AIME 2019 I · 第 7 题

Solution 1

Add the two equations to get that $\log x+\log y+2(\log(\gcd(x,y))+2(\log(\text{lcm}(x,y)))=630$. Then, we use the theorem $\log a+\log b=\log ab$ to get the equation, $\log (xy)+2(\log(\gcd(x,y))+\log(\text{lcm}(x,y)))=630$. Using the theorem that $\gcd(x,y) \cdot \text{lcm}(x,y)=x\cdot y$, along with the previously mentioned theorem, we can get the equation $3\log(xy)=630$. This can easily be simplified to $\log(xy)=210$, or $xy = 10^{210}$.

$10^{210}$ can be factored into $2^{210} \cdot 5^{210}$, and $m+n$ equals to the sum of the exponents of $2$ and $5$, which is $210+210 = 420$. Multiply by two to get $2m +2n$, which is $840$. Then, use the first equation ($\log x + 2\log(\gcd(x,y)) = 60$) to show that $x$ has to have lower degrees of $2$ and $5$ than $y$ (you can also test when $x>y$, which is a contradiction to the restrains we set before). Therefore, $\gcd(x,y)=x$. Then, turn the equation into $3\log x = 60$, which yields $\log x = 20$, or $x = 10^{20}$. Factor this into $2^{20} \cdot 5^{20}$, and add the two 20's, resulting in $m$, which is $40$. Add $m$ to $2m + 2n$ (which is $840$) to get $40+840 = \boxed{880}$.

~minor mistake fix by virjoy2001 ~minor mistake fix by oralayhan

Remark: You can obtain the contradiction by using LTE. If $\nu_2{(x)}\geq{\nu_2{(y)}}, \nu_2{(y^2x)}=60$. However, $\nu_2{(xy)}=210$ a contradiction. Same goes with taking $\nu_5{(x,y)}$

Easier Approach to Finish

After noting that $xy=10^{210},$ notice that we can let $x=10^a$ and $y=10^b.$ Thus, we have from the given equations (1) and (2) respectively, that:

$$a+2a=60$$ $$b+2b=570$$ Solving, we get $(a,b)=(20,190).$ This matches with our constraint that $xy=10^{210}$ (this constraint can actually be rederived by adding the two equations) so we finish from here.

$x=2^{20}\cdot 5^{20}, y=2^{190}\cdot 5^{190}.$ Thus, the answer desired is $3(20+20)+2(190+190)=880.$ ~mathboy282 (minor addition by Technodoggo)

Why only 10?

If you didn't understand the main solution 1 or you read the easier finish, a question that might still linger is why can we let $x=10^a$ and $y=10^b$?

Notice how if we set $x = 2^a$ and $y = 5^b$, in the first equation we obtain $a\log(2) =60$. However, This implies that $a$ must be a non-integer logarithmic value, which is not allowed as that implies $x$ is not an integer. The same applies for $y$. Thus, we cannot have $x = 2^a$ and $y = 5^b$.

~Pinotation

Solution 2

First simplifying the first and second equations, we get that

$$\log_{10}(x\cdot\text{gcd}(x,y)^2)=60$$ $$\log_{10}(y\cdot\text{lcm}(x,y)^2)=570$$ Thus, when the two equations are added, we have that

$$\log_{10}(x\cdot y\cdot\text{gcd}^2\cdot\text{lcm}^2)=630$$ When simplified, this equals

$$\log_{10}(x^3y^3)=630$$ so this means that

$$x^3y^3=10^{630}$$ so

$$xy=10^{210}.$$ Now, the following cannot be done on a proof contest but let's (intuitively) assume that $x and$x$and$y$are both powers of$10$. This means the first equation would simplify to

$$x^3=10^{60}$$ and

$$y^3=10^{570}.$$ Therefore, $x=10^{20}$ and $y=10^{190}$ and if we plug these values back, it works! $10^{20}$ has $20\cdot2=40$ total factors and $10^{190}$ has $190\cdot2=380$ so

$$3\cdot 40 + 2\cdot 380 = \boxed{880}.$$ Please remember that you should only assume on these math contests because they are timed; this would technically not be a valid solution.

Proof for WLOG

The following assumption made on solution 2 can be done with a generality proof, and is quite simple.

If we have $x = 10^a$ and $y = 10^b$, we know that $xy = 10^{a+b}$. Then, because

$$xy$$ is commutative and so are the powers, it doesn't matter which one of $a$ and $b$ is greater. Thus, we may safely assume $a < b$ (or vice versa), and continue.

Optionally, in a proof contest, you could also show that both $a < b$ and $a > b$ work as valid cases, and provide the same answer.

~Pinotation

Solution 3 (Easy Solution)

Let $x=10^a$ and $y=10^b$ and $a. Then the given equations become$3a=60$and$3b=570$. Therefore,$x=10^{20}=2^{20}\cdot5^{20}$and$y=10^{190}=2^{190}\cdot5^{190}$. Our answer is$3(20+20)+2(190+190)=\boxed{880}$.

