Difference between revisions of "2016 AMC 10A Problems/Problem 25"
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===Solution 1=== | ===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{ | + | 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{\textbf{(A)15}}</math>. |
===Solution 2=== | ===Solution 2=== |
Revision as of 20:24, 16 December 2017
Contents
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,
Solution 3 (Less Casework!)
As said in previous solutions, start by factoring and
. The prime factorizations are as follows:
What a great casework bash
To organize
and their respective LCMs in a simpler way, we can draw a triangle as follows such that
are the vertices and the LCMs are on the edges.
Uh oh, where da diagram?
Now we can split this triangle into three separate ones for each of the three different prime factors
.
Uh oh, where da diagram?
Analyzing for powers of
, is is quite obvious that
must have
as one of it's factors, since neither
can have a power of
exceeding
. Turning towards the vertices
, we know at least one of them must have
as its factors. Therefore, we have
ways for the powers of
for
since the only ones that satisfy the previous conditions are for ordered pairs
.
Powers of
.
Using the same logic as we did for powers of
, is becomes quite easy to note that
must have
as one of it's factors. Moving onto
, we can use the same logic to find the only ordered pairs
that will work are
.
The final and last case is the powers of
.
This is actually quite a simple case, since we know
must have
as part of its factorization while
cannot have a factor of
in their prime factorization.
Multiplying all the possible arrangements for prime factors , we get the answer:
.
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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