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2024 AMC 8 Problems/Problem 8: Difference between revisions

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<math>\textbf{(A) } 3\qquad\textbf{(B) } 4\qquad\textbf{(C) } 5\qquad\textbf{(D) } 6\qquad\textbf{(E) } 7</math>
<math>\textbf{(A) } 3\qquad\textbf{(B) } 4\qquad\textbf{(C) } 5\qquad\textbf{(D) } 6\qquad\textbf{(E) } 7</math>


==Solution 1 (BRUTE FORCE)==
==Solution 1 ==
How many values could be on the first day? Only <math>2</math> dollars. The second day, you can either add <math>3</math> dollars, or double, so you can have <math>5</math> dollars, or <math>4</math>. For each of these values, you have <math>2</math> values for each. For <math>5</math> dollars, you have <math>10</math> dollars or <math>8</math>, and for <math>4</math> dollars, you have <math>8</math> dollars or \$<math>7</math>. Now, you have <math>2</math> values for each of these. For <math>10</math> dollars, you have <math>13</math> dollars or <math>20</math>, for <math>8</math> dollars, you have <math>16</math> dollars or <math>11</math>, for <math>8</math> dollars, you have <math>16</math> dollars or <math>11</math>, and for <math>7</math> dollars, you have <math>14</math> dollars or <math>10</math>.
How many dollar values could be on the first day? Only <math>2</math> dollars. The second day, you can either add <math>3</math> dollars, or double, so you can have <math>5</math> dollars, or <math>4</math>. For each of these values, you have <math>2</math> values for each. For <math>5</math> dollars, you have <math>10</math> dollars or <math>8</math>, and for <math>4</math> dollars, you have <math>8</math> dollars or <math>7</math> dollars. Now, you have <math>2</math> values for each of these. For <math>10</math> dollars, you have <math>13</math> dollars or <math>20</math>, for <math>8</math> dollars, you have <math>16</math> dollars or <math>11</math>, for <math>8</math> dollars, you have <math>16</math> dollars or <math>11</math>, and for <math>7</math> dollars, you have <math>14</math> dollars or <math>10</math>.


On the final day, there are 11, 11, 16, and 16 repeating, leaving you with <math>8-2 = \boxed{\textbf{(D)\ 6}}</math> different values.
On the final day, there are 11, 11, 16, and 16 repeating, leaving you with <math>8-2 = \boxed{\textbf{(D)\ 6}}</math> different values.
Line 11: Line 11:
~ cxsmi (minor formatting edits)
~ cxsmi (minor formatting edits)


==Solution 2 (Slightly less BRUTE FORCE)==
==Solution 2 ==
Continue as in Solution 1 to get <math>7</math>, <math>8</math>, or <math>10</math> dollars by the 2nd day. The only way to get the same dollar amount occurring twice by branching (multiply by <math>2</math> or adding <math>3</math>) from here is if <math>7+3=10\cdot 2</math> or <math>7+3=8\cdot 2</math> which both aren't true. Hence our answer is <math>3\cdot2=\boxed{\textbf{(D)\ 6}}</math>.  
Continue as in Solution 1 to get <math>7</math>, <math>8</math>, or <math>10</math> dollars by the 2nd day. The only way to get the same dollar amount occurring twice by branching (multiply by <math>2</math> or adding <math>3</math>) from here is if <math>7+3=10\cdot 2</math> or <math>7+3=8\cdot 2</math> which both aren't true. Hence our answer is <math>3\cdot2=\boxed{\textbf{(D)\ 6}}</math>.  
   
   
~ Sahan Wijetunga
~ Sahan Wijetunga
==Solution 3 (Brute Force)==
Let <math>a</math> represent Taye adding <math>3</math> dollars and <math>d</math> represent Taye doubling his money.
We can list out all the possibilities:
<math>a</math> => <math>a</math> => <math>a</math> = <math>11</math>,
<math>a</math> => <math>a</math> => <math>d</math> = <math>16</math>,
<math>a</math> => <math>d</math> => <math>a</math> = <math>12</math>,
<math>a</math> => <math>d</math> => <math>d</math> = <math>20</math>,
<math>d</math> => <math>a</math> => <math>a</math> = <math>10</math>,
<math>d</math> => <math>a</math> => <math>d</math> = <math>14</math>,
<math>d</math> => <math>d</math> => <math>a</math> = <math>11</math>,
<math>d</math> => <math>d</math> => <math>d</math> = <math>16</math>.
The unique values are <math>10,11,12,14,16,20</math>. Thus, our desired answer is <math>\boxed{\textbf{(D)\ 6}}</math>
~Irfans123
==Video by MathTalks 😉==
https://youtu.be/9GVWXv9Pg1E?si=lhCKMjJ0wvfc_MfY
~rc1219
==Video Solution by Central Valley Math Circle(Goes through the full thought process)==
https://youtu.be/dC0inhFYaTw
~mr_mathman
==Video Solution by Math-X (First fully understand the problem!!!)==
https://youtu.be/BaE00H2SHQM?si=6wjacdxeAgtpc0fW&t=1762
~Math-X


