System of least common multiples

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va2010

1276 posts

va2010

#1 h Feb 3, 2016, 3:24 PM • 2 Y

Y by Adventure10, Mango247

How many ordered triples $(x, y, z)$ of positive integers satisfy $\text{lcm}(x, y) = 72$ , $\text{lcm}(x, z)= 600$ , and $\text{lcm}(y, z) = 900$ ?

$\textbf{(A) } 15 \qquad\textbf{(B) } 16 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 27 \qquad\textbf{(E) } 64$

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Royalreter1

1913 posts

Royalreter1

#2 h Feb 3, 2016, 3:25 PM • 2 Y

Y by Adventure10, Mango247

Also 2016 AMC 10A #25

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dongrae

14 posts

dongrae

#3 h Feb 3, 2016, 4:01 PM • 1 Y

Y by Adventure10

I think #22 is (A) 15

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DrMath

2130 posts

DrMath

#4 h Feb 3, 2016, 4:05 PM • 9 Y

Y by Not_a_Username, bestwillcui1, AstrapiGnosis, ThisIsASentence, W.Sun, DarkRunner, Bradygho, Adventure10, Mango247

Really badly misplaced for its difficulty.

Note $8\mid x$ and there are 5 ways to choose the powers of 2 that divide $y,z$ such that their maximum is 2. The answer must be divisible by 5 which automatically gives A.

If you want to check, there are 3 ways to choose the power of 3 and 1 way to choose the power of 5, confirming the answer is $5\cdot 3\cdot 1=15$ (A)

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eyzhang

150 posts

eyzhang

#5 h Feb 3, 2016, 4:09 PM • 2 Y

Y by Adventure10, Mango247

Really similar to 1987 AIME number 7!
I just recently did that prob^, so I skipped straight to this problem when I saw it

And yeah, I agree, badly misplaced for it's difficulty.

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Wave-Particle

3690 posts

Wave-Particle

#6 h Feb 3, 2016, 4:42 PM • 2 Y

Y by Not_a_Username, Adventure10

Just set up a bunch of max equations and you get 3*5=15

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bestwillcui1

2735 posts

bestwillcui1

#7 h Feb 3, 2016, 4:49 PM • 3 Y

Y by ACthinker, Adventure10, Mango247

yeah how is this a #22 or a #25....took 30 seconds.

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nosaj

2008 posts

nosaj

#8 h Feb 3, 2016, 5:01 PM • 1 Y

Y by Adventure10

After spending far too much time on #25 on the 2015 AMC 10B, I decided not to look at Problem 25 at all this year. Rip

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Mudkipswims42

8867 posts

Mudkipswims42

#9 h Feb 3, 2016, 6:32 PM • 1 Y

Y by Adventure10

Wait I am confused on how they got 5,3, and 1.... I got 27 by doing $2*3*6*\dfrac{3}{4}$ ,..

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DeathLlama9

1755 posts

DeathLlama9

#10 h Feb 3, 2016, 6:40 PM • 2 Y

Y by Adventure10, Mango247

BTW to make life easier just make $w$ equivalent to $\frac{z}{25}$ and you can rewrite stuff while ensuring everything's distinct

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acegikmoqsuwy2000

767 posts

acegikmoqsuwy2000

#11 h Feb 3, 2016, 6:46 PM • 1 Y

Y by Adventure10

DeathLlama9 wrote:

BTW to make life easier just make $w$ equivalent to $\frac{z}{25}$ and you can rewrite stuff while ensuring everything's distinct

yeah i did that then did casework on the powers of two and got 5 possibilities, and then noting that 5|15 and none of the other options we get A

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dhwang314

1046 posts

dhwang314

#12 h Feb 3, 2016, 6:48 PM • 1 Y

Y by Adventure10

This is the problem I could have gotten right, but the problem placement confused me
EDIT:
IncompleteSolution

This post has been edited 2 times. Last edited by dhwang314, Feb 3, 2016, 6:55 PM
Reason: edit

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smo

179 posts

smo

#14 h Feb 5, 2017, 5:38 AM • 1 Y

Y by Adventure10

DrMath wrote:

there are 5 ways to choose the powers of 2 that divide $y,z$ such that their maximum is 2.

I got 9 for this, could someone explain why it's 5?

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Wave-Particle

3690 posts

Wave-Particle

#15 h Feb 5, 2017, 6:51 AM • 2 Y

Y by Adventure10, Mango247

smo wrote:

DrMath wrote:

there are 5 ways to choose the powers of 2 that divide $y,z$ such that their maximum is 2.

I got 9 for this, could someone explain why it's 5?

$(2,2), (2,1), (1,2), (0,2), (2,0)$ . Have any others to add?

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smo

179 posts

smo

#16 h Feb 5, 2017, 7:10 AM • 2 Y

Y by Adventure10, Mango247

Ohh, my mistake was in thinking that both of them could be zero, so I had extras which wouldn't work. Thanks.

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618173

1751 posts

618173

#17 h Nov 6, 2020, 5:42 AM • 1 Y

Y by Mango247

I just solved an AMC 10 #25 I'm so cool lol

Solution

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samrocksnature

8791 posts

samrocksnature

#19 h Nov 6, 2020, 6:03 AM

Y by

DrMath wrote:

Really badly misplaced for its difficulty.

Note $8\mid x$ and there are 5 ways to choose the powers of 2 that divide $y,z$ such that their maximum is 2. The answer must be divisible by 5 which automatically gives A.

