Difference between revisions of "2024 AMC 10B Problems/Problem 25"
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~Failure.net | ~Failure.net | ||
− | ==Video Solution 1 by | + | ==Video Solution 1 by Chicken_Rice (In Less Than 5 Mins ⚡🚀)== |
− | https://youtu.be/ | + | https://youtu.be/kQGbyhze_mY?si=b2RhYMWt_RS8DnOa |
− | ~ | + | ~ Chicken Rice subcriber :) |
==Video Solution 2 by SpreadTheMathLove== | ==Video Solution 2 by SpreadTheMathLove== |
Revision as of 19:08, 20 November 2024
Contents
[hide]Problem
Each of bricks (right rectangular prisms) has dimensions , where , , and are pairwise relatively prime positive integers. These bricks are arranged to form a block, as shown on the left below. A th brick with the same dimensions is introduced, and these bricks are reconfigured into a block, shown on the right. The new block is unit taller, unit wider, and unit deeper than the old one. What is ?
Solution 1 (Less than 60 seconds)
The xx block has side lengths of . The xx block has side lengths of .
We can create the following system of equations, knowing that the new block has unit taller, deeper, and wider than the original:
Adding all the equations together, we get . Adding to both sides, we get . The question states that are all relatively prime positive integers. Therefore, our answer must be congruent to . The only answer choice satisfying this is . ~lprado
Solution 2
We will define the equations the same as solution 1. Solve equation 2 for c and substitute that value in for equation 3, giving us
Multiply 14 to the first equation and rearrange to get and multiply the third by 2 and rearrange to get Solve for b to get , substitute into equation 1 from the original to get , and lastly, substitute a into original equation 2 to get . Thus, . ~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
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.