If `N=a^(2)b^(4)` is divisible by 8 and 27. If a and b are positive integers not having any common factors except one,what is the minimum value of the least multiple of a and b?

Options

A. 6
B. 8
C. 9
D. 12

Answer

To solve the problem step by step, we need to find the minimum value of the least multiple of \( a \) and \( b \) given that \( N = a^2 b^4 \) is divisible by both 8 and 27, and that \( a \) and \( b \) are positive integers with no common factors other than 1.

### Step 1: Analyze the divisibility conditions

1. **Divisibility by 8**:
- The number 8 can be expressed as \( 2^3 \). Therefore, \( N \) must include at least \( 2^3 \) in its prime factorization.

2. **Divisibility by 27**:
- The number 27 can be expressed as \( 3^3 \). Therefore, \( N \) must include at least \( 3^3 \) in its prime factorization.

### Step 2: Express \( N \) in terms of \( a \) and \( b \)

Given \( N = a^2 b^4 \), we can express \( a \) and \( b \) in terms of their prime factors:
- Let \( a = 2^x \cdot 3^y \) (where \( x \) and \( y \) are non-negative integers).
- Let \( b = 2^z \cdot 3^w \) (where \( z \) and \( w \) are non-negative integers).

Then,
\[ N = (2^x \cdot 3^y)^2 \cdot (2^z \cdot 3^w)^4 = 2^{2x + 4z} \cdot 3^{2y + 4w} \]

### Step 3: Set up inequalities based on divisibility

From the divisibility conditions:
1. For \( N \) to be divisible by \( 2^3 \):
\[ 2x + 4z \geq 3 \]

2. For \( N \) to be divisible by \( 3^3 \):
\[ 2y + 4w \geq 3 \]

### Step 4: Find minimum values for \( a \) and \( b \)

#### Finding \( b \):
To satisfy \( 2y + 4w \geq 3 \):
- If we set \( w = 1 \) (the smallest integer that can satisfy the condition), then:
\[ 2y + 4(1) \geq 3 \implies 2y + 4 \geq 3 \implies 2y \geq -1 \implies y \geq 0 \]
The smallest value for \( y \) is 0. Thus, we can take \( b = 2^0 \cdot 3^1 = 3 \).

#### Finding \( a \):
To satisfy \( 2x + 4z \geq 3 \):
- If we set \( z = 1 \) (the smallest integer that can satisfy the condition), then:
\[ 2x + 4(1) \geq 3 \implies 2x + 4 \geq 3 \implies 2x \geq -1 \implies x \geq 0 \]
The smallest value for \( x \) is 0. Thus, we can take \( a = 2^0 \cdot 3^0 = 1 \).

### Step 5: Calculate the least multiple of \( a \) and \( b \)

Now, we have:
- \( a = 2^0 \cdot 3^0 = 1 \)
- \( b = 2^0 \cdot 3^1 = 3 \)

The least multiple of \( a \) and \( b \) is:
\[ \text{lcm}(a, b) = \text{lcm}(1, 3) = 3 \]

### Conclusion

The minimum value of the least multiple of \( a \) and \( b \) is \( 3 \).