Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
 
(13 intermediate revisions by 6 users not shown)
Line 12: Line 12:
 
</math>
 
</math>
  
−
==Solution 1==
+
==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>.
 
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 20: Line 20:
 
<cmath>3c+1=7a</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{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
 
~lprado
  
Line 38: Line 38:
 
<cmath>27b+15 = 42a</cmath>
 
<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>.
 
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 = 92</math>.
+
Thus, <math>a+b+c = 19+29+44 = \boxed{E(92)}</math>.
 
~Failure.net
 
~Failure.net
  
−
==Video Solution 1 by Pi Academy (In Less Than 3 Mins ⚡🚀)==
+
==Video Solution 1 by Pi Academy (In Less Than 5 Mins ⚡🚀)==
  
−
https://youtu.be/Xn1JLzT7mW4?feature=shared
+
https://youtu.be/Xn1JLzT7mW4?si=borSg8xrYLAz7mNY
  
−
~ Pi Academy
+
~ jj_empire10
  
 
==Video Solution 2 by SpreadTheMathLove==
 
==Video Solution 2 by SpreadTheMathLove==

Latest revision as of 18:11, 23 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 $a + b + c$?

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 (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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_10B_Problems/Problem_25&oldid=235370"