses four distinct edges of some unit cube?

$\textbf{(A)}~\frac{1}{54}\qquad\textbf{(B)}~\frac{7}{54}\qquad\textbf{(C)}~\frac{1}{6}\qquad\textbf{(D)}~\frac{5}{18}\qquad\textbf{(E)}~\frac{2}{5}$

Diagrams

~ Diagram by PaperMath

~ Solution diagram by juwushu

Solution 1

WLOG, assume that the first two moves are equal for all possible combinations, since the direction does not matter. The first move has a $\frac{6}{6}$ probability of being along one of the $8$ unit cubes around the origin, and the second move has a $\frac{4}{6}$ chance. Now, there are two cases. We are currently on one of the points of the $2$ by $2$ squares that are aligned with the axes. The first case is if the bee moves to the corner of a cube farthest away from the origin. Here, there is a $\frac{2}{6}$ chance of this happening and a $\frac{2}{6}$ chance of the fourth move remaining on one of the cubes. The second case is if the bee moves along the same plane of the $2$ by $2$ squares previously, ending up on a point 1 away from the origin. There is a $\frac{1}{6}$ chance of this happening and a $\frac{3}{6}$ chance of remaining on one of the cubes. Now, multiply and sum for the answer. $\frac{2}{3}\cdot(\frac{1}{3}\cdot\frac{1}{3}+\frac{1}{6}\cdot\frac{1}{2})=\frac{2}{3}\cdot(\frac{1}{9}+\frac{1}{12})$ Evaluating this gives you the answer of $\boxed{\textbf{(B) }\frac{7}{54}}$ . Solution by juwushu. Minor edits by andliu766

Solution 2

Remove the dice. It is not necessary for this problem. It is the same problem if you instead view this problem as the bee moving one up or down in the $x, y,$ and $z$ directions.

Count the total number of ways for one cube of the 8 total ones. It is not hard to see that after the first step (which you can travel to $(1,0,0), (0,1,0),$ or $(0,0,1)$ - 3 ways), you have a total of 8 moves to traverse a face of that 1x1 cube. Hence, for each cube, you would have $8 \cdot 3 = 24$ total moves.

There are a total of 8 cubes you could have traversed, though, so multiplying this by $8$ yields $8 \cdot 8 \cdot 3.$

We have to account for overcounting. For example, when we traversed the face in the $xz$ plane $(0,0,0), (0,0,1),(1,0,0),(1,0,1),$ the number of ways that we traverse that particular face got counted twice, once from the cube in the far upper right and once from the cube in the closer upper right.

For a given face, you can go clockwise or counterclockwise, hence yielding $2$ ways. There are a total of $12$ shared faces, each one shared by exactly two faces so no additonal PIE is needed. We simply subtract $2\cdot12$ from our original count now, yielding $8 \cdot 3 \cdot 8 - 2 \cdot 12 = 168.$

With no restrictions, there are a total of $6^4$ ways to move four steps. Your probability is $\frac{168}{6^4}=\frac{7}{54}.$

~mathboy282

Video Solution by Power Solve

https://www.youtube.com/watch?v=2uTWPiWzfB0

Video Solution by Pi Academy (Fast and Easy ⚡️🚀)

