10 Super Fun Math Riddles for Kids Ages 10+ (with Answers ... - Free Printable

We are given a puzzle:

> Find three different whole numbers such that their sum equals their product, and both equal some number $ D $.

That is:
$$
A + B + C = D \quad \text{and} \quad A \times B \times C = D
$$
So:
$$
A + B + C = A \times B \times C
$$

And we want three different whole numbers (i.e., non-negative integers, usually starting from 0 or 1).

---

Step 1: Understand the constraints

Let’s denote:
- $ A, B, C $: three different whole numbers.
- $ A + B + C = A \times B \times C $

Let’s try small values.

---

Try small numbers

Try including 1, because multiplying by 1 doesn’t change the product much.

#### Try: 1, 2, 3

Sum: $ 1 + 2 + 3 = 6 $
Product: $ 1 \times 2 \times 3 = 6 $

✔ Sum = Product = 6

And they are all different whole numbers.

So this works!

Let’s check if there are others.

---

#### Try: 1, 2, 4

Sum: $ 1+2+4 = 7 $
Product: $ 1×2×4 = 8 $ → Not equal

#### Try: 1, 2, 5

Sum: 8, Product: 10 → No

#### Try: 1, 3, 4

Sum: 8, Product: 12 → No

#### Try: 1, 2, 6

Sum: 9, Product: 12 → No

#### Try: 1, 3, 5

Sum: 9, Product: 15 → No

#### Try: 1, 1, 2 → not allowed (not different)

#### Try: 0, 1, something?

Let’s try including 0

If one of them is 0, then the product is 0.

So $ A × B × C = 0 $ → Then sum must also be 0.

But sum of three whole numbers (non-negative) can only be 0 if all are 0 → but they must be different, so impossible.

So 0 cannot be included.

Thus, all numbers ≥ 1.

Now, what about larger numbers?

Try: 1, 2, 3 → already works.

Try: 1, 2, 4 → sum=7, product=8 → too big

As numbers increase, product grows faster than sum.

So likely only small solutions.

Try: 1, 3, 3 → not different

Try: 2, 3, 4 → sum=9, product=24 → no

Try: 1, 4, 5 → sum=10, product=20 → no

Try: 1, 1, 4 → not different

So seems like only solution with distinct whole numbers is:

> 1, 2, 3

Check again:
- Sum: $ 1 + 2 + 3 = 6 $
- Product: $ 1 × 2 × 3 = 6 $
- All different ✔️
- Whole numbers ✔️

---

Are there any other solutions?

Let’s suppose $ A < B < C $, all ≥ 1.

Then $ A \geq 1 $, $ B \geq 2 $, $ C \geq 3 $

We know:
$$
A + B + C = A \times B \times C
$$

Let’s assume $ A = 1 $

Then equation becomes:
$$
1 + B + C = 1 \times B \times C = BC
\Rightarrow BC - B - C = 1
\Rightarrow BC - B - C + 1 = 2
\Rightarrow (B - 1)(C - 1) = 2
$$

Nice! Factorization trick.

So:
$$
(B - 1)(C - 1) = 2
$$

Possible factor pairs of 2:
- $ 1 \times 2 $
- $ 2 \times 1 $

So:
1. $ B - 1 = 1 $, $ C - 1 = 2 $ → $ B = 2 $, $ C = 3 $
2. $ B - 1 = 2 $, $ C - 1 = 1 $ → $ B = 3 $, $ C = 2 $ → But $ B < C $, so invalid

So only solution with $ A = 1 $: $ (1, 2, 3) $

Now try $ A = 2 $

Then $ A = 2 $, $ B \geq 3 $, $ C \geq 4 $

Equation:
$$
2 + B + C = 2BC
\Rightarrow 2BC - B - C = 2
$$

Try small values:

- $ B = 3 $: $ 2×3×C - 3 - C = 2 $ → $ 6C - 3 - C = 2 $ → $ 5C = 5 $ → $ C = 1 $ → But $ C > B > A $, so $ C \geq 4 $ → invalid
- $ B = 4 $: $ 2×4×C - 4 - C = 2 $ → $ 8C - 4 - C = 2 $ → $ 7C = 6 $ → $ C < 1 $ → invalid

As $ B $ increases, left side grows faster → no solution.

Try $ A = 3 $: even worse — product will be much larger than sum.

So no other solutions.

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✔ Final Answer:

The three different whole numbers are:
> 1, 2, and 3

Because:
- $ 1 + 2 + 3 = 6 $
- $ 1 × 2 × 3 = 6 $

They are different, whole numbers, and satisfy the condition.

✔️ Answer: 1, 2, 3