Math Riddle From A Kenyan Bank - Mind Your Decisions - Free Printable

Let's solve this math riddle step by step.

We are given a series of equations that do not follow standard arithmetic, so we must deduce a pattern or rule behind them:

---

Given:

```
4 + 2 = 26
8 + 1 = 79
6 + 5 = 111
7 + 3 = ?
```

And then:
Suppose $ 7 + 3 = z $.
Find a different pair $ x, y $ such that $ x + y = z $ (using the same rule).

---

Step 1: Look for a pattern

Let’s examine each equation and see if we can find a relationship between the left-hand side and the right-hand side.

---

#### First: 4 + 2 = 26

Try to break down 26.

- $ 4 \times 2 = 8 $
- $ 4 + 2 = 6 $
- But 26 is not obviously related.

Wait — maybe it's not addition or multiplication alone.

Try:
- $ 4 \times 2 = 8 $
- $ 4 + 2 = 6 $
- Is there a way to get 26 from 4 and 2?

What if we do:
- $ 4 \times 6 = 24 $, close to 26?
- Or $ 4 \times 5 + 6 = 20 + 6 = 26 $? Not helpful.

Try another idea.

Look at:
- $ 4 + 2 = 6 $ → but result is 26
- Maybe it's concatenation?

Wait — what if:
- $ 4 \times 2 = 8 $, but 26 is not 8.
- What if it's $ 4 \times 5 + 6 $? No.

Alternative idea: Try reverse digits or something.

Wait — look at the second equation:

#### Second: 8 + 1 = 79

Try:
- $ 8 \times 1 = 8 $
- $ 8 + 1 = 9 $
- But result is 79.

Hmm.

But notice: 79 — could be related to 8 and 1.

What if:
- $ 8 - 1 = 7 $
- $ 8 + 1 = 9 $
- Then concatenate: 7 and 9 → 79 ✔

Oh! That works!

So:
- $ 8 + 1 = 79 $ → $ (8 - 1) $ and $ (8 + 1) $ → 7 and 9 → 79

Let’s test this on the first one.

#### First: 4 + 2 = ?

- $ 4 - 2 = 2 $
- $ 4 + 2 = 6 $
- Concatenate: 2 and 6 → 26 ✔

Perfect!

Now third:

#### Third: 6 + 5 = 111

- $ 6 - 5 = 1 $
- $ 6 + 5 = 11 $
- Concatenate: 1 and 11 → 111 ✔

Yes! It matches.

So the pattern is:

> For $ a + b = ? $, compute:
> - $ a - b $
> - $ a + b $
> - Then concatenate the two results (as strings)

But note: order matters. We do $ a - b $ first, then $ a + b $, and concatenate.

Let’s verify:

- 4 + 2 → (4-2)=2, (4+2)=6 → "2" + "6" = "26" → 26 ✔
- 8 + 1 → (8-1)=7, (8+1)=9 → "7" + "9" = "79" ✔
- 6 + 5 → (6-5)=1, (6+5)=11 → "1" + "11" = "111" ✔

Perfect.

---

Now: 7 + 3 = ?

Apply the same rule:

- $ 7 - 3 = 4 $
- $ 7 + 3 = 10 $
- Concatenate: "4" + "10" = "410"

So:
> $ 7 + 3 = 410 $

Thus, $ z = 410 $

---

Final Part:

> Suppose $ 7 + 3 = z $. Find a different pair $ x, y $ such that $ x + y = z $, using the same rule.

So we need to find another pair $ x, y $ such that when we apply the rule:
- $ x - y $ and $ x + y $, concatenated, gives 410

So we want:
- $ (x - y) $ concatenated with $ (x + y) $ = 410

That means:
- The first part (difference) should be a number whose digits form the beginning of 410
- The second part (sum) forms the rest

Now, 410 has three digits.

Possibility 1:
- $ x - y = 4 $
- $ x + y = 10 $

Then concatenate: "4" + "10" = "410" → matches.

So let’s solve:
$$
x - y = 4 \\
x + y = 10
$$

Add equations:
$ 2x = 14 $ → $ x = 7 $, $ y = 3 $

This is the original pair: $ 7 + 3 $

But the question asks for a different pair.

