Difference between revisions of "2024 AMC 8 Problems/Problem 8"

Revision as of 20:50, 26 January 2024

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 (BRUTE FORCE)

How many 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 $ $7$ . 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 $10$ .

$ $11$ and $ $16$ repeat leaving you with $8-2 = \boxed{\textbf{(D)\ 6}}$ different values.

~ cxsmi (minor formatting edits)

Solution 2 (Slightly less BRUTE FORCE)

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

Video Solution 1 (easy to digest) by Power Solve

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

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

@@ Line 10: / Line 10: @@
 ~ cxsmi (minor formatting edits)
+==Solution 2 (Slightly less BRUTE FORCE)==
+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
 ==Video Solution 1 (easy to digest) by Power Solve==

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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