Difference between revisions of "2023 AMC 10B Problems/Problem 5"

Revision as of 18:17, 22 August 2025

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

Solution 3

If the list of numbers written on the board is $a_1, a_2, a_3, \ldots, a_n$ , then we can formulate two equations:

$3n + \sum_{i=1}^{n} a_i = 45$

$3 \sum_{i=1}^{n} a_i = 45$

We can rewrite the first equation by multiplying both sides by $3$ :

$3(3n + \sum_{i=1}^{n} a_i) = 3(45)$

$\Rightarrow 9n + 3 \sum_{i=1}^{n} a_i = 135$

Now, subtract the second equation from the first:

$(9n + 3 \sum_{i=1}^{n} a_i) - (3 \sum_{i=1}^{n} a_i) = 135 - 45$

$\Rightarrow 9n = 135 - 45$

$\Rightarrow 9n = 90$

$\Rightarrow n =\boxed{\textbf{(A) }10}$

~ $Shalomkeshet$

Solution 4

Notice how the problem tries to throw us off. We don't need to find the sum, but rather how many 3's do we need to remove to get to the sum.

We have that $ 3a + 3b + 3c + 3d + \dots = 45 $. Dividing by 3 gives us $ a + b + c + d + \dots = 15 $.

Notice how we can write Maddy's list like $ (a + b + c + d + \dots) + (3 + 3 + 3 + 3 + \dots) = 45 $. We know that $ a + b + c + d + \dots = 15 $, and we know that the definition of adding 3's repeatedly is just multiplying 3 by how many numbers it appears, $ l $, which is our list.

This gives us the equation $ 15 + 3l = 45 $, and solving for $ l $ gives us $\boxed{A. 10}$

This problem teaches us to look for patterns and use our elementary taught topics to our advantage.

~Pinotation

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)

Video Solution by Interstigation

https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L

@@ Line 58: / Line 58: @@
 Notice how we can write Maddy's list like \( (a + b + c + d + \dots) + (3 + 3 + 3 + 3 + \dots) = 45 \). We know that \( a + b + c + d + \dots = 15 \), and we know that the definition of adding 3's repeatedly is just multiplying 3 by how many numbers it appears, \( l \), which is our list.
-This gives us the equation \( 15 + 3l = 45 \), and solving for \( l \) gives us <math>\boxed(A. 10)</math>
+This gives us the equation \( 15 + 3l = 45 \), and solving for \( l \) gives us <math>\boxed{A. 10}</math>
 This problem teaches us to look for patterns and use our elementary taught topics to our advantage.

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

Page

Toolbox

Search