https://youtu.be/3Dk-vmdxbrw?si=G6TcyLN2Rfwd2YYJ

Video Solution by Innovative Minds

https://youtu.be/CxLGUPhYYDI

~i_am_suk_at_math_2

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=6SQ74nt3ynw

@@ Line 1: / Line 1: @@
-{{duplicate|[[2024 AMC 10A Problems/Problem 19|2024 AMC 10A #19]] and [[2024 AMC 12A Problems/Problem 12|2024 AMC 12A #12]]}}
 ==Problem==
-The first three terms of a geometric sequence are the integers <math>a,\,720,</math> and <math>b,</math> where <math>a<720<b.</math> What is the sum of the digits of the least possible value of <math>b?</math>
+A bee is moving in three-dimensional space. A fair six-sided die with faces labeled <math>A^{+}, A^{-}, B^{+}, B^{-}, C^{+}</math>, and <math>C^{-}</math> is rolled. Suppose the bee occupies the point <math>(a, b, c)</math>. If the die shows <math>A^{+}</math>, then the bee moves to the point <math>(a + 1, b, c)</math> and if the die shows <math>A^{-}</math>, then the bee moves to the point <math>(a - 1, b, c)</math>. Analogous moves are made with the other four outcomes. Suppose the bee starts at the point <math>(0, 0, 0)</math> and the die is rolled four times. What is the probability that the bee traverses four distinct edges of some unit cube?
-<math>\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 21</math>
-==Solution 1==
-For a geometric sequence, we have <math>ab=720^2=2^8 3^4 5^2</math>, and we can test values for <math>b</math>. We find that <math>b=768</math> and <math>a=675</math> works, and we can test multiples of <math>5</math> in between the two values. Finding that none of the multiples of 5 divide <math>720^2</math> besides <math>720</math> itself, we know that the answer is <math>7+6+8=\boxed{\textbf{(E) } 21}</math>.
-(Note: To find the value of <math>b</math> without bashing, we can observe that <math>2^8=256</math>, and that multiplying it by <math>3</math> gives us <math>768</math>, which is really close to <math>720</math>. ~ YTH)
+<math>\textbf{(A)}~\frac{1}{54}\qquad\textbf{(B)}~\frac{7}{54}\qquad\textbf{(C)}~\frac{1}{6}\qquad\textbf{(D)}~\frac{5}{18}\qquad\textbf{(E)}~\frac{2}{5}</math>
-Note: The reason why <math>ab=720^2</math> is because <math>b/720 = 720/a</math>. Rearranging this gives <math>ab = 720^2</math>
+== Diagrams ==
+[[Image:Diagrams.jpeg|200px]]
+[[File:Bee total.png|600px|right]]
+~ Diagram by PaperMath
-~eevee9406
+[[Image:2024amc10a24solution.jpg|300px]]
-Note: Another reason that <math>ab=720^2</math> is because the <math>\sqrt{ab}=720</math> (as the middle term in a geometric series is always the geometric mean [the geometric mean is the square root of the product of the first and last terms of the series]) and squaring on both sides results in <math>ab=720^2</math>.
+~ Solution diagram by juwushu
-~ThatPrimePunnyGuy
+== Solution 1 ==
+WLOG, assume that the first two moves are equal for all possible combinations, since the direction does not matter. The first move has a <math>\frac{6}{6}</math> probability of being along one of the <math>8</math> unit cubes around the origin, and the second move has a <math>\frac{4}{6}</math> chance.
+Now, there are two cases. We are currently on one of the points of the <math>2</math> by <math>2</math> squares that are aligned with the axes.
+The first case is if the bee moves to the corner of a cube farthest away from the origin.
+Here, there is a <math>\frac{2}{6}</math> chance of this happening and a <math>\frac{2}{6}</math> chance of the fourth move remaining on one of the cubes.
+The second case is if the bee moves along the same plane of the <math>2</math> by <math>2</math> squares previously, ending up on a point 1 away from the origin. There is a <math>\frac{1}{6}</math> chance of this happening and a <math>\frac{3}{6}</math> chance of remaining on one of the cubes.
+Now, multiply and sum for the answer.
+<cmath>\frac{2}{3}\cdot(\frac{1}{3}\cdot\frac{1}{3}+\frac{1}{6}\cdot\frac{1}{2})=\frac{2}{3}\cdot(\frac{1}{9}+\frac{1}{12})</cmath>
+Evaluating this gives you the answer of <math>\boxed{\textbf{(B) }\frac{7}{54}}</math>.
+Solution by [[User:Juwushu|juwushu]].
+Minor edits by andliu766
 ==Solution 2==
-We have <math>720 = 2^4 \cdot 3^2 \cdot 5</math>. We want to find factors <math>x</math> and <math>y</math> where <math>y>x</math> such that <math>\frac{y}{x}</math> is minimized, as <math>720 \cdot \frac{y}{x}</math> will then be the least possible value of <math>b</math>. After experimenting, we see this is achieved when <math>y=16</math> and <math>x=15</math>, which means our value of <math>b</math> is <math>720 \cdot \frac{16}{15} = 768</math>, so our sum is <math>7+6+8=\boxed{\textbf{(E) } 21}</math>.
+Remove the dice. It is not necessary for this problem. It is the same problem if you instead view this problem as the bee moving one up or down in the <math>x, y,</math> and <math>z</math> directions.
-~i_am_suk_at_math_2
-== Solution 3 (Similar to previous solutions) ==
+Count the total number of ways for one cube of the 8 total ones. It is not hard to see that after the first step (which you can travel to <math>(1,0,0), (0,1,0),</math> or <math>(0,0,1)</math> -  3 ways), you have a total of 8 moves to traverse a face of that 1x1 cube. Hence, for each cube, you would have <math>8 \cdot 3 = 24</math> total moves.
