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Difference between revisions of "2024 AMC 10B Problems/Problem 25"
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==Problem==
 
==Problem==
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You made it! If you're good with asymptote, please insert an image along with the rest of the problem!
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Each of <math>27</math> bricks (right rectangular prisms) has dimensions <math>a \times b \times c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are pairwise relatively prime positive integers. These bricks are arranged to form a <math>3 \times 3 \times 3</math> block, as shown on the left below. A <math>28</math>th brick with the same dimensions is introduced, and these bricks are reconfigured into a <math>2 \times 2 \times 7</math> block, shown on the right. The new block is <math>1</math> unit taller, <math>1</math> unit wider, and <math>1</math> unit deeper than the old one. What is <math>a + b + c</math>?
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(diagram pls)
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<math>
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\textbf{(A) }88 \qquad
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\textbf{(B) }89 \qquad
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\textbf{(C) }90 \qquad
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\textbf{(D) }91 \qquad
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\textbf{(E) }92 \qquad
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</math>
  
 
==Solution 1==
 
==Solution 1==

Revision as of 09:09, 14 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 $a + b + c$?

(diagram pls)

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

Solution 1

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 $\boxed{92}$. ~lprado

See also

2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 20 		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

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