Difference between revisions of "2016 AMC 10A Problems/Problem 25"
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So <math>x=8,24</math>. | So <math>x=8,24</math>. | ||
− | <math>y</math> can be <math>9,18,36</math> in both cases of <math>x</math> but NOT <math>72</math> because <math>\lcm{y,z}=900</math> and <math>72\nmid 900</math>. | + | <math>y</math> can be <math>9,18,36</math> in both cases of <math>x</math> but NOT <math>72</math> because <math>\text{lcm}{y,z}=900</math> and <math>72\nmid 900</math>. |
So there are six sets of <math>x,y</math> and we will list all possible values of <math>z</math> based on those. | So there are six sets of <math>x,y</math> and we will list all possible values of <math>z</math> based on those. | ||
− | <math>25\mid z</math> because <math>z</math> must source all powers of <math>5</math>. <math>z\in\{25,50,75,100,150,300\}</math>. <math>z\ | + | <math>25\mid z</math> because <math>z</math> must source all powers of <math>5</math>. <math>z\in\{25,50,75,100,150,300\}</math>. <math>z\ne\{200,225\}</math> because of <math>\text{lcm}</math> restrictions. |
By different sourcing of powers of <math>2</math> and <math>3</math>, | By different sourcing of powers of <math>2</math> and <math>3</math>, |
Revision as of 22:22, 5 February 2016
Contents
Problem
How many ordered triples of positive integers satisfy and ?
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 $\text{lcm}(x,z}=600$ (Error compiling LaTeX. ! Extra }, or forgotten $.) 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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