Divisional MO 2015
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Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
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How many pairs of natural number can be formed whose LCM will be 9800?
- Kazi_Zareer
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Re: Divisional MO 2015
See, $9800 = 2^3 \times 5^2 \times 7^2$
Let the two numbers will be $p = 2^a \times 5^b \times 7^c$ and $q = 2^x \times 5^y \times 7^z$.
Now consider the pairs of $(a,x)$ which are possible: $(0,3), (1,3),(2,3),(3,3), (3,0), (3,1), (3,2),(1,2),(2,1)$ , numbers of pairs is $9$.
Again, similarly the pairs of $(b,y)$ possible is: $(0,2).(1,2),(2,2),(2,1),(1,1)$ which is $5$ pairs
and the last pairs of $(c,z)$ possible is same as the numbers of pairs of $(b,y)$ which is also $5$.
So, the total number of pairs of natural number is $: 9 \times 5 \times 5 = 225$
I hope it's correct.
Let the two numbers will be $p = 2^a \times 5^b \times 7^c$ and $q = 2^x \times 5^y \times 7^z$.
Now consider the pairs of $(a,x)$ which are possible: $(0,3), (1,3),(2,3),(3,3), (3,0), (3,1), (3,2),(1,2),(2,1)$ , numbers of pairs is $9$.
Again, similarly the pairs of $(b,y)$ possible is: $(0,2).(1,2),(2,2),(2,1),(1,1)$ which is $5$ pairs
and the last pairs of $(c,z)$ possible is same as the numbers of pairs of $(b,y)$ which is also $5$.
So, the total number of pairs of natural number is $: 9 \times 5 \times 5 = 225$
I hope it's correct.
Last edited by Kazi_Zareer on Tue Jan 24, 2017 8:00 pm, edited 1 time in total.
We cannot solve our problems with the same thinking we used when we create them.
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Re: Divisional MO 2015
but if we try this for LCM of 6 it says that there are 9 pairs...but actually there are 5 pairs whose LCM is 6...
Re: Divisional MO 2015
You are overcounting ( by a lot)Kazi_Zareer wrote:See, $9800 = 2^3 \times 5^2 \times 7^2$
Let the two numbers will be $p = 2^a \times 5^b \times 7^c$ and $q = 2^x \times 5^y \times 7^z$.
Now consider the pairs of $(a,x)$ which are possible: $(0,3), (1,3),(2,3),(3,3), (3,0), (3,1), (3,2),(1,2),(2,1)$ , numbers of pairs is $9$.
Again, similarly the pairs of $(b,y)$ possible is: $(0,2).(1,2),(2,2),(2,1),(1,1)$ which is $5$ pairs
and the last pairs of $(c,z)$ possible is same as the numbers of pairs of $(b,y)$ which is also $5$.
So, the total number of pairs of natural number is $: 9 \times 5 \times 5 = 225$
I hope it's correct.
See that $(p,q)$ are same when $p = 2^1 \times 5^1 \times 7^1$, $q = 2^3 \times 5^2 \times 7^2$ and
$p = 2^3 \times 5^2 \times 7^2$, $q = 2^1 \times 5^1 \times 7^1$.
I think the answer is $225 - 64 - 8 - 16 - 16 + 1 = 122$. I got this by substracting the over-counted pairs, but I am not sure whether it is the correct answer or not. Someone please clarify.
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- Charles Caleb Colton
- Kazi_Zareer
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- Joined:Thu Aug 20, 2015 7:11 pm
- Location:Malibagh,Dhaka-1217
Re: Divisional MO 2015
Sorry. I did some miscalculates. I didn't even notice it! What the ****Kazi_Zareer wrote:See, $9800 = 2^3 \times 5^2 \times 7^2$
Let the two numbers will be $p = 2^a \times 5^b \times 7^c$ and $q = 2^x \times 5^y \times 7^z$.
Now consider the pairs of $(a,x)$ which are possible: $(0,3), (1,3),(2,3),(3,3), (3,0), (3,1), (3,2),(1,2),(2,1)$ , numbers of pairs is $9$.
Again, similarly the pairs of $(b,y)$ possible is: $(0,2).(1,2),(2,2),(2,1),(1,1)$ which is $5$ pairs
and the last pairs of $(c,z)$ possible is same as the numbers of pairs of $(b,y)$ which is also $5$.
So, the total number of pairs of natural number is $: 9 \times 5 \times 5 = 225$
I hope it's correct.
Actually the pairs of $(a,x)$ which are possible: $(3,0), (3,1),(3,2),(3,3), (0,3), (1,3), (2,3)$, so $7$ in number.
Also, the pairs of $(b,y)$ possible is: $(0,2).(1,2),(2,2),(2,1)$, it's $4.$ the last pairs of $(c,z)$, is also $4$.
$7 \times 4 \times 4 = 112$. Now, it's okay.
We cannot solve our problems with the same thinking we used when we create them.