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

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==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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[[File:AMC10B2024 P25.png]]
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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>
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==Solution 1 (Less than 60 seconds)==
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The <math>3</math>x<math>3</math>x<math>3</math> block has side lengths of <math>3a, 3b, 3c</math>. The <math>2</math>x<math>2</math>x<math>7</math> block has side lengths of <math>2b, 2c, 7a</math>.
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We can create the following system of equations, knowing that the new block has <math>1</math> unit taller, deeper, and wider than the original:
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<cmath>3a+1 = 2b</cmath>
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<cmath>3b+1=2c</cmath>
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<cmath>3c+1=7a</cmath>
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Adding all the equations together, we get <math>b+c+3 = 4a</math>. Adding <math>a-3</math> to both sides, we get <math>a+b+c = 5a-3</math>. The question states that <math>a,b,c</math> are all relatively prime positive integers. Therefore, our answer must be congruent to <math>2 \pmod{5}</math>. The only answer choice satisfying this is <math>\boxed{E(92)}</math>.
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~lprado
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==Solution 2==
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We will define the equations the same as solution 1.
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<cmath>3a+1 = 2b</cmath>
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<cmath>3b+1=2c</cmath>
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<cmath>3c+1=7a</cmath>
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Solve equation 2 for c and substitute that value in for equation 3, giving us
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<cmath>3a+1 = 2b</cmath>
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<cmath>\frac{3b+1}{2}=c</cmath>
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<cmath>\frac{3*(3b+1)}{2}+1=7a</cmath>
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Multiply 14 to the first equation and rearrange to get
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<cmath>42a = 28b-14</cmath>
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and multiply the third by 2 and rearrange to get
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<cmath>27b+15 = 42a</cmath>
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Solve for b to get <math>b = 29</math>, substitute into equation 1 from the original to get <math>a = 19</math>, and lastly, substitute a into original equation 2 to get <math>c = 44</math>.
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Thus, <math>a+b+c = 19+29+44 = \boxed{E(92)}</math>.
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~Failure.net
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==Video Solution 1 by Pi Academy (In Less Than 5 Mins ⚡🚀)==
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https://www.youtube.com/watch?v=XI7jmtVchZ0
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 +
~ jj_empire10
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==Video Solution 2 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=24EZaeAThuE
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==See also==
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{{AMC10 box|year=2024|ab=B|num-b=24|after=Last Problem}}
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{{MAA Notice}}

Latest revision as of 16:20, 7 December 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 $a + b + c$?

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://www.youtube.com/watch?v=XI7jmtVchZ0

~ jj_empire10

Video Solution 2 by SpreadTheMathLove

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

See also

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

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