==Video Solution (A Clever Explanation You’ll Get Instantly)==
==Video Solution (A Clever Explanation You’ll Get Instantly)==
Line 52: Line 82:


https://youtu.be/ruYzBZYaWSA
https://youtu.be/ruYzBZYaWSA
==Video Solution by WhyMath==
https://youtu.be/uvizv1hjSps
== Video solution by TheNeuralMathAcademy ==
https://youtu.be/f63MY1T2MgI&t=625s


==See Also==
==See Also==
{{AMC8 box|year=2024|num-b=7|num-a=9}}
{{AMC8 box|year=2024|num-b=7|num-a=9}}
{{MAA Notice}}
{{MAA Notice}}
[[Category:Introductory Combinatorics Problems]]

Latest revision as of 20:16, 22 August 2025

Problem

On Monday, Taye has $\$2$. Every day, he either gains $\$3$ or doubles the amount of money he had on the previous day. How many different dollar amounts could Taye have on Thursday, $3$ days later?

$\textbf{(A) } 3\qquad\textbf{(B) } 4\qquad\textbf{(C) } 5\qquad\textbf{(D) } 6\qquad\textbf{(E) } 7$

Solution 1

How many dollar values could be on the first day? Only $2$ dollars. The second day, you can either add $3$ dollars, or double, so you can have $5$ dollars, or $4$. For each of these values, you have $2$ values for each. For $5$ dollars, you have $10$ dollars or $8$, and for $4$ dollars, you have $8$ dollars or $7$ dollars. Now, you have $2$ values for each of these. For $10$ dollars, you have $13$ dollars or $20$, for $8$ dollars, you have $16$ dollars or $11$, for $8$ dollars, you have $16$ dollars or $11$, and for $7$ dollars, you have $14$ dollars or $10$.

On the final day, there are 11, 11, 16, and 16 repeating, leaving you with $8-2 = \boxed{\textbf{(D)\ 6}}$ different values.

~ cxsmi (minor formatting edits)

Solution 2

Continue as in Solution 1 to get $7$, $8$, or $10$ dollars by the 2nd day. The only way to get the same dollar amount occurring twice by branching (multiply by $2$ or adding $3$) from here is if $7+3=10\cdot 2$ or $7+3=8\cdot 2$ which both aren't true. Hence our answer is $3\cdot2=\boxed{\textbf{(D)\ 6}}$.

~ Sahan Wijetunga

Solution 3 (Brute Force)

Let $a$ represent Taye adding $3$ dollars and $d$ represent Taye doubling his money. We can list out all the possibilities: $a$ => $a$ => $a$ = $11$, $a$ => $a$ => $d$ = $16$, $a$ => $d$ => $a$ = $12$, $a$ => $d$ => $d$ = $20$, $d$ => $a$ => $a$ = $10$, $d$ => $a$ => $d$ = $14$, $d$ => $d$ => $a$ = $11$, $d$ => $d$ => $d$ = $16$. The unique values are $10,11,12,14,16,20$. Thus, our desired answer is $\boxed{\textbf{(D)\ 6}}$

~Irfans123

Video by MathTalks 😉

https://youtu.be/9GVWXv9Pg1E?si=lhCKMjJ0wvfc_MfY 

~rc1219

Video Solution by Central Valley Math Circle(Goes through the full thought process)

https://youtu.be/dC0inhFYaTw

~mr_mathman

Video Solution by Math-X (First fully understand the problem!!!)

https://youtu.be/BaE00H2SHQM?si=6wjacdxeAgtpc0fW&t=1762 

~Math-X

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/5ZIFnqymdDQ?si=4oMd4d0cnZ-nghTe&t=873

~hsnacademy

Video Solution 1 (easy to digest) by Power Solve

https://youtu.be/16YYti_pDUg?si=5kw0dc_bZwASNiWm&t=121

Video Solution by Math-X (First fully understand the problem!!!)

https://youtu.be/BaE00H2SHQM?si=6wjacdxeAgtpc0fW&t=1762 

~Math-X

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=V-xN8Njd_Lc

~NiuniuMaths

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=L83DxusGkSY

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=alDY4yhaEEg

Video Solution by Interstigation

https://youtu.be/ktzijuZtDas&t=791

Video Solution by Daily Dose of Math (Certified, Simple, and Logical)

https://youtu.be/8GHuS5HEoWc

~Thesmartgreekmathdude

Video Solution by Dr. David (Tree Method)

https://youtu.be/ruYzBZYaWSA

Video Solution by WhyMath

https://youtu.be/uvizv1hjSps

Video solution by TheNeuralMathAcademy

https://youtu.be/f63MY1T2MgI&t=625s

See Also

2024 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions

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