So we need another pair $ x, y $ such that:
- $ x - y $ and $ x + y $, when concatenated, give 410

Are there other ways to split "410" into two parts?

Let’s consider possible splits:

1. $ x - y = 4 $, $ x + y = 10 $ → already used → (7,3)
2. $ x - y = 41 $, $ x + y = 0 $ → impossible (sum can't be 0 if difference is 41)
3. $ x - y = 410 $, $ x + y = $ nothing → invalid

No other valid split.

Wait — could it be:
- $ x - y = 41 $, $ x + y = 0 $? → no, sum can’t be zero unless both zero.

Or:
- $ x - y = 4 $, $ x + y = 10 $ → only way.

But wait — what if we interpret the concatenation differently?

Could $ x - y = 41 $, $ x + y = 0 $? → no.

Alternatively, is there a case where $ x - y = 4 $, $ x + y = 10 $, but different values?

No — that system has unique solution: $ x=7, y=3 $

So only one pair gives this result?

But the problem says: find a different pair $ x, y $ such that $ x + y = z $, meaning under the same rule, the output is 410.

So unless there's another way to write 410 as a concatenation of $ (x - y) $ and $ (x + y) $, we’re stuck.

But perhaps the concatenation doesn’t require the numbers to be written in full?

Wait — what if $ x - y = 4 $, $ x + y = 10 $ → 4 and 10 → 410

Is there any other possibility?

Suppose $ x - y = 41 $, $ x + y = 0 $? → no

Or $ x - y = 410 $, $ x + y = $ nothing → invalid

No.

But here’s an idea: Could $ x - y = 4 $, $ x + y = 10 $ be interpreted differently?

Wait — what if $ x $ and $ y $ are not integers?

But the examples use integers.

Alternatively, maybe we can have $ x - y = 4 $, $ x + y = 10 $, but $ x $ and $ y $ are not 7 and 3?

No — solving:
- $ x - y = 4 $
- $ x + y = 10 $
→ $ x = 7 $, $ y = 3 $ → only solution.

So no other integer pair satisfies this.

But the problem says: find a different pair $ x, y $ such that $ x + y = z $

So maybe we misinterpreted the rule?

Wait — could the rule be applied differently?

Let’s double-check the pattern.

Wait — in all cases:
- $ a + b $ → $ (a - b) $ concatenated with $ (a + b) $

But what if the order is reversed? Let’s check.

In 4 + 2 = 26:
- If we did $ a + b = 6 $, $ a - b = 2 $, then concatenate 6 and 2 → 62 ≠ 26 → no

So order is: difference first, then sum

So: $ (a - b) $ + $ (a + b) $ → concatenated

So for $ x + y $ to produce 410, we need:
- $ x - y $ = 4
- $ x + y $ = 10
→ only solution: $ x = 7 $, $ y = 3 $

So no other pair?

But the problem says: “Find a different pair” — implying there is one.

So either:
1. Our pattern is wrong
2. There's another interpretation
3. Or perhaps the rule is not always $ a - b $ then $ a + b $

Wait — let’s re-express the rule more carefully.

Let’s suppose the operation $ a + b $ produces:
- $ (a - b) $ concatenated with $ (a + b) $

So the result is a string formed by writing $ a - b $, then $ a + b $

Now, can we get 410 in another way?

Suppose:
- $ x - y = 41 $
- $ x + y = 0 $ → impossible

Or:
- $ x - y = 4 $
- $ x + y = 10 $ → only way

But wait — what if $ x - y = 4 $, $ x + y = 10 $, but $ x $ and $ y $ are not integers?

Say $ x = 7.5 $, $ y = 3.5 $ → $ x - y = 4 $, $ x + y = 11 $ → 4 and 11 → 411 ≠ 410

No.

Alternatively, maybe the rule is different.

Wait — let’s try another idea.

Is it possible that the rule is:
- $ a + b = (a \times b) $ concatenated with $ (a + b) $?

Check:
- 4 + 2: $ 4×2=8 $, $ 4+2=6 $ → 86 ≠ 26 → no

Or $ (a + b) $ and $ (a × b) $?

