Difference between revisions of "2019 AIME I Problems/Problem 7"
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Let <math>x=10^a</math> and <math>y=10^b</math> and <math>a<b</math>. Then the given equations become <math>3a=60</math> and <math>3b=570</math>. Therefore, <math>x=10^{20}=2^{20}\cdot5^{20}</math> and <math>y=10^{190}=2^{190}\cdot5^{190}</math>. Our answer is <math>3(20+20)+2(190+190)=\boxed{880}</math>. | Let <math>x=10^a</math> and <math>y=10^b</math> and <math>a<b</math>. Then the given equations become <math>3a=60</math> and <math>3b=570</math>. Therefore, <math>x=10^{20}=2^{20}\cdot5^{20}</math> and <math>y=10^{190}=2^{190}\cdot5^{190}</math>. Our answer is <math>3(20+20)+2(190+190)=\boxed{880}</math>. | ||
==Solution 4 == | ==Solution 4 == | ||
− | We will use the notation <math>(a, b)</math> for <math>\gcd(a, b)</math> and <math>[a, b]</math> as <math>\text{lcm}(a, b)</math>. We can start a | + | We will use the notation <math>(a, b)</math> for <math>\gcd(a, b)</math> and <math>[a, b]</math> as <math>\text{lcm}(a, b)</math>. |
+ | We can start with a similar way to Solution 1. We have, by logarithm properties, <math>\log_{10}{x}+\log_{10}{(x, y)^2}=60</math> or <math>x(x, y)^2=10^{60}</math>. We can do something similar to the second equation and our two equations become <cmath>x(x, y)^2=10^{60}</cmath> <cmath>y[x, y]^2=10^{570}</cmath>. Adding the two equations gives us <math>xy(x, y)^2[x, y]^2=10^{630}</math>. Since we know that <math>(a, b)\cdot[a, b]=ab</math>, <math>x^3y^3=10^{630}</math>, or <math>xy=10^{210}</math>. We can express <math>x</math> as <math>2^a5^b</math> and <math>y</math> as <math>2^c5^d</math>. Another way to express <math>(x, y)</math> is now <math>2^{min(a, c)}5^{min(b, d)}</math>, and <math>[x, y]</math> is now <math>2^{max(a, c)}5^{max(b, d)}</math>. We know that <math>x<y</math>, and thus, <math>a<c</math>, and <math>b<d</math>. Our equations for <math>lcm</math> and <math>gcd</math> now become <cmath>2^a5^b(2^a5^a)^2=10^{60}</cmath> or <math>a=b=20</math>. Doing the same for the <math>lcm</math> equation, we have <math>c=d=190</math>, and <math>190+20=210</math>, which satisfies <math>xy=210</math>. Thus, <math>3m+2n=3(20+20)+2(190+190)=\boxed{880}</math>. | ||
~awsomek | ~awsomek | ||
Revision as of 19:50, 31 May 2020
Contents
Problem 7
There are positive integers and
that satisfy the system of equations
Let
be the number of (not necessarily distinct) prime factors in the prime factorization of
, and let
be the number of (not necessarily distinct) prime factors in the prime factorization of
. Find
.
Solution 1
Add the two equations to get that .
Then, we use the theorem
to get the equation,
.
Using the theorem that
, along with the previously mentioned theorem, we can get the equation
.
This can easily be simplified to
, or
.
can be factored into
, and
equals to the sum of the exponents of 2 and 5, which is
.
Multiply by two to get
, which is
.
Then, use the first equation (
) to show that x has to have lower degrees of 2 and 5 than y. Therefore, making the
. Then, turn the equation into
, which yields
, or
.
Factor this into
, and add the two 20's, resulting in m, which is 40.
Add
to
(which is 840) to get
.
Solution 2 (Crappier Solution)
First simplifying the first and second equations, we get that
Thus, when the two equations are added, we have that
When simplified, this equals
so this means that
so
Now, the following cannot be done on a proof contest but let's (intuitively) assume that and
and
are both powers of
. This means the first equation would simplify to
and
Therefore,
and
and if we plug these values back, it works!
has
total factors and
has
so
Please remember that you should only assume on these math contests because they are timed; this would technically not be a valid solution.
Solution 3 (Easy Solution)
Let and
and
. Then the given equations become
and
. Therefore,
and
. Our answer is
.
Solution 4
We will use the notation for
and
as
.
We can start with a similar way to Solution 1. We have, by logarithm properties,
or
. We can do something similar to the second equation and our two equations become
. Adding the two equations gives us
. Since we know that
,
, or
. We can express
as
and
as
. Another way to express
is now
, and
is now
. We know that
, and thus,
, and
. Our equations for
and
now become
or
. Doing the same for the
equation, we have
, and
, which satisfies
. Thus,
.
~awsomek
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.