How many ordered pairs (a, b) exist such that LCM of a and b is 23 57 1113 (a, b ∈ N)?2460283526452840

To determine how many ordered pairs $(a, b)$ exist such that $\text{LCM}(a, b) = 235711$, we will proceed with the following steps:

1. Break Down the Problem

First, note that $235711$ is the product of the first few prime numbers: $235711 = 2^1 \cdot 3^1 \cdot 5^1 \cdot 7^1 \cdot 11^1 \cdot 13^1$ We need to find natural numbers $a$ and $b$ such that the least common multiple is equal to $235711$.

2. Relevant Concepts

By the properties of LCM and GCD, we have: $\text{LCM}(a, b) = \frac{a \cdot b}{\text{GCD}(a, b)}$ If we denote the prime factorization of $a$ and $b$ as follows: $a = 2^{x_1} \cdot 3^{y_1} \cdot 5^{z_1} \cdot 7^{w_1} \cdot 11^{u_1} \cdot 13^{v_1}$ $b = 2^{x_2} \cdot 3^{y_2} \cdot 5^{z_2} \cdot 7^{w_2} \cdot 11^{u_2} \cdot 13^{v_2}$

3. Analysis and Detail

To achieve $\text{LCM}(a,b) = 235711$, for each prime factor, the maximum exponent in $a$ or $b$ must equal the respective exponent in the LCM:

1. For the prime $2$ (exponent $1$): $\max(x_1, x_2) = 1$

   Possible pairs:

   • $(1, 0)$
   • $(0, 1)$

   Total combinations = 2

2. For the prime $3$ (exponent $1$): $\max(y_1, y_2) = 1$

   Possible pairs:

   • $(1, 0)$
   • $(0, 1)$

   Total combinations = 2

3. For the prime $5$ (exponent $1$): $\max(z_1, z_2) = 1$

   Possible pairs:

   • $(1, 0)$
   • $(0, 1)$

   Total combinations = 2

4. For the prime $7$ (exponent $1$): $\max(w_1, w_2) = 1$

   Possible pairs:

   • $(1, 0)$
   • $(0, 1)$

   Total combinations = 2

5. For the prime $11$ (exponent $1$): $\max(u_1, u_2) = 1$

   Possible pairs:

   • $(1, 0)$
   • $(0, 1)$

   Total combinations = 2

6. For the prime $13$ (exponent $1$): $\max(v_1, v_2) = 1$

   Possible pairs:

   • $(1, 0)$
   • $(0, 1)$

   Total combinations = 2

4. Verify and Summarize

Since each prime contributes $2$ combinations, we multiply the total combinations: $\text{Total combinations} = 2^6 = 64$

Final Answer

The total number of ordered pairs $(a, b)$ such that $\text{LCM}(a, b) = 235711$ is $64$.