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 ______.

Every 2-digit number having greatest common divisor $$6$$ with another 2-digit number must itself be divisible by $$6$$, because any common divisor of the two numbers divides their gcd.
Hence write the numbers as $$m = 6a$$ and $$n = 6b$$ with $$a,b \in \mathbb{N}$$.

Because $$m$$ and $$n$$ are 2-digit numbers,

$$10 \le m = 6a \le 99 \quad \text{and} \quad 10 \le n = 6b \le 99$$

which gives

$$\frac{10}{6} \le a \le \frac{99}{6}, \qquad \frac{10}{6} \le b \le \frac{99}{6}$$

or, after taking integer values,

$$2 \le a \le 16, \qquad 2 \le b \le 16.$$

The condition $$\gcd(m,n)=6$$ becomes $$\gcd(6a,6b)=6,$$ i.e.

$$\gcd(a,b)=1.$$ Thus we must count ordered pairs $$(a,b)$$ such that

$$2 \le a \lt b \le 16 \quad \text{and} \quad \gcd(a,b)=1.$$ Call such pairs “coprime pairs”.

List each value of $$a$$ and count suitable $$b$$ values individually:

$$\begin{aligned} a=2 &: b=3,5,7,9,11,13,15 &\Rightarrow 7\\ a=3 &: b=4,5,7,8,10,11,13,14,16 &\Rightarrow 9\\ a=4 &: b=5,7,9,11,13,15 &\Rightarrow 6\\ a=5 &: b=6,7,8,9,11,12,13,14,16 &\Rightarrow 9\\ a=6 &: b=7,11,13 &\Rightarrow 3\\ a=7 &: b=8,9,10,11,12,13,15,16 &\Rightarrow 8\\ a=8 &: b=9,11,13,15 &\Rightarrow 4\\ a=9 &: b=10,11,13,14,16 &\Rightarrow 5\\ a=10&: b=11,13 &\Rightarrow 2\\ a=11&: b=12,13,14,15,16 &\Rightarrow 5\\ a=12&: b=13 &\Rightarrow 1\\ a=13&: b=14,15,16 &\Rightarrow 3\\ a=14&: b=15 &\Rightarrow 1\\ a=15&: b=16 &\Rightarrow 1 \end{aligned}$$

Adding these counts:

$$7+9+6+9+3+8+4+5+2+5+1+3+1+1 = 64.$$

Therefore there are $$64$$ ordered pairs $$(a,b)$$, and hence $$64$$ ordered pairs $$(m,n)$$ with $$m \lt n$$ and $$\gcd(m,n)=6$$ where both $$m$$ and $$n$$ are 2-digit numbers.

Final answer: $$64$$.