The number of ordered pairs of positive integers $(a,b)$ such that LCM of a and b is $2^{3}5^{7}11^{13}$

Question

I started by taking two numbers such as $2^{2}5^{7}11^{13}$ and $2^{3}5^{7}11^{13}$. The LCM of those two numbers is $2^{3}5^{7}11^{13}$. Similarly, If I take two numbers like $2^{3-x}5^{7-y}11^{13-z}$ and $2^{3}5^{7}11^{13}$, the LCM is $2^{3}5^{7}11^{13}$. Here $0\le x\le3$, $0\le y\le7$, $0\le z\le13$

The number of ways of choosing 3 numbers $x,y,z$ is $^4C_1\cdot^8C_1\cdot^{14}C_1$

The above value is only for $b=2^{3}5^{7}11^{13}$.But now I have to consider another value of $b$ and the the solution becomes lengthy. Is this the correct approach?

Answer 1

Let $a = 2^{x_{1}}\cdot 5^{y_{1}}\cdot 11^{z_{1}}$ and $b = 2^{x_{2}}\cdot 5^{y_{2}}\cdot 11^{z_{2}}\;,$ Then Given $\bf{LCM(a,b)} = 2^{3}\cdot 5^{7}\cdot 11^{13}$

So Here $0\leq x_{1},x_{2}\leq 3$ and $0\leq y_{1},y_{2}\leq 7$ and $0\leq z_{1},z_{2}\leq 13.$

Here ordered pairs of $(x_{1},x_{2}) = \left\{(0,3),(1,3),(2,3),(3,3),((3,0),(3,1),(3,2)\right\}$

So Total $7$ ordered pairs

Similarly ordered pairs for $(y_{1},y_{2}),$ We get $15$ ordered pairs..

Similarly ordered pairs for $(z_{1},z_{2}),$ We get $27$ ordered pairs..

So Total ordered pairs of $(a,b) = 7\times 15 \times 27$

Answer 2

Hint:

• One of $a$ and $b$ must have 3 factors two, the other less.
• One of $a$ and $b$ must have 7 factors five, the other less.
• One of $a$ and $b$ must have 13 factors eleven, the other less.
• No other prime factors.