Animated Solution for Mathematics - Permutations and Combinations: Let $m$ and $n,(m<n)$ be two 2-digit numbers. Then the total numbers of pairs $(m,n)$, such that $\gcd (m,n)=6$, is ________

Imagine you are standing at the threshold of a vast, numerical landscape. You are tasked with finding pairs of 2-digit numbers $(m,n)$ that share a specific, elegant property: their greatest common divisor, or $\gcd (m,n)$, is exactly $6$.

This isn't just a math problem; it's a puzzle of structure and constraints. Let's break it down together, step by step.

When we are told that $\gcd (m,n)=6$, we are being given a secret key. It tells us that both $m$ and $n$ are built from the same fundamental block: $6$.

We can express them as $m=6a$ and $n=6b$. By doing this, we have effectively 'extracted' the common factor.

If we substitute these back into our $\gcd$ condition, we get:

Dividing both sides by $6$, we arrive at a beautiful simplification: $\gcd (a,b)=1$. This means $a$ and $b$ must be co-prime—they share no common DNA other than the number $1$.

This is our new, simplified objective: find pairs $(a,b)$ such that $\gcd (a,b)=1$.

Now, we must respect the boundaries. The problem tells us that $m$ and $n$ are 2-digit numbers.

The smallest 2-digit number is $10$, and the largest is $99$. So, we have the inequality $10\le m<n\le 99$.

Substituting our expressions $m=6a$ and $n=6b$, we get:

Dividing this entire inequality by $6$, we find the range for our variables: $1.66\le a<b\le 16.5$.

Since $a$ and $b$ must be integers, our search space is clearly defined: $2\le a<b\le 16$. We are no longer looking for arbitrary numbers; we are looking for integers in a very specific, manageable box.

How do we count these pairs without missing any? The most robust strategy is to fix the larger variable, $b$, and iterate through its possible values from $3$ to $16$.

For each $b$, we count how many valid $a$ values exist such that $2\le a<b$ and $\gcd (a,b)=1$.

For $b=3$, $a$ can only be $2$. That is $1$ pair.

For $b=4$, $a$ can be $3$ (since $\gcd (3,4)=1$). That is $1$ pair.

For $b=5$, $a$ can be $2,3,4$. That is $3$ pairs.

For $b=6$, $a$ can only be $5$. That is $1$ pair.

Summing these up, we have $1+1+3+1=6$ pairs.

Moving to $b=7$ to $10$: - For $b=7$, $a$ can be $2,3,4,5,6$ ($5$ pairs). - For $b=8$, $a$ can be $3,5,7$ ($3$ pairs). - For $b=9$, $a$ can be $2,4,5,7,8$ ($5$ pairs). - For $b=10$, $a$ can be $3,7,9$ ($3$ pairs).

Summing these, we get $5+3+5+3=16$ pairs.

Continuing to $b=11$ to $13$: - For $b=11$, $a$ can be $2,3,\dots ,10$ ($9$ pairs). - For $b=12$, $a$ can be $5,7,11$ ($3$ pairs). - For $b=13$, $a$ can be $2,3,\dots ,12$ ($11$ pairs).

Summing these, we get $9+3+11=23$ pairs.

Finally, for $b=14$ to $16$: - For $b=14$, $a$ can be $3,5,9,11,13$ ($5$ pairs). - For $b=15$, $a$ can be $2,4,7,8,11,13,14$ ($7$ pairs). - For $b=16$, $a$ can be $3,5,7,9,11,13,15$ ($7$ pairs).

Summing these, we get $5+7+7=19$ pairs.

We have methodically explored every possibility. Now, we simply add our subtotals: $6+16+23+19$.

The result is $64$.

By breaking a complex condition into smaller, manageable pieces, we have navigated the entire landscape and found exactly $64$ valid pairs. This is the power of systematic thinking—it turns a daunting problem into a series of small, satisfying victories.