Difference between revisions of "2024 AMC 10B Problems/Problem 25"
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| + | ==Solution 1== | ||
| + | The <math>3x3x3</math> block has side lengths of <math>3a, 3b, 3c</math>. The <math>2x2x7</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 1 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 = 4a-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 | ||
Revision as of 00:26, 14 November 2024
Solution 1
The 
block has side lengths of 
. The 
block has side lengths of 
.
We can create the following system of equations, knowing that the new block has 1 unit taller, deeper, and wider than the original:
![\[3a+1 = 2b\]](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
