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 1== | + | == Solution == |
+ | |||
+ | ===Solution 1=== | ||
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=\boxed{15 \text{(A)}}</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=\boxed{15 \text{(A)}}</math>. | ||
− | ==Solution 2== | + | ===Solution 2=== |
It is well known that if the <math>\text{lcm}(a,b)=c</math> and <math>c</math> can be written as <math>p_1^ap_2^bp_3^c\dots</math>, then the highest power of all prime numbers <math>p_1,p_2,p_3\dots</math> must divide into either <math>a</math> and/or <math>b</math>. Or else a lower <math>c_0=p_1^{a-\epsilon}p_2^{b-\epsilon}p_3^{c-\epsilon}\dots</math> is the <math>\text{lcm}</math>. | It is well known that if the <math>\text{lcm}(a,b)=c</math> and <math>c</math> can be written as <math>p_1^ap_2^bp_3^c\dots</math>, then the highest power of all prime numbers <math>p_1,p_2,p_3\dots</math> must divide into either <math>a</math> and/or <math>b</math>. Or else a lower <math>c_0=p_1^{a-\epsilon}p_2^{b-\epsilon}p_3^{c-\epsilon}\dots</math> is the <math>\text{lcm}</math>. |
Revision as of 18:49, 7 February 2016
Problem
How many ordered triples of positive integers satisfy and ?
Solution
Solution 1
We prime factorize and . The prime factorizations are , and , respectively. Let , and . We know that 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 .
Solution 2
It is well known that if the and can be written as , then the highest power of all prime numbers must divide into either and/or . Or else a lower is the .
Start from : so or or both. But because and . So .
can be in both cases of but NOT because and .
So there are six sets of and we will list all possible values of based on those.
because must source all powers of . . because of restrictions.
By different sourcing of powers of and ,
is "enabled" by sourcing the power of . is uncovered by sourcing all powers of . And is uncovered by and both at full power capacity.
Counting the cases,
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 |
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