2024 AMC 10B Problems/Problem 25

Revision as of 18:11, 23 November 2024 by Jj empire10 (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 $a + b + c$?

AMC10B2024 P25.png

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

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

See also

2024 AMC 10B (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png