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Difference between revisions of "2023 AMC 10B Problems/Problem 5"
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~Math-X
 
~Math-X
  
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==Video Solution==
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https://youtu.be/-yk7ozNRrtQ
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2023|ab=B|num-b=4|num-a=6}}
 
{{AMC10 box|year=2023|ab=B|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 01:58, 20 November 2023

Problem

Maddy and Lara see a list of numbers written on a blackboard. Maddy adds $3$ to each number in the list and finds that the sum of her new numbers is $45$. Lara multiplies each number in the list by $3$ and finds that the sum of her new numbers is also $45$. How many numbers are written on the blackboard?

$\textbf{(A) }10\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }8\qquad\textbf{(E) }9$

Solution

Let there be $n$ numbers in the list of numbers, and let their sum be $S$. Then we have the following

\[S+3n=45\]

\[3S=45\]

From the second equation, $S=15$. So, $15+3n=45$ $\Rightarrow$ $n=\boxed{\textbf{(A) }10}.$


~Mintylemon66 (formatted atictacksh)

Solution 2

Let $x_1,x_2,x_3,...,x_n$ where $x_n$ represents the $n$th number written on the board. Lara's multiplied each number by $3$, so her sum will be $3x_1+3x_2+3x_3+...+3x_n$. This is the same as $3\cdot (x_1+x_2+x_3+...+x_n)$. We are given this quantity is equal to $45$, so the original numbers add to $\frac{45}{3}=15$. Maddy adds $3$ to each of the $n$ terms which yields, $x_1+3+x_2+3+x_3+3+...+x_n+3$. This is the same as the sum of the original series plus $3 \cdot n$. Setting this equal to $45$, $15+3n=45 \Rightarrow n =\boxed{\textbf{(A) }10}.$

~vsinghminhas

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=SUnhwbA5_So

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

https://youtu.be/EuLkw8HFdk4?si=6dyj2QxkbBuNk6j7&t=951

~Math-X

Video Solution

https://youtu.be/-yk7ozNRrtQ

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See also

2023 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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 AMC 10 Problems and Solutions

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

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