Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AMC 8 Problems/Problem 22"

(→‎Problem)
(→‎Solution 2)
Line 4: Line 4:
 
<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 10</math>
 
<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 10</math>
  
−
==Solution 2==
+
==Solution 1==
 
In this solution, we will use trial and error to solve.
 
In this solution, we will use trial and error to solve.
 
<math>4000</math> can be expressed as <math>200 \times 20</math>. We divide <math>200</math> by <math>20</math> and get <math>10</math>, divide <math>20</math> by <math>10</math> and get <math>2</math>, and divide <math>10</math> by <math>2</math> to get <math>\boxed{\textbf{(D)}\ 5}</math>. No one said that they have to be in ascending order!
 
<math>4000</math> can be expressed as <math>200 \times 20</math>. We divide <math>200</math> by <math>20</math> and get <math>10</math>, divide <math>20</math> by <math>10</math> and get <math>2</math>, and divide <math>10</math> by <math>2</math> to get <math>\boxed{\textbf{(D)}\ 5}</math>. No one said that they have to be in ascending order!
  
 
Solution by [[User:ILoveMath31415926535|ILoveMath31415926535]] and clarification edits by apex304
 
Solution by [[User:ILoveMath31415926535|ILoveMath31415926535]] and clarification edits by apex304
 +
 +
 +
==Solution 2==
 +
Consider the first term is <math>a</math> and the second term is <math>b</math>. Then, the following term will be <math>ab</math>, <math>ab^2</math>, <math>a^2b^3</math> and <math>a^3b^5</math>. Notice that <math>4000=2^5\times 5^3</math>, then we obtain <math>a=5</math> and <math>b=\boxed{\textbf{(D)}\ 5}</math>.
 +
 +
Solution by [[User:xana233|xana233]]
  
 
==Video Solution (THINKING CREATIVELY!!!)==
 
==Video Solution (THINKING CREATIVELY!!!)==

Revision as of 00:29, 4 March 2024

Problem

In a sequence of positive integers, each term after the second is the product of the previous two terms. The sixth term is $4000$. What is the first term?

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 10$

Solution 1

In this solution, we will use trial and error to solve. $4000$ can be expressed as $200 \times 20$. We divide $200$ by $20$ and get $10$, divide $20$ by $10$ and get $2$, and divide $10$ by $2$ to get $\boxed{\textbf{(D)}\ 5}$. No one said that they have to be in ascending order!

Solution by ILoveMath31415926535 and clarification edits by apex304


Solution 2

Consider the first term is $a$ and the second term is $b$. Then, the following term will be $ab$, $ab^2$, $a^2b^3$ and $a^3b^5$. Notice that $4000=2^5\times 5^3$, then we obtain $a=5$ and $b=\boxed{\textbf{(D)}\ 5}$.

Solution by xana233

Video Solution (THINKING CREATIVELY!!!)

https://youtu.be/LAeSj372-UQ

~Education, the Study of Everything

Video Solution by Math-X (Smart and Simple)

https://youtu.be/Ku_c1YHnLt0?si=uptT6DExGvKiatZK&t=4952 ~Math-X

Video Solution 1 (Using Diophantine Equations)

https://youtu.be/SwPcIZxp_gY

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=ms4agKn7lqc

Animated Video Solution

https://youtu.be/tnv1XzSOagA

~Star League (https://starleague.us)

Video Solution by Magic Square

https://youtu.be/-N46BeEKaCQ?t=2649

Video Solution by Interstigation

https://youtu.be/DBqko2xATxs&t=3007

Video Solution by WhyMath

https://youtu.be/RCYRD7OLSLc

~savannahsolver

Video Solution by harungurcan

https://www.youtube.com/watch?v=Ki4tPSGAapU&t=1249s

~harungurcan

See Also

2023 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_8_Problems/Problem_22&oldid=216443"