-To minimize the value of <math>b</math>, where it has to be an integer, and it has to be greater than 720, we can express the common ratio as <math>\dfrac{n+1}{n}</math>, where the value has to be greater than <math>1</math>, and <math>n</math>, and <math>n+1</math> have to be factors of 720. Since the bigger the denominator gets, the smaller the value of the fraction, we essentially have to find the biggest value for <math>n</math>, where itself and <math>n+1</math> are factors of <math>720</math>. From here, we can check whether <math>n(n+1) = 720</math> yields an integer root, which it doesn't. So, then we check the next biggest factor of <math>720</math>, which is <math>360</math>. <math>n(n+1) = 360</math>, this doesn't have an integer root either. So, then we check the next biggest factor which is <math>240</math>, <math>n(n+1) = 240</math>, which we get <math>15</math> as a root. This means the common ration is <math>\dfrac{15}{16}</math>. We then multiply <math>\dfrac{15}{16}</math> times <math>720</math> and add up the digits getting <math>\boxed{\textbf{(E) } 21}</math>.
+There are a total of 8 cubes you could have traversed, though, so multiplying this by <math>8</math> yields <math>8 \cdot 8 \cdot 3.</math>
-~yuvag
+We have to account for overcounting. For example, when we traversed the face in the <math>xz</math> plane <math>(0,0,0), (0,0,1),(1,0,0),(1,0,1),</math> the number of ways that we traverse that particular face got counted twice, once from the cube in the far upper right and once from the cube in the closer upper right.
-== Solution 4 ==
+For a given face, you can go clockwise or counterclockwise, hence yielding <math>2</math> ways. There are a total of <math>12</math> shared faces, each one shared by exactly two faces so no additonal PIE is needed. We simply subtract <math>2\cdot12</math> from our original count now, yielding <math>8 \cdot 3 \cdot 8 - 2 \cdot 12 = 168.</math>
-Let the common ratio of the geometric sequence be <math>\frac{x}{y}</math>, with <math>x>y</math>. This means that <math>720(\frac{x}{y})</math> and <math>720(\frac{y}{x})</math> must both be integers, therefore <math>x</math> and <math>y</math> are both factors of <math>720</math>. We would achieve the smallest ratio <math>\frac{x}{y}</math> if <math>x</math> and <math>y</math> are consecutive, so by listing out the factors of <math>720</math>, we find that <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, <math>6</math>, <math>8</math>, <math>9</math>, <math>15</math>, <math>16</math> are the only consecutive factors (any factors larger than these would result in the ratio simplifying, making it larger than just using consecutive integers). <math>15</math> and <math>16</math> are the largest, so we find the common ratio to be <math>\frac{16}{15}</math>, making <math>b=720(\frac{16}{15})</math> giving us <math>768</math>. The sum of its digits is <math>\boxed{\textbf{(E) } 21}</math>.
+With no restrictions, there are a total of <math>6^4</math> ways to move four steps. Your probability is <math>\frac{168}{6^4}=\frac{7}{54}.</math>
-~lisztepos
+~mathboy282
-== Solution 5 (Quick but not much recommended) ==
-To find the smallest possible value of <math>b</math> such that <cmath>a<720<b</cmath>, we need to find the <math>2</math> factors of <math>720</math> that are closest to each other. After a little bit of light bash we find that those <math>2</math> numbers are <math>48</math> and <math>45</math>. Note that the <math>2</math> numbers don't necessarily have to multiply to <math>2024</math>. You could have used prime factorization to find this. Next all we do is find <math>b</math> which is just: <cmath>\dfrac{48}{45} \cdot 720 = \boxed{\textbf{(E) } 21}</cmath>.
--jb2015007
-==Solution 6 (similar to other but even faster ig) ==
-We look for the smallest possible increasing ratio in the geometric sequence for <math>a</math>, <math>720</math>, and <math>b</math>.
-First, we find the prime factorization of <math>720</math>, which is really easy (I'm not typing it). Then, we look for the closest possible fraction-ratio out of the primes, and after a quick look, its easy to see it is <math>16</math> and <math>15</math>. Now that we found our ratio, we multiply <math>720</math> with <math>16/15</math> to get <math>768</math>, so the sum of the digits = is <math>21</math>.
-~hashbrown2009 (trash latex ik sorry)
 == Video Solution by Power Solve ==
-https://www.youtube.com/watch?v=B0JOMiiCtAo
+https://www.youtube.com/watch?v=2uTWPiWzfB0
-==Video Solution by SF==
+== Video Solution by Pi Academy (Fast and Easy ⚡️🚀) ==
-https://www.youtube.com/watch?v=JoF4-fo7W1w&pp=ygULYW1jIDEwIDIwMjQ%3D
-==Detailed Video Solution by Scholars Foundation==
+https://youtu.be/3Dk-vmdxbrw?si=G6TcyLN2Rfwd2YYJ
-https://www.youtube.com/watch?v=JoF4-fo7W1w&pp=ygULYW1jIDEwIDIwMjQ%3D
+==Video Solution by Innovative Minds==
+https://youtu.be/CxLGUPhYYDI
-==Video Solution by Pi Academy==
+~i_am_suk_at_math_2
-https://youtu.be/fW7OGWee31c?si=oq7toGPh2QaksLHE
 ==Video Solution by SpreadTheMathLove==
@@ Line 67: / Line 58: @@
 ==See also==
 {{AMC10 box|year=2024|ab=A|num-b=18|num-a=20}}
-{{AMC12 box|year=2024|ab=A|num-b=11|num-a=13}}
 {{MAA Notice}}

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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