Animated Solution for Mathematics - Permutations and Combinations: If the LCM of $p,q$ is $r^2{s}^4{t}^2$, where $r,s,t$ are prime numbers and $p,q$ are the positive integers then the number of ordered pair $(p,q)$ is

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are peeling back the curtain on the fundamental structure of integers.

When we look at the Least Common Multiple (LCM), we often see it as a simple tool for adding fractions. But in the realm of JEE Advanced, the LCM is a gateway to understanding the combinatorial DNA of numbers. Let us embark on this journey.

The Prime Factorization

Imagine any positive integer $p$ as a container. Inside this container, we have a collection of prime building blocks.

We are told that $LCM(p,q)=r^2{s}^4{t}^2$, where $r,s,$ and $t$ are prime numbers. This is our universe.

If $p$ or $q$ contained any prime factor other than $r,s,$ or $t$, that prime would inevitably appear in the LCM. Since the LCM is restricted to these three primes, we can confidently state that $p$ and $q$ are composed solely of these primes.

We can express them as:

Here, the exponents $a_i,b_i,c_i$ are non-negative integers. This is the first step of our master plan: reducing the complexity of the integers into the simplicity of their exponents.

The Max Rule

How do we construct the LCM from these exponents? The rule is elegant and absolute: the exponent of a prime in the LCM is the maximum of the exponents of that prime in the individual numbers.

Thus, for our primes $r,s,$ and $t$, we have the following constraints:

This is the heart of the problem. We are no longer dealing with large integers; we are playing a game of choosing exponents such that their maximum hits a specific target.

The Combinatorial Game

Let us focus on the prime $r$. We need $\max (a_1,a_2)=2$. Let's list the possibilities for the ordered pair $(a_1,a_2)$:

1. If $a_2=2$, then $a_1$ can be $0,1,$ or $2$. (3 pairs: $(0,2),(1,2),(2,2)$)

2. If $a_1=2$, then $a_2$ can be $0$ or $1$. (2 pairs: $(2,0),(2,1)$)

Notice that $(2,2)$ is counted in both scenarios, so we must be careful not to double-count. The distinct pairs are $(0,2),(1,2),(2,2),(2,1),(2,0)$.

That is exactly $5$ pairs. Observe the pattern: $2(2)+1=5$.

This is not a coincidence! For any exponent $n$, the number of pairs $(x,y)$ such that $\max (x,y)=n$ is $2n+1$.

The Grand Synthesis

We apply this logic to all primes. For $s$, where the maximum exponent is $4$, the number of pairs is $2(4)+1=9$.

For $t$, where the maximum exponent is $2$, the number of pairs is $2(2)+1=5$. Because the choices for the exponents of $r,s,$ and $t$ are independent, we use the Fundamental Principle of Counting.

We multiply the possibilities:

There we have it. Two hundred and twenty-five ordered pairs. It is a beautiful result, born from the simple realization that the LCM is just a collection of maximums.

Keep this logic in your toolkit; it is a powerful weapon for any number theory challenge you might face. The final answer is 225.