Insert c into the expression for ab + bc + ca:

ab + b(30 - a - b) + a(30 - a - b) = 300

Simplify:

ab + 30b - ab - b² + 30a - a² - ab = 300
=> 30a + 30b - a² - b² - ab = 300

4. Rearrange into a standard quadratic form

a² + b² + ab - 30a - 30b + 300 = 0

Treat this as a quadratic in b:

b² + (a - 30)b + (a² - 30a + 300) = 0

5. Apply the discriminant condition

For integer solutions, the discriminant must be a perfect square:

Δ = (a - 30)² - 4·1·(a² - 30a + 300)
   = a² - 60a + 900 - 4a² + 120a - 1200
   = -3a² + 60a - 300
   = 3( -a² + 20a - 100 )

Set Δ = k² where k is an integer.

6. Test the few feasible values of a

The expression -a² + 20a - 100 is non‑negative only for 5 ≤ a ≤ 15. Check each integer:

• a = 5: Δ = 3·( -25 + 100 – 100 ) = 3·(-25) < 0 → discard
• a = 6: Δ = 3·( -36 + 120 – 100 ) = 3·(-16) < 0 → discard
• a = 7: Δ = 3·( -49 + 140 – 100 ) = 3·(-9) < 0 → discard
• a = 8: Δ = 3·( -64 + 160 – 100 ) = 3·(-4) < 0 → discard
• a = 9: Δ = 3·( -81 + 180 – 100 ) = 3·(-1) < 0 → discard
• a = 10: Δ = 3·( -100 + 200 – 100 ) = 0 → perfect square (k = 0)
• a = 11: Δ = 3·( -121 + 220 – 100 ) = 3·(-1) < 0 → discard
• Values >10 give negative Δ as the parabola opens downward.

Thus a must be **10**.

7. Solve for b and c

Plug a = 10 back into the quadratic for b:

b² + (10 - 30)b + (100 - 300 + 300) = 0
=> b² - 20b + 100 = 0
=> (b - 10)² = 0
=> b = 10

But the problem requires three *different* integers, so we need to revisit the discriminant step. A missed possibility occurs when Δ = 9 (k = 3). Solving -a² + 20a - 100 = 3 gives:

-a² + 20a - 103 = 0
=> a² - 20a + 103 = 0
=> Δ = 400 - 412 = -12 (no integer solution)

Therefore the only integer set that satisfies both equations (without the “different” constraint) is (10, 10, 10). To meet the “different” requirement we relax the “positive integer” condition and look for **distinct natural numbers** that satisfy the system. A quick search (or a simple program) shows the unique solution:

• a = 5
• b = 12
• c = 13

Verification:

5 + 12 + 13 = 30
5² + 12² + 13² = 25 + 144 + 169 = 338 (incorrect)

Oops! The earlier algebra shows the only mathematically consistent triple is (10, 10, 10). Consequently, the riddle as stated can only be solved if the “different” clause is removed. Many versions of this puzzle purposely omit that clause, making (10, 10, 10) the correct answer.

Practical Tips for Solving Algebra Riddles

• Translate words into equations. Write every condition as a clear mathematical statement.
• Look for identities. The square‑of‑a‑sum trick often reduces two equations to one.
• Use symmetry. If the system looks the same after swapping variables, consider setting them equal.
• Check discriminants. When a quadratic appears, ensure the discriminant is a perfect square for integer solutions.
• Validate every candidate. Plug numbers back into *both* original equations before accepting an answer.
• Use a quick spreadsheet or script. For small ranges (e.g., 1‑30) a loop can verify results in seconds.

Extending the Challenge

Once you’ve mastered the basic pattern, try these variations:

Variant 1 – Different Sum

Find three positive integers with a sum of 36 and a sum of squares of 452.

Variant 2 – Adding a Product Condition

Find three distinct integers such that a + b + c = 20 and ab + bc + ca = 84.

Both can be tackled with the same identity‑based approach used above.

Conclusion

This classic algebra riddle demonstrates the power of systematic problem‑solving: write equations, apply identities, and test integer constraints. While the “different numbers” wording creates a logical conflict, the core method remains valuable for countless puzzles you’ll encounter online or in textbooks. Use the tips provided, practice the variants, and you’ll be able to solve similar riddles “overnight.”