Let m and n, m < n be two 2-digit numbers. Then the total number… — Solution | JEE Mathematics

MCQMediumJEE 2025Applications of P&C

Let $m$ and $n$ , $m < n$ be two $2$ -digit numbers. Then the total number of pairs $(m, n)$ such that $\gcd(m, n) = 6$ , is _____

Answer

Correct answer:A

Standard Method

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

Find: The total number of pairs $(m, n)$ satisfying these conditions.

Let

m = 6a, \quad n = 6b

where $a$ and $b$ are co-prime numbers.

Since $m$ and $n$ are two-digit numbers,

10 \leq m \leq 99 \quad \text{and} \quad 10 \leq n \leq 99

So,

10 \leq 6a \leq 99 \Rightarrow 2 \leq a \leq 16

and

10 \leq 6b \leq 99 \Rightarrow 2 \leq b \leq 16

Counting Co-prime Pairs

Thus, we need to count pairs $(a, b)$ such that $2 \leq a < b \leq 16$ and $\gcd(a, b) = 1$ .

The valid pairs listed in the solution are:

Answer from Listed Count

Adding the number of valid choices of $b$ for each admissible $a$ gives the total count stated in the solution.

Therefore, the total number of such pairs is $64$ , so the correct option is A.

Assuming $\gcd(m, n) = 6$ only means both numbers are divisible by $6$ . This is incomplete because after writing $m = 6a$ and $n = 6b$ , we must also require $\gcd(a, b) = 1$ . Otherwise the gcd of $m$ and $n$ would be greater than $6$ .
Including values of $a$ or $b$ outside the range $2$ to $16$ . This is wrong because $m$ and $n$ are two-digit numbers, so dividing by $6$ gives the restricted range for $a$ and $b$ .
Counting pairs with $a = b$ or with $a > b$ . This violates the condition $m < n$ , which becomes $a < b$ after dividing both numbers by $6$ .

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