2023 AMC 10B Problems/Problem 5

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

Page

Toolbox

Search

2023 AMC 10B Problems/Problem 5

Problem

Solution

Solution 2