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

Latest revision as of 19:08, 20 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 Chicken_Rice (In Less Than 5 Mins ⚡🚀)

https://youtu.be/kQGbyhze_mY?si=b2RhYMWt_RS8DnOa

~ Chicken Rice subcriber :)

Video Solution 2 by SpreadTheMathLove

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

@@ Line 1: / Line 1: @@
 ==Problem==
-You made it! If you're good with asymptote, please insert an image along with the rest of the 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>?
-==Solution 1==
+[[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>.
@@ Line 10: / Line 20: @@
 <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{92}</math>.
+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==
+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 <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>.
+Thus, <math>a+b+c = 19+29+44 = \boxed{E(92)}</math>.
+~Failure.net
+==Video Solution 1 by Chicken_Rice (In Less Than 5 Mins ⚡🚀)==
+https://youtu.be/kQGbyhze_mY?si=b2RhYMWt_RS8DnOa
+~ Chicken Rice subcriber :)
+==Video Solution 2 by SpreadTheMathLove==
+https://www.youtube.com/watch?v=24EZaeAThuE
 ==See also==
-{{AMC10 box|year=2024|ab=B|num-b=20|after=Last Problem}}
+{{AMC10 box|year=2024|ab=B|num-b=24|after=Last Problem}}
 {{MAA Notice}}

2024 AMC 10B (Problems • Answer Key • Resources)
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