Math riddle with answer | Number And Math Puzzle - BrainFans - Free Printable

We are given a pattern:

```
398 = 964
118 = 164
356 = 936
423 = ???
```

We need to find the rule that transforms the left-hand number into the right-hand number.

---

Let’s break down each number digit by digit and look for a pattern.

First example: 398 → 964

Digits of 398:
- Hundreds digit: 3
- Tens digit: 9
- Units digit: 8

Result: 964

Let’s try squaring each digit:

- 3² = 9
- 9² = 81
- 8² = 64

Hmm, if we concatenate the squares: 9, 81, 64 → 98164 — too long.

But notice: 964 has 3 digits. So maybe we take only the last digit of each square?

- 3² = 9 → last digit: 9
- 9² = 81 → last digit: 1
- 8² = 64 → last digit: 4

That gives 914, but we need 964. Not matching.

Wait — what if we take the square of each digit, but then concatenate the results as two-digit numbers? But 3²=09, 9²=81, 8²=64 → 098164 — still too long.

Alternative idea: Maybe it's square of first digit, square of second digit, square of third digit, but written as single digits? That doesn’t work for 9²=81.

Wait — let’s look at the result 964.

What if we do:

- First digit of result = (first digit)² → 3² = 9 ✔
- Second digit of result = (second digit)² → 9² = 81 → take tens digit? 8? But result is 6 ✘
- Or maybe units digit of square? 9²=81 → 1, but result has 6 ✘

Not working.

---

Try another approach.

Look at all three examples together.

Example 1: 398 → 964

Example 2: 118 → 164

Digits: 1, 1, 8

Squares: 1, 1, 64 → if we take last digit: 1,1,4 → 114 ≠ 164

But 164:
- 1 → 1² = 1
- 6 → ?
- 4 → 8² = 64 → last digit 4

So maybe middle digit is something else.

Wait — what if the result is:

> (first digit)² , (second digit)² , (third digit)² — but written as two-digit numbers concatenated, and then take only the last three digits?

Try for 398:

- 3² = 09
- 9² = 81
- 8² = 64

Concatenate: 098164 → last 3 digits: 164 — but we need 964 ✘

Wait — maybe first digit squared, third digit squared, second digit squared? Let’s test.

For 398:
- 3² = 9
- 8² = 64
- 9² = 81
→ Concatenate: 9, 64, 81 → 96481 — last 3 digits? 481 ✘

But wait — 964 appears in the middle!

In 098164, if we take digits 2-4: 9,8,1 → 981 ✘

Alternatively — what if we take square of each digit, and then take the units digit of each square, but in a different order?

Try this:

For 398 → 964

- 3² = 9 → use 9
- 8² = 64 → use 6 (tens digit?) or 4 (units digit?)
- 9² = 81 → use 8 or 1?

If we do:

- First digit of result = (first digit)² → 3² = 9 ✔
- Second digit of result = (third digit)² → 8² = 64 → take 6 (tens digit) ✔
- Third digit of result = (second digit)² → 9² = 81 → take 4? No, 1? But result is 4.

Wait — 964: third digit is 4, which is units digit of 8².

So maybe:

> Result = [ (first digit)² ] [ (third digit)² tens digit ] [ (second digit)² units digit ]

Let’s test:

398 → 964

- 3² = 9 → first digit: 9 ✔
- 8² = 64 → tens digit: 6 ✔
- 9² = 81 → units digit: 1 ✘ — but we need 4.

Nope.

Wait — what if it’s:

> Result = [ (first digit)² ] [ (second digit)² tens digit ] [ (third digit)² units digit ]

For 398:

- 3² = 9
- 9² = 81 → tens digit = 8
- 8² = 64 → units digit = 4 → gives 984 ✘ (we need 964)

Close! 984 vs 964 — off by one digit.

Wait — what if it’s:

> Result = [ (first digit)² ] [ (third digit)² tens digit ] [ (second digit)² units digit ]

398:

- 3² = 9
- 8² = 64 → tens = 6
- 9² = 81 → units = 1 → gives 961 ✘ (need 964)

Still not.

---

Let me try a completely different angle.

What if the result is formed by:

> Square of first digit, square of second digit, square of third digit — but each square is written as a single digit by taking modulo 10 (i.e., last digit), and then arranged in a specific order.

So:

398 → 3²=9, 9²=81→1, 8²=64→4 → so digits: 9,1,4 → 914 ≠ 964

But 964 — maybe it’s 9, 6, 4.

Where does 6 come from? 8² = 64 → 6 is the tens digit.

What if the pattern is:

> Result = [ (first digit)² ] [ (third digit)² tens digit ] [ (third digit)² units digit ]

For 398:

- 3² = 9
- 8² = 64 → tens=6, units=4 → 964 ✔

Oh! That works for the first one.

Let’s test on second example: 118 → 164

Digits: 1, 1, 8

Apply same rule:

- First digit squared: 1² = 1
- Third digit squared: 8² = 64 → tens digit = 6, units digit = 4
→ Result: 164 ✔

Perfect!

Third example: 356 → 936

Digits: 3, 5, 6

- First digit squared: 3² = 9
- Third digit squared: 6² = 36 → tens digit = 3, units digit = 6
→ Result: 936 ✔

Yes! This matches all three.

---

So the rule is:

> Given a 3-digit number ABC (where A, B, C are digits),
>
> The result is:
> (A²) (tens digit of C²) (units digit of C²)

In other words:

- First digit of result = square of the first digit of input.
- Second and third digits of result = the two digits of the square of the third digit of input.

Note: We don't use the middle digit (B) at all!

---

Now apply to 423

Digits: A=4, B=2, C=3

- A² = 4² = 16 → but we only want the value, not the digits? Wait — in previous examples, we used single digits.

Wait — in 398: 3²=9 → single digit.

In 118: 1²=1 → single digit.

In 356: 3²=9 → single digit.

But 4²=16 — that’s two digits. How do we handle this?

In all previous cases, the first digit was 1, 3 — whose squares are single-digit.

But 4²=16 — two digits. So how do we represent it?

Looking back at the examples — the result is always 3 digits.

In 398 → 964 (all single digits)

So if 4²=16, we can’t put “16” as the first digit — that would make the result 4 digits.

Therefore, perhaps we take only the units digit of A²?

Let’s test that hypothesis with previous examples.

For 398: A=3, 3²=9 → units digit=9 ✔

118: A=1, 1²=1 → 1 ✔

356: A=3, 3²=9 → 9 ✔

So even though 3²=9 is single digit, we’re effectively taking units digit of A².

Similarly, for 4²=16 → units digit = 6

Then:

C = 3 → C² = 9 → but wait, 3²=9 → which is 09? So tens digit=0, units digit=9?

But in previous examples, when C=8, 8²=64 → tens=6, units=4 → used both.

When C=6, 6²=36 → tens=3, units=6.

So for C=3, 3²=9 → we should write it as 09, so tens digit=0, units digit=9.

Thus, for 423:

- A=4 → A²=16 → units digit = 6
- C=3 → C²=9 → written as 09 → tens digit = 0, units digit = 9

So result = 6 0 9 → 609

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Final Answer: 609

Let’s verify the pattern holds:

- 398 → A=3→9, C=8→64 → 964 ✔
- 118 → A=1→1, C=8→64 → 164 ✔
- 356 → A=3→9, C=6→36 → 936 ✔
- 423 → A=4→6 (units of 16), C=3→09 → 609 ✔

Perfect.

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✔ Answer: 609