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

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Maddy and Lara see a list of numbers written on a blackboard. Maddy adds <math>3</math> to each number in the list and finds that the sum of her new numbers is <math>45</math>. Lara multiplies each number in the list by <math>3</math> and finds that the sum of her new numbers is also <math>45</math>. How many numbers are written on the blackboard?
 
Maddy and Lara see a list of numbers written on a blackboard. Maddy adds <math>3</math> to each number in the list and finds that the sum of her new numbers is <math>45</math>. Lara multiplies each number in the list by <math>3</math> and finds that the sum of her new numbers is also <math>45</math>. How many numbers are written on the blackboard?
  
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<math>\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10</math>
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<math>\textbf{(A) }10\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }8\qquad\textbf{(E) }9</math>
  
 
==Solution==
 
==Solution==
  
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Let there be <math>x</math> numbers in the list of numbers, and let their sum be <math>S</math>. Then we have the following
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Let there be <math>n</math> numbers in the list of numbers, and let their sum be <math>S</math>. Then we have the following
  
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<math>S+3x=45</math>
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<cmath>S+3n=45</cmath>
  
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<math>3S=45</math>
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<cmath>3S=45</cmath>
  
−
From the second equation, <math>S=15</math>, so <math>15+3x=45</math>. Solving, we find <math>x=\boxed{\textbf{(E) }10}.</math>
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From the second equation, <math>S=15</math>. So, <math>15+3n=45</math> <math>\Rightarrow</math> <math>n=\boxed{\textbf{(A) }10}.</math>
  
  
−
~Mintylemon66
+
~Mintylemon66 (formatted atictacksh)
  
 
==Solution 2==
 
==Solution 2==
  
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Let <math>x_1,x_2,x_3,...,x_n</math> where <math>x_n</math> represents the <math>n</math>th number written on the board. Lara's multiplied each number by <math>3</math>, so her sum will be <math>3x_1+3x_2+3x_3+...+3x_n</math>. This is the same as <math>3\cdot (x_1+x_2+x_3+...+x_n)</math>. We are given this quantity is equal to <math>45</math>, so the original numbers add to <math>\frac{45}{3}=15</math>. Maddy adds <math>3</math> to each of the <math>n</math> terms which yields, <math>x_1+3+x_2+3+x_3+3+...+x_n+3</math>. This is the same as the sum of the original series plus <math>3 \cdot n</math>. Setting this equal to <math>45</math>, <math>15+3n=45 \Rightarrow n =\boxed{\textbf{(E) }10}.</math>
+
Let <math>x_1,x_2,x_3,...,x_n</math> where <math>x_n</math> represents the <math>n</math>th number written on the board. Lara's multiplied each number by <math>3</math>, so her sum will be <math>3x_1+3x_2+3x_3+...+3x_n</math>. This is the same as <math>3\cdot (x_1+x_2+x_3+...+x_n)</math>. We are given this quantity is equal to <math>45</math>, so the original numbers add to <math>\frac{45}{3}=15</math>. Maddy adds <math>3</math> to each of the <math>n</math> terms which yields, <math>x_1+3+x_2+3+x_3+3+...+x_n+3</math>. This is the same as the sum of the original series plus <math>3 \cdot n</math>. Setting this equal to <math>45</math>, <math>15+3n=45 \Rightarrow n =\boxed{\textbf{(A) }10}.</math>
  
 
~vsinghminhas
 
~vsinghminhas
 +
 +
==Solution 3==
 +
 +
If the list of numbers written on the board is <math>a_1, a_2, a_3, \ldots, a_n</math>, then we can formulate two equations:
 +
 +
<cmath>3n + \sum_{i=1}^{n} a_i = 45</cmath>
 +
 +
<cmath>3 \sum_{i=1}^{n} a_i = 45</cmath>
 +
 +
We can rewrite the first equation by multiplying both sides by <math>3</math>:
 +
 +
<math>3(3n + \sum_{i=1}^{n} a_i) = 3(45)</math>
 +
 +
<math>\Rightarrow 9n + 3 \sum_{i=1}^{n} a_i = 135</math>
 +
 +
Now, subtract the second equation from the first:
 +
 +
<cmath>(9n + 3 \sum_{i=1}^{n} a_i) - (3 \sum_{i=1}^{n} a_i) = 135 - 45</cmath>
 +
 +
<cmath>\Rightarrow 9n = 135 - 45</cmath>
 +
 +
<cmath>\Rightarrow 9n = 90</cmath>
 +
 +
<cmath>\Rightarrow n =\boxed{\textbf{(A) }10}</cmath>
 +
 +
~ <math>shalomkeshet</math>
 +
 +
==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
 +
 +
==See also==
 +
{{AMC10 box|year=2023|ab=B|num-b=4|num-a=6}}
 +
{{MAA Notice}}

Latest revision as of 11:46, 5 November 2024

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:

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

\[\Rightarrow 9n = 135 - 45\]

\[\Rightarrow 9n = 90\]

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

~ $shalomkeshet$

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

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

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