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

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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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Revision as of 18:23, 15 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?

$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10$

On my copy of the AMC 10B, the order of the answers is different:

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

Solution

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

$S+3x=45$

$3S=45$

From the second equation, $S=15$, so $15+3x=45$. Solving, we find $x=\boxed{\textbf{(E) }10}.$


~Mintylemon66

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 $45$, $15+3n=45 \Rightarrow n =\boxed{\textbf{(E) }10}.$

~vsinghminhas

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