4+2: 6 and 8 → 68 ≠ 26 → no

Back to original.

Another thought: in 6 + 5 = 111

- $ 6 - 5 = 1 $
- $ 6 + 5 = 11 $
- Concatenate: "1" + "11" = "111" → yes

But what if we write the sum first? "11" + "1" = "111" → same!

Wait — is it possible that the order is sum then difference?

Let’s test:

For 4 + 2:
- Sum = 6, diff = 2 → 62 ≠ 26 → no

For 8 + 1:
- Sum = 9, diff = 7 → 97 ≠ 79 → no

So no.

So only diff then sum works.

So back to: $ 7 + 3 = 410 $

We need a different pair $ x, y $ such that applying the same rule gives 410.

But only (7,3) gives $ x - y = 4 $, $ x + y = 10 $

Unless...

Wait — what if $ x - y = 4 $, $ x + y = 10 $, but $ x $ and $ y $ are negative?

Try $ x = 7 $, $ y = 3 $ → only solution

Try $ x = -3 $, $ y = -7 $ → $ x - y = 4 $, $ x + y = -10 $ → then concat: "4" + "-10" = "4-10"? Not 410

Not valid.

So seems like only one pair gives 410.

But the problem says: “Find a different pair”

So maybe we misunderstood the meaning of $ x + y = z $

Wait — it says:

> Suppose $ 7 + 3 = z $. Find a different pair $ x, y $ where $ x + y = z $

But in the context of the rule, $ x + y $ means applying the same pattern, not actual addition.

So $ z = 410 $ (the result of the rule)

We need another pair $ x, y $ such that when we apply the rule:
- $ x - y $ concatenated with $ x + y $ = 410

So again, we need:
- $ x - y = 4 $
- $ x + y = 10 $

Only solution: $ x = 7 $, $ y = 3 $

So no other pair?

But maybe there’s another way to get 410.

Wait — what if:
- $ x - y = 41 $
- $ x + y = 0 $ → impossible

Or:
- $ x - y = 4 $, $ x + y = 10 $ → only

But what if:
- $ x - y = 4 $, $ x + y = 10 $ → 410
- But $ x $ and $ y $ are not 7 and 3? No, they are.

Unless we allow non-integers?

Try:
- $ x - y = 4 $
- $ x + y = 10 $
→ $ x = 7 $, $ y = 3 $ — only solution

So no other pair

But the problem says “very few can solve this”, and “find a different pair”

So maybe our rule is wrong

Let’s reconsider.

Alternative idea: Maybe the rule is:

$ a + b = (a \times b) $ concatenated with $ (a + b) $

Try:
- 4 + 2: $ 4×2=8 $, $ 4+2=6 $ → 86 ≠ 26 → no

No.

Or $ (a + b) $ and $ (a × b) $? 6 and 8 → 68 ≠ 26

No.

Wait — what if it’s:
- $ a \times (b + 1) $ or something?

4 + 2 = 26:
- $ 4 × 6 = 24 $, close to 26
- $ 4 × 5 + 6 = 20 + 6 = 26 $? Not clear

Another idea: Look at the results:

- 4 + 2 = 26 → 2 and 6 → 26
- 8 + 1 = 79 → 7 and 9 → 79
- 6 + 5 = 111 → 1 and 11 → 111

Notice: In each case, the first digit(s) is $ a - b $, second is $ a + b $

So for 7 + 3:
- $ 7 - 3 = 4 $
- $ 7 + 3 = 10 $
- So result = 410

So $ z = 410 $

Now, the challenge is: find a different pair $ x, y $ such that when you apply the same rule, you get 410

So we need:
- $ x - y $ concatenated with $ x + y $ = 410

Now, how can we split 410 into two parts?

Possibility 1: "4" and "10" → $ x - y = 4 $, $ x + y = 10 $ → $ x = 7 $, $ y = 3 $ → original

Possibility 2: "41" and "0" → $ x - y = 41 $, $ x + y = 0 $ → $ x = 20.5 $, $ y = -20.5 $ → but $ x + y = 0 $, so concat would be "41" + "0" = "410" → is that allowed?

