Divisional MO 2015

[Jarin Tasnim]

How many pairs of natural number can be formed whose LCM will be 9800?

[Kazi_Zareer]

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.

[Jarin Tasnim]

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...

[dshasan]

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.

You are overcounting ( by a lot)

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.

[Kazi_Zareer]

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.

Sorry. I did some miscalculates. I didn't even notice it! What the ****

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.