Difference between revisions of "2024 AMC 10B Problems/Problem 25"
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Solve for c: <cmath>3(29)+1=2c</cmath> <cmath>87+1=2c</cmath> <cmath>c=44</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> | + | Thus, <math>a+b+c = 19+29+44 = \boxed{E(92)}</math> |
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~newly2056 | ~newly2056 | ||
Latest revision as of 22:54, 12 August 2025
Contents
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 x
x
block has side lengths of
. The
x
x
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
Solution 3
We will define the equations the same as solution 1 and 2.
Multiply 1st equation by 1.5 to get
Add 1 to get
Substitute to get
Multiply this one by 1.5 as well, getting
Adding 1 again to get
Replace
with
getting
Solve for
getting
Solve for b:
Solve for c:
Thus,
~newly2056
Solution 4 (No Fractions)
As stated in solutions 1, 2, 3, the equations are as follows:
(1)
(2)
(3)
Multiply equation (2) by 2 to get equation (4):
(4)
and substitute (1) into (4):
(5)
(3)
Multiplying (5) by 3 and (3) by 4 gives:
from which we get
. Substituting
into (1) and (3) yields
and
. Therefore,
~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
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.