If you want to check, there are 3 ways to choose the power of 3 and 1 way to choose the power of 5, confirming the answer is $5\cdot 3\cdot 1=15$ (A)

I'm not quite sure what this aopser means by "5 ways to choose the powers of 2 that divide $y,z$ such that their maximum is 2" and why the answer is divisible by $5$

EDIT: nvm I get the divisible by $5$ part, but not the other

This post has been edited 1 time. Last edited by samrocksnature, Nov 6, 2020, 6:04 AM

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Pleaseletmewin

1574 posts

Pleaseletmewin

#20 h Nov 6, 2020, 6:04 AM

Y by

Alright, I'll type this up again because why not.
We have that $\text{lcm}(x,y)=2^3\cdot 3^2, \text{lcm}(x,z)=2^3\cdot 3\cdot 5^2,$ and $\text{lcm}(y,z)=2^2\cdot 3^2\cdot 5^2$ .
Focus on the powers of two first. $\max\{x,y\}=2^3, \max\{x,z\}=2^3$ , and $\max\{y,z\}=2^2$ . It's easy to see that $x=2^3$ . If not, then, one of $y$ and $z$ must be $2^3$ , which is a contradiction based on $\max\{y,z\}=2^2$ . Now, we just have to choose what $y$ and $z$ are. One of them must be $2^2$ and the other can be anything lower or equal to this. Simple counting shows that there are $5$ ways for this one.
Moving to the power of $3$ , we see that $\max\{x,y\}=3^2, \max\{x,z\}=3$ , and $\max\{y,z\}=2^2$ . By the same logic, we see that $y=2^2$ . It's also not hard to see that there are $3$ options for $x$ and $z$
Finally, for $5$ we see that $x,y$ have no power of $5$ and $z$ must be $5^2$ so there is only $1$ option for this case.
Hence, our answer is $3\cdot5=15$ as desired.

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Pleaseletmewin

1574 posts

Pleaseletmewin

#21 h Nov 6, 2020, 6:05 AM

Y by

samrocksnature wrote:

DrMath wrote:

Really badly misplaced for its difficulty.

Note $8\mid x$ and there are 5 ways to choose the powers of 2 that divide $y,z$ such that their maximum is 2. The answer must be divisible by 5 which automatically gives A.

If you want to check, there are 3 ways to choose the power of 3 and 1 way to choose the power of 5, confirming the answer is $5\cdot 3\cdot 1=15$ (A)

I'm not quite sure what this aopser means by "5 ways to choose the powers of 2 that divide $y,z$ such that their maximum is 2" and why the answer is divisible by $5$

What do you not understand?
You know that the maximum power of $2$ of $y$ and $z$ must be $2$ .
Just brute force it.
$(y,z)=(2^2,2^2), (2^2, 2^1), (2^2,2^0), (2^0,2^2),(2^1,2^2)$ .
Done.
@2above, I'm pretty sure if you understand why there are 5 ways to choose this, then you understand what DrMath is saying.

This post has been edited 1 time. Last edited by Pleaseletmewin, Nov 6, 2020, 6:07 AM

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samrocksnature

8791 posts

samrocksnature

#22 h Nov 6, 2020, 6:09 AM • 1 Y

Y by Mango247

Oh I see, but is there an alternative to brute forcing it?

EDIT: $2$ ways to choose the one that has $2^2$ and $0, 1, 2$ choices for the other? Doesn't that get $6$ ways?

This post has been edited 1 time. Last edited by samrocksnature, Nov 6, 2020, 6:10 AM

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Pleaseletmewin

1574 posts

Pleaseletmewin

#23 h Nov 6, 2020, 6:10 AM • 1 Y

Y by Mango247

samrocksnature wrote:

Oh I see, but is there an alternative to brute forcing it?

My dude, there are literally 5 choices.
I mean, I can do this constructively for bigger numbers but my point was, it's not hard to see what the solution are.

@above, you overcounted when both of them are $2^2$ .

This post has been edited 2 times. Last edited by Pleaseletmewin, Nov 6, 2020, 6:11 AM

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samrocksnature

8791 posts

samrocksnature

#24 h Nov 6, 2020, 6:12 AM

Y by

Thx so much I need to study some more

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SenorIncongito

307 posts

SenorIncongito

#25 h Nov 6, 2020, 6:12 AM

Y by

samrocksnature wrote:

Oh I see, but is there an alternative to brute forcing it?

EDIT: $2$ ways to choose the one that has $2^2$ and $0, 1, 2$ choices for the other? Doesn't that get $6$ ways?

Sometimes brute force is the only option you have... There isn't always a "clever" solution.

This post has been edited 1 time. Last edited by SenorIncongito, Nov 6, 2020, 6:21 AM

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samrocksnature

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samrocksnature

#26 h Nov 6, 2020, 6:13 AM

Y by

~~As I can tell by your pfp~~

Thanks for the tip though! I'm pretty new to all this mathing

@below thx

This post has been edited 2 times. Last edited by samrocksnature, Nov 6, 2020, 6:20 AM

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Pleaseletmewin

1574 posts

Pleaseletmewin

#27 h Nov 6, 2020, 6:16 AM

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samrocksnature wrote:

~~As I can tell by your pfp~~

Thanks for the tip though! I'm pretty new to all this mathing

I mean, if you really want me to, here we go.
Choose two nonnegative integers $x$ and $y$ such that at least one of them is $n$ and the other is less than or equal to $n$ .
First, discard the case when both are $n$ . Then, simply, we have $2(n-1)+1=2n-1$ ways to do this.

There you go. No brute force. Satisfied?
The reason I just listed out the solutions for $y$ and $z$ was because $3$ was such a small number, it wouldn't do any harm.

This post has been edited 4 times. Last edited by Pleaseletmewin, Nov 6, 2020, 6:21 AM
Reason: i can't grammar

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