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

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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

We are looking for the number of pairs of two-digit numbers $(m, n)$ such that $m < n$ and the greatest common divisor $\gcd(m, n) = 6$.

Step 1: Set Up the Conditions

Since $\gcd(m, n) = 6$, we can write:

$m = 6a$
$n = 6b$
$\gcd(a, b) = 1$ (they must be coprime)

Since $m$ and $n$ are 2-digit numbers and $m < n$:

$10 \le 6a < 6b \le 99$

Dividing the inequality by 6, we get:

$1.67 \le a < b \le 16.5$

Because $a$ and $b$ are integers, the range is: $2 \le a < b \le 16$.

Step 2: Systematic Counting of Coprime Pairs $(a, b)$

We iterate through all possible values of $a$ from 2 to 15 and find the number of values for $b$ in the range $[a+1, 16]$ that are coprime to $a$:

$a = 2$: $b \in \{3, 5, 7, 9, 11, 13, 15\}$ (7 values)
$a = 3$: $b \in \{4, 5, 7, 8, 10, 11, 13, 14, 16\}$ (9 values)
$a = 4$: $b \in \{5, 7, 9, 11, 13, 15\}$ (6 values)
$a = 5$: $b \in \{6, 7, 8, 9, 11, 12, 13, 14, 16\}$ (9 values)
$a = 6$: $b \in \{7, 11, 13\}$ (3 values)
$a = 7$: $b \in \{8, 9, 10, 11, 12, 13, 15, 16\}$ (8 values)
$a = 8$: $b \in \{9, 11, 13, 15\}$ (4 values)
$a = 9$: $b \in \{10, 11, 13, 14, 16\}$ (5 values)
$a = 10$: $b \in \{11, 13\}$ (2 values)
$a = 11$: $b \in \{12, 13, 14, 15, 16\}$ (5 values)
$a = 12$: $b \in \{13\}$ (1 value)
$a = 13$: $b \in \{14, 15, 16\}$ (3 values)
$a = 14$: $b \in \{15\}$ (1 value)
$a = 15$: $b \in \{16\}$ (1 value)

Step 3: Calculate the Total

Total pairs = $7 + 9 + 6 + 9 + 3 + 8 + 4 + 5 + 2 + 5 + 1 + 3 + 1 + 1$

Total pairs = $64$

Final Answer

The total number of pairs $(m, n)$ is 64.

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