Difference between revisions of "2024 AMC 10B Problems/Problem 25"
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+ | ==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== | ||
+ | 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 | ||
+ | |||
+ | ==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 | ||
+ | |||
+ | ==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://www.youtube.com/watch?v=24EZaeAThuE | ||
+ | |||
+ | ==See also== | ||
+ | {{AMC10 box|year=2024|ab=B|num-b=24|after=Last Problem}} | ||
+ | {{MAA Notice}} |
Latest revision as of 03:55, 20 February 2025
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 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
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://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.