But $ x + y = 0 $, which is just "0", so concatenation is "41" + "0" = "410"

So if we allow that, then:

- $ x - y = 41 $
- $ x + y = 0 $

Solve:
- Add: $ 2x = 41 $ → $ x = 20.5 $
- $ y = -20.5 $

Then:
- $ x - y = 20.5 - (-20.5) = 41 $
- $ x + y = 0 $
- Concatenate: "41" + "0" = "410"

So the result is 410 — same as $ 7 + 3 $

So this is a different pair: $ x = 20.5 $, $ y = -20.5 $

But is this acceptable? The original examples used integers, but the problem doesn't specify integers.

Also, the problem says "a different pair", so this could work.

But is concatenating "41" and "0" valid? Yes — because $ x + y = 0 $, which is "0", so concatenation is "41" + "0" = "410"

So this is a valid alternative.

Similarly, could we have:
- $ x - y = 410 $, $ x + y = $ empty? No.

Or $ x - y = 4 $, $ x + y = 10 $ — only one solution.

So only two possibilities:
1. $ x - y = 4 $, $ x + y = 10 $ → $ x=7, y=3 $
2. $ x - y = 41 $, $ x + y = 0 $ → $ x=20.5, y=-20.5 $

So the different pair is $ x = 20.5 $, $ y = -20.5 $

But maybe the problem expects integers.

Wait — is there an integer pair?

Try $ x - y = 4 $, $ x + y = 10 $ → only (7,3)

No other integer solutions.

So perhaps the intended answer is that no other pair exists, but the problem says "find a different pair", so likely the non-integer one is acceptable.

But maybe there’s a better interpretation.

Wait — another idea: Maybe the rule is:

$ a + b = (a \times b) $ concatenated with $ (a + b) $, but no — 4×2=8, 4+2=6 → 86 ≠ 26

No.

Wait — what if it's:
- $ (a + b) $ concatenated with $ (a \times b) $? 6 and 8 → 68 ≠ 26

No.

Another idea: Perhaps the result is $ (a \times b) + (a + b) $? 8 + 6 = 14 ≠ 26

No.

Back to original.

Perhaps the concatenation is of $ a - b $ and $ a + b $, but written without leading zeros, etc.

But in 6 + 5 = 111: $ a - b = 1 $, $ a + b = 11 $ → "1" + "11" = "111" → correct

So for $ x - y = 4 $, $ x + y = 10 $ → "4" + "10" = "410"

Now, is there a way to get 410 with different numbers?

Suppose $ x - y = 41 $, $ x + y = 0 $ → as above → "41" + "0" = "410"

So yes.

But maybe the problem allows that.

Alternatively, perhaps the rule is symmetric — but no.

Another possibility: Maybe the rule is:

$ a + b = (a \times (b + 1)) $ concatenated with something?

4 + 2: $ 4 × 3 = 12 $, not 26

No.

I think our original rule is correct.

So final answer:

- $ 7 + 3 = 410 $
- $ z = 410 $
- A different pair $ x, y $ such that $ x + y = z $ (under the same rule) is:
- $ x = 20.5 $, $ y = -20.5 $
- Because $ x - y = 41 $, $ x + y = 0 $, so concatenate "41" and "0" = "410"

But this is messy.

Wait — is there an integer pair?

Try $ x - y = 4 $, $ x + y = 10 $ → only (7,3)

No.

But what if we reverse the roles?

Suppose $ x - y = 4 $, $ x + y = 10 $ → only one solution.

So no.

Unless the rule is not $ a - b $ then $ a + b $, but something else.

Wait — look at 6 + 5 = 111

- $ 6 - 5 = 1 $
- $ 6 + 5 = 11 $
- "1" + "11" = "111"

But what if we do $ a + b $ then $ a - b $? "11" + "1" = "111" → same result!

So for 6 + 5, both orders give 111

But for 4 + 2:
- $ a - b = 2 $, $ a + b = 6 $ → "2" + "6" = "26"
- $ a + b = 6 $, $ a - b = 2 $ → "6" + "2" = "62" ≠ 26

So only one order works.

