Minimum sum of three positive integers with a fixed product

Expert intro: When a product is fixed, the smallest sum usually comes from making the factors as balanced as possible. For integer triples, the trick is to test divisor patterns near the cube root instead of trying everything blindly.

Detailed walkthrough

We want positive integers $a,b,c$ such that

$abc=2025=3^4\cdot 5^2$

and the sum $a+b+c$ is as small as possible.

Step 1: Look for balanced factors

The numbers should be close to each other. Try factor triples near $\sqrt[3]{2025}\approx 12.7$.

A very natural candidate is:

$9\cdot 15\cdot 15=2025$

and the sum is

$9+15+15=39$

Step 2: Check nearby triples

A few nearby factorizations are:

• $3,15,45 \rightarrow 63$
• $5,9,45 \rightarrow 59$
• $9,9,25 \rightarrow 43$
• $1,45,45 \rightarrow 91$

None of these beats 39.

So the minimum possible value is

$\boxed{39}$

💡 Pitfall guide

The usual mistake is forcing one factor to be 1 too early. That makes the sum much larger. Another trap is ignoring that the factors must be integers, not just positive real numbers. The closest-real-number intuition still helps, but you have to finish with divisor checking.

🔄 Real-world variant

If the problem asked for nonnegative integers instead of positive integers, the answer would still be 39, because using a zero factor would make the product impossible. If the product changed to a nearby number like 2024 or 2026, the best triple could shift, so the balanced-factor check would need to be repeated.

🔍 Related terms

prime factorization, integer factors, minimum sum