2024 AMC 8 Problems/Problem 8

Revision as of 17:24, 25 January 2024 by Tigertan (talk | contribs) (Solution 1 (BRUTE FORCE))

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. On the second day, you can either add $3$ dollars, or double so that 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 $10$.

There is a total of $2x2x2=8$ combinations, and $$11$ and $$16$ repeat. Leaving you with $8-2 = \boxed{\textbf{(C)} 6}$ different values.

~cxsmi (minor formatting edits)

Video Solution 1(easy to digest) by Power Solve

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