Problem on permutation and combination

Question

What are the number of ordered pairs of positive integers $(a,b)$ such that the $\operatorname{lcm}(a,b) = 2^3 \cdot 5^7 \cdot 11^{13}$.

Well, i tried to use the formula of exponent of a prime in a number, but I am unable to find out the number.

Please help.

Answer 1

Hint 1: show that $a$ can be written as $2^{x_2} 5^{x_5} 11^{x_{11}}$ for some nonnegative integers $x_2, x_5, x_{11}$, and similarly $b$ can be written as $2^{y_2} 5^{y_5} 11^{y_{11}}$ for some nonnegative integers $y_2, y_5, y_{11}$.

Hint 2: show that the LCM of $a$ and $b$ is $2^{\max(x_2, y_2)} 5^{\max(x_5, y_5)} 11^{\max(x_{11}, y_{11})}$, so you just need to count how many ways you can get $\max(x_2,y_2)=3$, $\max(x_5,y_5)=7$, and $\max(x_{11}, y_{11})=13$.

Answer 2

Let a= (2^x)•(5^y)•(11^z) and b=(2^s)•(5^t)•(11^u)

So, max(x,s)=3; max(y,t)=7; max(z,u)=13. Well, I found out:

1) max(x,s)=3 has 7 possibilities(3 alternate pairs and 1 same in both cases).-{(0,3),(1,3),(2,3),(3,2),(3,1),(3,0) and (3,3)×2times}

2)Similarly, max(y,t)=7 has 15 possiblities(7 alternate pairs and 1 same in both cases).

3)And, max(z,u)=13 has 27 possibilities(13 alt. + 1 same).

This gives 7×15×27 = 2835.