Let m and n (m

Question

(D) Let $m = 6a$ and $n = 6b$. Since $m < n$,we have $a < b$. Since $m$ and $n$ are $2$-digit numbers,$10 \leq 6a \leq 99$ and $10 \leq 6b \leq 99$,wh

Vedclass · Accepted Answer

$64$

Vedclass · Answer

(D) Let $m = 6a$ and $n = 6b$. Since $m Since $m$ and $n$ are $2$-digit numbers,$10 \leq 6a \leq 99$ and $10 \leq 6b \leq 99$,which implies $2 \leq a Also,$\operatorname{gcd}(m, n) = 6$ implies $\operatorname{gcd}(a, b) = 1$. We count the pairs $(a, b)$ such that $2 \leq a For $a=2: b \in \{3, 5, 7, 9, 11, 13, 15\}$ ($7$ pairs) For $a=3: b \in \{4, 5, 7, 8, 10, 11, 13, 14, 16\}$ ($9$ pairs) For $a=4: b \in \{5, 7, 9, 11, 13, 15\}$ ($6$ pairs) For $a=5: b \in \{6, 7, 8, 9, 11, 12, 13, 14, 16\}$ ($9$ pairs) For $a=6: b \in \{7, 11, 13\}$ ($3$ pairs) For $a=7: b \in \{8, 9, 10, 11, 12, 13, 15, 16\}$ ($8$ pairs) For $a=8: b \in \{9, 11, 13, 15\}$ ($4$ pairs) For $a=9: b \in \{10, 11, 13, 14, 16\}$ ($5$ pairs) For $a=10: b \in \{11, 13\}$ ($2$ pairs) For $a=11: b \in \{12, 13, 14, 15, 16\}$ ($5$ pairs) For $a=12: b \in \{13\}$ ($1$ pair) For $a=13: b \in \{14, 15, 16\}$ ($3$ pairs) For $a=14: b \in \{15\}$ ($1$ pair) For $a=15: b \in \{16\}$ ($1$ pair) Total number of pairs = $7 + 9 + 6 + 9 + 3 + 8 + 4 + 5 + 2 + 5 + 1 + 3 + 1 + 1 = 64$.

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

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