Difference between revisions of "2024 AMC 10B Problems/Problem 25"

Latest revision as of 18:11, 23 November 2024

Problem

Each of $27$ bricks (right rectangular prisms) has dimensions $a \times b \times c$ , where $a$ , $b$ , and $c$ are pairwise relatively prime positive integers. These bricks are arranged to form a $3 \times 3 \times 3$ block, as shown on the left below. A $28$ th brick with the same dimensions is introduced, and these bricks are reconfigured into a $2 \times 2 \times 7$ block, shown on the right. The new block is $1$ unit taller, $1$ unit wider, and $1$ unit deeper than the old one. What is $a + b + c$ ?

$\textbf{(A) }88 \qquad \textbf{(B) }89 \qquad \textbf{(C) }90 \qquad \textbf{(D) }91 \qquad \textbf{(E) }92 \qquad$

Solution 1 (Less than 60 seconds)

The $3$ x $3$ x $3$ block has side lengths of $3a, 3b, 3c$ . The $2$ x $2$ x $7$ block has side lengths of $2b, 2c, 7a$ .

We can create the following system of equations, knowing that the new block has $1$ unit taller, deeper, and wider than the original: $3a+1 = 2b$ $3b+1=2c$ $3c+1=7a$

Adding all the equations together, we get $b+c+3 = 4a$ . Adding $a-3$ to both sides, we get $a+b+c = 5a-3$ . The question states that $a,b,c$ are all relatively prime positive integers. Therefore, our answer must be congruent to $2 \pmod{5}$ . The only answer choice satisfying this is $\boxed{E(92)}$ . ~lprado

Solution 2

We will define the equations the same as solution 1. $3a+1 = 2b$ $3b+1=2c$ $3c+1=7a$ Solve equation 2 for c and substitute that value in for equation 3, giving us $3a+1 = 2b$ $\frac{3b+1}{2}=c$ $\frac{3*(3b+1)}{2}+1=7a$

Multiply 14 to the first equation and rearrange to get $42a = 28b-14$ and multiply the third by 2 and rearrange to get $27b+15 = 42a$ Solve for b to get $b = 29$ , substitute into equation 1 from the original to get $a = 19$ , and lastly, substitute a into original equation 2 to get $c = 44$ . Thus, $a+b+c = 19+29+44 = \boxed{E(92)}$ . ~Failure.net

Video Solution 1 by Pi Academy (In Less Than 5 Mins ⚡🚀)

https://youtu.be/Xn1JLzT7mW4?si=borSg8xrYLAz7mNY

~ jj_empire10

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=24EZaeAThuE

@@ Line 41: / Line 41: @@
 ~Failure.net
-==Video Solution 1 by Chicken_Rice (In Less Than 5 Mins ⚡🚀)==
+==Video Solution 1 by Pi Academy (In Less Than 5 Mins ⚡🚀)==
-https://youtu.be/Xn1JLzT7mW4?feature=shared
+https://youtu.be/Xn1JLzT7mW4?si=borSg8xrYLAz7mNY
-https://www.youtube.com/watch?v=kQGbyhze_mY
-https://youtu.be/kQGbyhze_mY?si=b2RhYMWt_RS8DnOa
-~ Pi Asademy
+~ jj_empire10
 ==Video Solution 2 by SpreadTheMathLove==

2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Problem
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