Note from Wiselion: This solution also uses the assumptions as solution 2. In theory x and y don/t have to be power of 10. Although having it not a power of 10 suggests that the sums won't be integers, if the gcd and lcm logs match up respectively with the remains of log x and y then the logs could add up to an integer. (We'd have to prove it) (PS never wrote the solution)

Solution 4

We will use the notation $(a, b)$ for $\gcd(a, b)$ and $[a, b]$ as $\text{lcm}(a, b)$. We can start with a similar way to Solution 1. We have, by logarithm properties, $\log_{10}{x}+\log_{10}{(x, y)^2}=60$ or $x(x, y)^2=10^{60}$. We can do something similar to the second equation and our two equations become

$$x(x, y)^2=10^{60}$$ $$y[x, y]^2=10^{570}$$ Adding the two equations gives us $xy(x, y)^2[x, y]^2=10^{630}$. Since we know that $(a, b)\cdot[a, b]=ab$, $x^3y^3=10^{630}$, or $xy=10^{210}$. We can express $x$ as $2^a5^b$ and $y$ as $2^c5^d$. Another way to express $(x, y)$ is now $2^{min(a, c)}5^{min(b, d)}$, and $[x, y]$ is now $2^{max(a, c)}5^{max(b, d)}$. We know that $x, and thus,$a, and $b. Our equations for$lcm$and$gcd$ now become

$$2^a5^b(2^a5^a)^2=10^{60}$$ or $a=b=20$. Doing the same for the $lcm$ equation, we have $c=d=190$, and $190+20=210$, which satisfies $xy=210$. Thus, $3m+2n=3(20+20)+2(190+190)=\boxed{880}$. ~awsomek

Solution 5

Let $x=d\alpha, y=d\beta, (\alpha, \beta)=1$. Simplifying, $d^3\alpha=10^{60}, d^3\alpha^2\beta^3=10^{510} \implies \alpha\beta^3 = 10^{510}=2^{510} \cdot 5^{510}$. Notice that since $\alpha, \beta$ are coprime, and $\alpha < 5^{90}$(Prove it yourself !) , $\alpha=1, \beta = 10^{170}$. Hence, $x=10^{20}, y=10^{190}$ giving the answer $\boxed{880}$.

(Solution by Prabh1512)

Solution 6 (Official MAA)

The two equations are equivalent to $x(\gcd(x,y))^2=10^{60}$ and $y(\operatorname{lcm}(x,y))^2=10^{570}$ respectively. Multiplying corresponding sides of the equations leads to $xy(\gcd(x,y)\operatorname{lcm}(x,y))^2=(xy)^3=10^{630}$, so $xy=10^{210}$. It follows that there are nonnegative integers $a,\,b,\,c,$ and $d$ such that $(x,y)=(2^a5^b,2^c5^d)$ with $a+c=b+d=210$. Furthermore,

$$\frac{(\operatorname{lcm}(x,y))^2}{x}=\frac{y(\operatorname{lcm}(x,y))^2}{xy}=\frac{10^{570}}{10^{210}}=10^{360}.$$ Thus $\max(2a,2c)-a=\max(2b,2d)-b=360.$ Because neither $2a-a$ nor $2b-b$ can equal $360$ when $a+c=b+d=210,$ it follows that $2c-a=2d-b=360$. Hence $(a,b,c,d)=(20,20,190,190)$, so the prime factorization of $x$ has $20+20=40$ prime factors, and the prime factorization of $y$ has $190+190=380$ prime factors. The requested sum is $3\cdot40+2\cdot380=880.$

Solution 7

Add the two equations and use the fact that $\gcd\left(x,y\right)\cdot\mathrm{lcm}\left(x,y\right)=xy$ to find that $xy=10^{210}$. So let $x=2^a5^b$ and $y=2^{210-a}5^{210-b}$ for $0\leq a,b\leq210$. If $a\geq105$ then the exponent of $2$ in $x\cdot\gcd\left(x,y\right)^2=10^{60}$ is $a+2\left(210-a\right)=420-a$, so $a=360$, contradiction. So $a<105$. Then the exponent of $2$ in $x\cdot\gcd\left(x,y\right)^2$ is $a+2a=3a$, so $a=20$. Similarly, $b=20$. Then $3m+2n=3\left(a+b\right)+2\left(420-a-b\right)=\boxed{880}$ as desired.

~from trumpeter in the AoPS Forums Contest Discussion

Solution 8

We can simplify the equations step by step. The first equation simplifies to $log($(x)($(gcd(x,y))^2$)$)=60$. The second equation simplifies to log($(y)$($(lcm(x,y)^2$)$)=570$. Up to here, we used the exponent and addition log identities.

Now before we move on to the next few simplification steps, we must remember that $gcd(a,b)$*$lcm(a,b)$=$a*b$.

Due to the fact that these log's are in base $10$, this makes the first and second equation equal to $10^{60}$, $10^{570}$ respectively. In this step, we switched the log's into exponential form. Now we multiply both equations to get $x*y$*$(xy)^2$=($x^3$)($y^3$)=$10^{630}$. Now we take the cube root of both sides to get $xy=10^{210}$.

We've now gotten to the crucial part of this equation. Though this wouldn't pass for full points in a proof-based contest, this is AIME. So, we assume that $x. We also let$x$=$10^a$and$y$=$10^b$That means that$gcd(x,y)$is$x$and the$lcm(x,y)$is$y$due to the fact that we are also assuming that both$x,y$are$10^a$,$10^b$ respectively.

If we put our last few insights together into the first and second equation, we see that $x$=$10^{20}$. We also see that $y$=$10^{190}$. We could check these if wanted (don't worry they work), but if you were very limited on time for this question, just assume these values work and move on.

Now $10^{20}$ factors as $2^{20}$*$5^{20}$. This has $40$ prime factors. $2$, $20$ times and $5$, $20$ times. $10^{190}$ factors as $2^{190}$*$5^{190}$. This has $380$ prime factors. $2$, $190$ times and $5$, $190$ times. Now it's just $40*3+380*2=880$ as our final answer.

-Schintalpati

Video Solution(Pretty Straightforward)

https://www.youtube.com/watch?v=NOLk9-A4eDo Remember to subscribe!

~North America math Contest Go Go Go