For 8 + 1:
- $ a - b = 7 $, $ a + b = 9 $ → "7" + "9" = "79"
- "9" + "7" = "97" ≠ 79

So only difference first, then sum.

So no symmetry.

Therefore, only one way to get 410: $ x - y = 4 $, $ x + y = 10 $ → $ x=7, y=3 $

But the problem says "find a different pair"

So perhaps the intended answer is that no other pair exists, but that contradicts the instruction.

Wait — maybe the rule is different.

Another idea: Maybe it's $ a + b = (a \times b) $ concatenated with $ (a + b) $, but adjusted.

4 + 2 = 26:
- $ 4 \times 6 = 24 $, close to 26
- $ 4 \times 5 + 6 = 20 + 6 = 26 $? Not clear

No.

Perhaps it's:
- $ a + b = (a \times (b + 1)) $ concatenated with $ (a + b) $?

4 + 2: $ 4 × 3 = 12 $, $ 4+2=6 $ → 126 ≠ 26

No.

I think our initial rule is correct.

Perhaps the "different pair" is meant to be in the same format, but with different numbers that yield the same result.

But mathematically, only one pair gives $ x - y = 4 $, $ x + y = 10 $

Unless we consider $ x = 3 $, $ y = 7 $ — but $ x - y = -4 $, $ x + y = 10 $ → concat: "-4" + "10" = "-410" ≠ 410

Not valid.

So only (7,3)

But wait — what if $ x - y = 4 $, $ x + y = 10 $, but $ x $ and $ y $ are not 7 and 3? Impossible.

So perhaps the problem has a typo, or we're missing something.

Wait — look at the last line:

> Find a different pair of numbers $ x, y $ where $ x + y = z $

But $ z = 410 $, and $ x + y $ is not addition, but the operation.

So we need $ x + y $ (operation) = 410

So only one pair.

But maybe the intended answer is that $ x = 4 $, $ y = 10 $, but that doesn't make sense.

Another idea: Maybe the rule is:

$ a + b = (a + b) \times (a - b) $ or something?

4 + 2: $ (6) \times (2) = 12 $ ≠ 26

No.

I think the only logical conclusion is:

- $ 7 + 3 = 410 $
- $ z = 410 $
- A different pair that gives the same result is $ x = 20.5 $, $ y = -20.5 $, since:
- $ x - y = 41 $
- $ x + y = 0 $
- Concatenate: "41" + "0" = "410"

So even though it's not integer, it satisfies the rule.

Alternatively, perhaps the problem allows that.

But maybe there's a better interpretation.

Wait — what if the rule is:

$ a + b = (a \times b) + (a + b) $?

4 + 2: $ 8 + 6 = 14 $ ≠ 26

No.

After careful analysis, I believe the only possible answer is:

Final Answer:

- $ 7 + 3 = 410 $
- $ z = 410 $
- A different pair $ x, y $ such that $ x + y = z $ (under the same rule) is $ x = 20.5 $, $ y = -20.5 $

Because:
- $ x - y = 20.5 - (-20.5) = 41 $
- $ x + y = 0 $
- Concatenate: "41" + "0" = "410"

So it works.

If only integers are allowed, then no other pair exists, but the problem implies there is one.

Perhaps the intended answer is that $ x = 4 $, $ y = 10 $, but that doesn't fit.

Given the information, I'll go with the non-integer pair.

But let's check if there's a simpler integer pair.

Wait — what if $ x = 10 $, $ y = 6 $:
- $ x - y = 4 $
- $ x + y = 16 $
- Concatenate: "4" + "16" = "416" ≠ 410

No.

Or $ x = 5 $, $ y = 1 $:
- $ x - y = 4 $
- $ x + y = 6 $
- "4" + "6" = "46" ≠ 410

No.

Only (7,3) and (20.5, -20.5)

So I think that's it.

✔ Solution:

- $ 7 + 3 = 410 $
- $ z = 410 $
- A different pair is $ x = 20.5 $, $ y = -20.5 $

Because:
- $ x - y = 41 $
- $ x + y = 0 $
- Concatenated: "41" + "0" = "410"

Thus, $ x + y = z $ under the same rule.

Answer: $ x = 20.5 $, $ y = -20.5 $