We need the number of pairs $(m,n)$ such that:

• $m,n$ are two-digit numbers,
• $m<n$,
• $\gcd(m,n)=6$.

If $\gcd(m,n)=6$, then both $m$ and $n$ must be multiples of $6$. So write $m=6a, \qquad n=6b$ where $\gcd(a,b)=1.$

Since $m,n$ are two-digit numbers, $10 \le 6a \le 99, \qquad 10 \le 6b \le 99.$ Dividing by $6$, $\frac{10}{6} \le a,b \le \frac{99}{6}.$ Hence $2 \le a,b \le 16.$

Also, since $m<n$, we need $a<b.$

Therefore the problem reduces to counting pairs $(a,b)$ such that $2 \le a<b\le 16, \qquad \gcd(a,b)=1.$

Count for each $b$ the number of integers $a$ with $2\le a<b$ and $\gcd(a,b)=1$.

For a fixed $b$, the count of integers $1\le a<b$ coprime to $b$ is $\varphi(b)$. Since $a=1$ is included in $\varphi(b)$ but not allowed here, we subtract $1$.

So required count is $\varphi(b)-1$ for each $b=3,4,\dots,16$.

Now compute: [ \begin{aligned} \varphi(3)-1 &= 2-1=1 \ \varphi(4)-1 &= 2-1=1 \ \varphi(5)-1 &= 4-1=3 \ \varphi(6)-1 &= 2-1=1 \ \varphi(7)-1 &= 6-1=5 \ \varphi(8)-1 &= 4-1=3 \ \varphi(9)-1 &= 6-1=5 \ \varphi(10)-1 &= 4-1=3 \ \varphi(11)-1 &= 10-1=9 \ \varphi(12)-1 &= 4-1=3 \ \varphi(13)-1 &= 12-1=11 \ \varphi(14)-1 &= 6-1=5 \ \varphi(15)-1 &= 8-1=7 \ \varphi(16)-1 &= 8-1=7 \end{aligned} ]

Sum them: $1+1+3+1+5+3+5+3+9+3+11+5+7+7=64.$

Hence the total number of pairs $(m,n)$ is $\boxed{64}.$

Comparison with stored answer:

• Derived answer: $64$
• Stored correct answer: $64$
• They match.