2024 AMC 10B Problems/Problem 25: Difference between revisions

Latest revision as of 07:07, 9 November 2025

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 (Solution 1 but longer)

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

Solution 3

We will define the equations the same as solution 1 and 2. $3a+1 = 2b$ $3b+1=2c$ $3c+1=7a$ Multiply 1st equation by 1.5 to get $4.5a+1.5=3b$ Add 1 to get $4.5a+2.5=3b+1$ Substitute to get $4.5a+2.5=2c$ Multiply this one by 1.5 as well, getting $6.75a+3.75=3c$ Adding 1 again to get $6.75a+4.75=3c+1$ Replace $3c+1$ with $7a$ getting $6.75a+4.75=7a$ Solve for $a$ getting $0.25a=4.75$ $a=19$

Solve for b: $3(19)+1=2b$ $57+1=2b$ $b=29$

Solve for c: $3(29)+1=2c$ $87+1=2c$ $c=44$ Thus, $a+b+c = 19+29+44 = \boxed{E(92)}$

~newly2056

Solution 4 (No Fractions)

As stated in solutions 1, 2, 3, the equations are as follows: (1) $3a+1=2b$ (2) $3b+1=2c$ (3) $3c+1=7a$ Multiply equation (2) by 2 to get equation (4): (4) $6b+2=4c$ and substitute (1) into (4): (5) $9a+5=4c$ (3) $7a-1=3c$ Multiplying (5) by 3 and (3) by 4 gives: $3(9a+5)=4(7a-1)$ from which we get $a=19$ . Substituting $a$ into (1) and (3) yields $b=29$ and $c=44$ . Therefore, $a+b+c = \boxed{E(92)}$

~mathwizard123123

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://youtu.be/b-BBUKAVgeI?si=GAN4hho24927eJKX

@@ Line 1: / Line 1: @@
+==Problem==
+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>?
+[[File:AMC10B2024 P25.png]]
+<math>
+\textbf{(A) }88 \qquad
+\textbf{(B) }89 \qquad
+\textbf{(C) }90 \qquad
+\textbf{(D) }91 \qquad
+\textbf{(E) }92 \qquad
+</math>
+==Solution 1 (Less than 60 seconds)==
+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>.
+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:
+<cmath>3a+1 = 2b</cmath>
+<cmath>3b+1=2c</cmath>
+<cmath>3c+1=7a</cmath>
+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>.
+~lprado
+==Solution 2 (Solution 1 but longer)==
+We will define the equations the same as solution 1.
+<cmath>3a+1 = 2b</cmath>
+<cmath>3b+1=2c</cmath>
+<cmath>3c+1=7a</cmath>
+Solve equation 2 for c and substitute that value in for equation 3, giving us
+<cmath>3a+1 = 2b</cmath>
+<cmath>\frac{3b+1}{2}=c</cmath>
+<cmath>\frac{3*(3b+1)}{2}+1=7a</cmath>
+Multiply 14 to the first equation and rearrange to get
+<cmath>42a = 28b-14</cmath>
+and multiply the third by 2 and rearrange to get
+<cmath>27b+15 = 42a</cmath>
+Solve for b to get <imath>b = 29</imath>, substitute into equation 1 from the original to get <imath>a = 19</imath>, and lastly, substitute a into original equation 2 to get <imath>c = 44</imath>.
+Thus, <imath>a+b+c = 19+29+44 = \boxed{E(92)}</imath>.
+~Failure.net
+==Solution 3==
+We will define the equations the same as solution 1 and 2.
+<cmath>3a+1 = 2b</cmath>
+<cmath>3b+1=2c</cmath>
+<cmath>3c+1=7a</cmath>
+Multiply 1st equation by 1.5 to get
+<cmath>4.5a+1.5=3b</cmath>
+Add 1 to get
+<cmath>4.5a+2.5=3b+1</cmath>
+Substitute to get
+<cmath>4.5a+2.5=2c</cmath>
+Multiply this one by 1.5 as well, getting
+<cmath>6.75a+3.75=3c</cmath>
+Adding 1 again to get
+<cmath>6.75a+4.75=3c+1</cmath>
+Replace <cmath>3c+1</cmath> with <cmath>7a</cmath> getting <cmath>6.75a+4.75=7a</cmath>
+Solve for <cmath>a</cmath> getting <cmath>0.25a=4.75</cmath> <cmath>a=19</cmath>
+Solve for b: <cmath>3(19)+1=2b</cmath> <cmath>57+1=2b</cmath> <cmath>b=29</cmath>
+Solve for c: <cmath>3(29)+1=2c</cmath> <cmath>87+1=2c</cmath> <cmath>c=44</cmath>
+Thus, <math>a+b+c = 19+29+44 = \boxed{E(92)}</math>
+~newly2056
+==Solution 4 (No Fractions)==
+As stated in solutions 1, 2, 3, the equations are as follows:
+(1) <cmath>3a+1=2b</cmath>
+(2) <cmath>3b+1=2c</cmath>
+(3) <cmath>3c+1=7a</cmath>
+Multiply equation (2) by 2 to get equation (4):
+(4) <cmath>6b+2=4c</cmath>
+and substitute (1) into (4):
+(5) <cmath>9a+5=4c</cmath>
+(3) <cmath>7a-1=3c</cmath>
+Multiplying (5) by 3 and (3) by 4 gives:
+<cmath>3(9a+5)=4(7a-1)</cmath>
+from which we get <math>a=19</math>. Substituting <math>a</math> into (1) and (3) yields <math>b=29</math> and <math>c=44</math>. Therefore, <math>a+b+c = \boxed{E(92)}</math>
+~mathwizard123123
+==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://youtu.be/b-BBUKAVgeI?si=GAN4hho24927eJKX
+==See also==
+{{AMC10 box|year=2024|ab=B|num-b=24|after=Last Problem}}
+{{MAA Notice}}

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