Difference between revisions of "2024 AMC 10B Problems/Problem 25"
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| + | ==Problem== | ||
| + | You made it! If you're good with asymptote, please insert an image along with the rest of the problem! | ||
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==Solution 1== | ==Solution 1== | ||
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>. | ||
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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{92}</math>. | ||
~lprado | ~lprado | ||
| + | |||
| + | ==See also== | ||
| + | {{AMC10 box|year=2024|ab=B|num-b=20|after=Last Problem}} | ||
| + | {{MAA Notice}} | ||
Revision as of 01:35, 14 November 2024
Problem
You made it! If you're good with asymptote, please insert an image along with the rest of the problem!
Solution 1
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:
![\[3a+1 = 2b\]](http://latex.artofproblemsolving.com/3/1/3/313d38b3f4e9711c2c6d9176d8b2f695b126ca2a.png)
![\[3b+1=2c\]](http://latex.artofproblemsolving.com/f/0/6/f064fed198a80e2d4df1412c2d682645ca4c5d37.png)
![\[3c+1=7a\]](http://latex.artofproblemsolving.com/c/8/2/c82d94c0eb58f9202d41b78c5a585a2e52b3abc5.png)
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
See also
| 2024 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
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. 
