Question

The number of ordered pairs of positive integers $(m, n)$ satisfying $m \le 2n \le 60, n \le 2m \le 60$ is

• A. 240
• B. 480
• C. 900
• D. none of these

Answer 1

Answer: (B)

480

Answer 2

The problem asks us to find the number of ordered pairs of positive integers $(m, n)$ satisfying the given inequalities:

1. $m \le 2n \le 60$
2. $n \le 2m \le 60$

Let's break down these inequalities:

From $m \le 2n \le 60$:

• $m \le 2n$
• $2n \le 60 \implies n \le 30$

From $n \le 2m \le 60$:

• $n \le 2m$
• $2m \le 60 \implies m \le 30$

Since $m$ and $n$ are positive integers, we have $m \ge 1$ and $n \ge 1$. Combining all conditions, we need to find the number of pairs $(m, n)$ such that:

(A) $1 \le m \le 30$ (B) $1 \le n \le 30$ (C) $m \le 2n$ (D) $n \le 2m$

Notice that the conditions are symmetric with respect to $m$ and $n$. If $(m, n)$ is a solution, then $(n, m)$ is also a solution. This suggests we can count solutions for $m=n$, $m<n$, and $m>n$ separately.

Case 1: $m=n$

Substitute $n=m$ into the inequalities: $m \le 2m \le 60$

The condition $m \le 2m$ is true for all positive integers $m$.

The condition $2m \le 60$ implies $m \le 30$.

So, for $m=n$, any integer $m$ from 1 to 30 will satisfy the conditions. The pairs are $(1,1), (2,2), \dots, (30,30)$. Number of pairs for $m=n$ is 30.

Case 2: $m<n$

We need to satisfy $m < n$, along with conditions (A), (B), (C), (D). Since $m < n$ and $m, n$ are positive integers, $m \ge 1 \implies n \ge 2$.

Condition (C) $m \le 2n$: Since $m < n$, and $n$ is positive, $m \le n < 2n$ is always true. So $m \le 2n$ is automatically satisfied.

Thus, for $m<n$, we only need to satisfy:

(A) $1 \le m \le 30$ (B) $m < n \le 30$ (D) $n \le 2m$

Combining (B) and (D), we need $m < n \le \min(2m, 30)$. We will iterate through possible values of $m$ from 1 to 30.

Subcase 2.1: $2m \le 30 \implies m \le 15$. For $m \in \{1, 2, \dots, 15\}$, the range for $n$ is $m < n \le 2m$. The number of possible values for $n$ is $2m - (m+1) + 1 = m$.

• If $m=1$, $1 < n \le 2$. So $n=2$. (1 pair: $(1,2)$)
• If $m=2$, $2 < n \le 4$. So $n \in \{3,4\}$. (2 pairs: $(2,3), (2,4)$)
• ...
• If $m=15$, $15 < n \le 30$. So $n \in \{16, \dots, 30\}$. (15 pairs: $(15,16), \dots, (15,30)$)

The total number of pairs for this subcase is $\sum_{m=1}^{15} m = \frac{15 \times (15+1)}{2} = \frac{15 \times 16}{2} = 15 \times 8 = 120$.

Subcase 2.2: $2m > 30 \implies m > 15$. For $m \in \{16, 17, \dots, 30\}$, the range for $n$ is $m < n \le 30$. (Since $2m > 30$, $\min(2m, 30) = 30$). The number of possible values for $n$ is $30 - (m+1) + 1 = 30 - m$.

• If $m=16$, $16 < n \le 30$. So $n \in \{17, \dots, 30\}$. Number of pairs $= 30-16 = 14$.
• If $m=17$, $17 < n \le 30$. So $n \in \{18, \dots, 30\}$. Number of pairs $= 30-17 = 13$.
• ...
• If $m=29$, $29 < n \le 30$. So $n=30$. Number of pairs $= 30-29 = 1$.
• If $m=30$, $30 < n \le 30$. No possible values for $n$. Number of pairs $= 0$.

The total number of pairs for this subcase is $\sum_{m=16}^{29} (30-m)$. Let $k = 30-m$. When $m=16, k=14$. When $m=29, k=1$. So the sum is $\sum_{k=1}^{14} k = \frac{14 \times (14+1)}{2} = \frac{14 \times 15}{2} = 7 \times 15 = 105$.

Total number of pairs for $m<n$ is $120 + 105 = 225$.

Case 3: $m>n$

Due to the symmetry of the conditions, the number of pairs for $m>n$ is the same as the number of pairs for $m<n$. Number of pairs for $m>n$ is 225.

Total Number of Ordered Pairs

Total pairs = (pairs for $m=n$) + (pairs for $m<n$) + (pairs for $m>n$) Total pairs = $30 + 225 + 225 = 30 + 450 = 480$.