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

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

@@ Line 1: / Line 1: @@
+==Problem==
+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?
+<math>\textbf{(A) }10\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }8\qquad\textbf{(E) }9</math>
+==Solution==
+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
+<cmath>S+3n=45</cmath>
+<cmath>3S=45</cmath>
+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 (formatted atictacksh)
+==Solution 2==
+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
+==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}}

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