Consider two positive integer a and b. Find the least possible value of the product ab if `a^(b)b^(a)` is divisible by 2000

Answer

To solve the problem, we need to find the least possible value of the product \( ab \) such that \( a^b \cdot b^a \) is divisible by 2000. Let's go through the steps systematically.

### Step 1: Factorize 2000
First, we need to factor 2000 into its prime factors:
\[ 2000 = 2^4 \times 5^3 \]

### Step 2: Understand the divisibility condition
For \( a^b \cdot b^a \) to be divisible by \( 2000 \), it must satisfy:
\[ a^b \cdot b^a \text{ must be of the form } 2^m \cdot 5^n \text{ where } m \geq 4 \text{ and } n \geq 3 \]

### Step 3: Express \( a \) and \( b \)
Let us express \( a \) and \( b \) in terms of their prime factors:
\[ a = 2^x \cdot 5^y \quad \text{and} \quad b = 2^z \cdot 5^w \]
where \( x, y, z, w \) are non-negative integers.

### Step 4: Calculate \( a^b \cdot b^a \)
Now we can express \( a^b \cdot b^a \):
\[ a^b = (2^x \cdot 5^y)^{2^z \cdot 5^w} = 2^{x \cdot 2^z \cdot 5^w} \cdot 5^{y \cdot 2^z \cdot 5^w} \]
\[ b^a = (2^z \cdot 5^w)^{2^x \cdot 5^y} = 2^{z \cdot 2^x \cdot 5^y} \cdot 5^{w \cdot 2^x \cdot 5^y} \]
Thus,
\[ a^b \cdot b^a = 2^{x \cdot 2^z \cdot 5^w + z \cdot 2^x \cdot 5^y} \cdot 5^{y \cdot 2^z \cdot 5^w + w \cdot 2^x \cdot 5^y} \]

### Step 5: Set up inequalities for \( m \) and \( n \)
To satisfy the divisibility by \( 2000 \):
1. \( x \cdot 2^z \cdot 5^w + z \cdot 2^x \cdot 5^y \geq 4 \)
2. \( y \cdot 2^z \cdot 5^w + w \cdot 2^x \cdot 5^y \geq 3 \)

### Step 6: Choose values for \( a \) and \( b \)
To minimize \( ab \), we can try small values for \( a \) and \( b \).

Let's try \( a = 4 \) and \( b = 5 \):
\[ a = 4 = 2^2 \quad (x = 2, y = 0) \]
\[ b = 5 = 5^1 \quad (z = 0, w = 1) \]
Now calculate \( ab \):
\[ ab = 4 \cdot 5 = 20 \]

### Step 7: Verify divisibility
Now we need to check if \( a^b \cdot b^a \) is divisible by \( 2000 \):
\[ a^b = 4^5 = 1024 = 2^{10} \]
\[ b^a = 5^4 = 625 = 5^4 \]
Thus,
\[ a^b \cdot b^a = 2^{10} \cdot 5^4 \]
Now check the powers:
- \( 10 \geq 4 \) (satisfied)
- \( 4 \geq 3 \) (satisfied)

### Conclusion
The least possible value of the product \( ab \) such that \( a^b \cdot b^a \) is divisible by 2000 is:
\[ \boxed{20} \]