Difference between revisions of "2016 AMC 10A Problems/Problem 25"
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<math>\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64</math> | <math>\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64</math> | ||
| − | + | ==Solution== | |
We prime factorize <math>72,600,</math> and <math>900</math>. The prime factorizations are <math>2^3\times 3^2</math>, <math>2^3\times 3\times 5^2</math> and <math>2^2\times 3^2\times 5^2</math>, respectively. Let <math>x=2^a\times 3^b\times 5^c</math>, <math>y=2^d\times 3^e\times 5^f</math> and <math>z=2^g\times 3^h\times 5^i</math>. We know that <cmath>\max(a,d)=3</cmath> <cmath>\max(b,e)=2</cmath> <cmath>\max(a,g)=3</cmath> <cmath>\max(b,h)=1</cmath> <cmath>\max(c,i)=2</cmath> <cmath>\max(d,g)=2</cmath> <cmath>\max(e,h)=2</cmath> and <math>c=f=0</math> since <math>\text{lcm}(x,y)</math> isn't a multiple of 5. Since <math>\max(d,g)=2</math> we know that <math>a=3</math>. We also know that since <math>\max(b,h)=1</math> that <math>e=2</math>. So now some equations have become useless to us...let's take them out. <cmath>\max(b,h)=1</cmath> <cmath>\max(d,g)=2</cmath> are the only two important ones left. We do casework on each now. If <math>\max(b,h)=1</math> then <math>(b,h)=(1,0),(0,1)</math> or <math>(1,1)</math>. Similarly if <math>\max(d,g)=2</math> then <math>(d,g)=(2,0),(2,1),(2,2),(1,2),(0,2)</math>. Thus our answer is <math>5\times 3=15</math>. | We prime factorize <math>72,600,</math> and <math>900</math>. The prime factorizations are <math>2^3\times 3^2</math>, <math>2^3\times 3\times 5^2</math> and <math>2^2\times 3^2\times 5^2</math>, respectively. Let <math>x=2^a\times 3^b\times 5^c</math>, <math>y=2^d\times 3^e\times 5^f</math> and <math>z=2^g\times 3^h\times 5^i</math>. We know that <cmath>\max(a,d)=3</cmath> <cmath>\max(b,e)=2</cmath> <cmath>\max(a,g)=3</cmath> <cmath>\max(b,h)=1</cmath> <cmath>\max(c,i)=2</cmath> <cmath>\max(d,g)=2</cmath> <cmath>\max(e,h)=2</cmath> and <math>c=f=0</math> since <math>\text{lcm}(x,y)</math> isn't a multiple of 5. Since <math>\max(d,g)=2</math> we know that <math>a=3</math>. We also know that since <math>\max(b,h)=1</math> that <math>e=2</math>. So now some equations have become useless to us...let's take them out. <cmath>\max(b,h)=1</cmath> <cmath>\max(d,g)=2</cmath> are the only two important ones left. We do casework on each now. If <math>\max(b,h)=1</math> then <math>(b,h)=(1,0),(0,1)</math> or <math>(1,1)</math>. Similarly if <math>\max(d,g)=2</math> then <math>(d,g)=(2,0),(2,1),(2,2),(1,2),(0,2)</math>. Thus our answer is <math>5\times 3=15</math>. | ||
Revision as of 20:11, 3 February 2016
Problem
How many ordered triples 
of positive integers satisfy 
and 
?
Solution
We prime factorize 
and 
. The prime factorizations are 
, 
and 
, respectively. Let 
, 
and 
. We know that ![\[\max(a,d)=3\]](http://latex.artofproblemsolving.com/5/4/1/54144303e6de4ae84935ad1df9a2e569c7041808.png)
![\[\max(b,e)=2\]](http://latex.artofproblemsolving.com/b/7/d/b7de5a5aefbaccf608fb441e29df3ac33370880c.png)
![\[\max(a,g)=3\]](http://latex.artofproblemsolving.com/8/4/f/84f6de12d6a6405690950d990a3222e90f0b561b.png)
![\[\max(b,h)=1\]](http://latex.artofproblemsolving.com/a/8/3/a83271852abe6343879fad35f475114d4bdbb656.png)
![\[\max(c,i)=2\]](http://latex.artofproblemsolving.com/4/0/0/4006861c44f3e57be3b6e3c63512dd130924b51a.png)
![\[\max(d,g)=2\]](http://latex.artofproblemsolving.com/9/4/7/947289a810fdb4c7fec0f66f8d18c74ef47cd61c.png)
![\[\max(e,h)=2\]](http://latex.artofproblemsolving.com/9/e/e/9eef758965cfbad23241d45c675de9a45f148d0e.png)
and 
since 
isn't a multiple of 5. Since 
we know that 
. We also know that since 
that 
. So now some equations have become useless to us...let's take them out. are the only two important ones left. We do casework on each now. If then 
or 
. Similarly if then 
. Thus our answer is 
.
See Also
| 2016 AMC 10A (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 | ||
| 2016 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 21 |
Followed by Problem 23 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
