Question

For two positive integers $a$ and $b$, if $(a+b)^{(a+b)}$ is divisible by $500$, then the least possible value of $a\times b$ is:

• A. 8
• B. 9
• C. 10
• D. 12
• E. None of the above

Hint:

When $n^n$ must be divisible by a prime power $p^k$, check $n..... v_p(n)\ge k$. After fixing $n=a+b$, remember: for a fixed sum, the product is minimized at the extreme split (e.g., $1$ and $S-1$).

Answer 1

The Correct Option is B

Step 1: Convert the divisibility condition.
We want \(n^n\) to be divisible by \(500\). Factorize: \[ 500 = 2^2 \cdot 5^3. \] Let \(n = a+b\). For a prime \(p\), the exponent of \(p\) in \(n^n\) is: \[ v_p(n^n) = n \cdot v_p(n). \] Hence we require: \[ n \cdot v_2(n) \;\geq\; 2, \qquad n \cdot v_5(n) \;\geq\; 3. \]

Step 2: Find the least \(n\) satisfying both conditions.
From the \(5\)-condition: we need \(v_5(n)\geq 1 \;\Rightarrow\; 5 \mid n\). Then \(n \cdot v_5(n) \geq n \geq 5 \geq 3\), so every multiple of 5 works for the \(5\)-condition.

Check small multiples of 5:

• \(n=5\): \(v_2(n)=0 \;\Rightarrow\; n\cdot v_2(n)=0 < 2\) (fails).
• \(n=10\): \(v_5(n)=1\), \(v_2(n)=1\). Then \[ n \cdot v_5(n) = 10 \geq 3, \quad n \cdot v_2(n) = 10 \geq 2. \] ✅ Works.

Thus the least possible value is: \[ n = a+b = \boxed{10}. \]

Step 3: Minimize \(ab\) given \(a+b=10\), \(a,b \in \mathbb{Z}_{>0}\).
For a fixed positive sum, the product \(ab\) is minimized when the numbers are farthest apart. So the minimum occurs at \((a,b)=(1,9)\) or \((9,1)\): \[ ab = 1 \cdot 9 = \boxed{9}. \]

Final Answer:

\[ \boxed{9} \]