Math maze activity where students multiply and divide by powers of 10 to reach the highest possible value.

Let’s solve this maze step by step to find the path that gives the biggest value at the end.

We start with 5.

At each step, we can choose to go right or down (based on arrows), and we multiply or divide by powers of 10. Our goal is to get the largest possible number at the END box.

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Step-by-step exploration:

We’ll try different paths from START to END and calculate the result for each. Since there are many paths, we’ll focus on ones that mostly multiply (especially by big numbers like ×1000) and avoid dividing unless necessary.

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#### Path 1: Try going DOWN first (×100)

Start: 5

→ Down: ×100 → 5 × 100 = 500

Now from here, we can go right (÷100) or down (÷10). Let’s try both.

##### Subpath 1a: Go right (÷100)

500 ÷ 100 = 5

Then from here, we can go down (×100) → 5 × 100 = 500

Then right (×10) → 500 × 10 = 5000

Then down (×10) → 5000 × 10 = 50,000

Then left (÷100) → 50,000 ÷ 100 = 500

Then left (÷10) → 500 ÷ 10 = 50

Ends at 50 — not great.

##### Subpath 1b: From 500, go down (÷10)

500 ÷ 10 = 50

Then right (×100) → 50 × 100 = 5000

Then down (÷100) → 5000 ÷ 100 = 50

Then left (÷10) → 50 ÷ 10 = 5

Ends at 5 — worse.

So starting with ×100 doesn’t seem best.

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#### Path 2: Start by going RIGHT (×10)

Start: 5

→ Right: ×10 → 5 × 10 = 50

Now from here, we can go right (÷100) or down (÷10).

##### Subpath 2a: Go right (÷100)

50 ÷ 100 = 0.5

That’s small — probably not good. But let’s see where it leads.

From 0.5, go down (×1000) → 0.5 × 1000 = 500

Then left (×100) → 500 × 100 = 50,000

Then down (÷100) → 50,000 ÷ 100 = 500

Then left (×10) → 500 × 10 = 5000

Then down (÷100) → 5000 ÷ 100 = 50

Then left (÷10) → 50 ÷ 10 = 5

Ends at 5 — bad.

##### Subpath 2b: From 50, go down (÷10)

50 ÷ 10 = 5

Back to 5 — not helpful.

Wait — maybe we missed a better path?

Let’s try another approach: look for paths that include ×1000 early and avoid division as much as possible.

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#### Path 3: Start → Right (×10) → then Down (÷10)? No, that cancels out.

What if we go:

Start → Right (×10) → then Right again? Wait, no arrow directly right from second box — only down or back left.

Actually, looking at the grid layout:

The maze is arranged in a 4x4 grid of boxes (including START and END). Arrows show allowed directions.

Let me map the positions:

Label rows 1 to 4 (top to bottom), columns A to D (left to right).

- START is at (1,A) = 5
- END is at (4,D)

Arrows:

From (1,A):
- Right → (1,B) via ×10
- Down → (2,A) via ×100

From (1,B):
- Left ← (1,A)
- Right → (1,C) via ÷100
- Down → (2,B) via ÷10

From (1,C):
- Left ← (1,B)
- Down → (2,C) via ×1000

From (2,A):
- Up ↑ (1,A)
- Right → (2,B) via ÷100
- Down → (3,A) via ÷10

From (2,B):
- Left ← (2,A)
- Right → (2,C) via ×100
- Down → (3,B) via ×100

From (2,C):
- Left ← (2,B)
- Up ↑ (1,C)
- Down → (3,C) via ÷100

From (3,A):
- Up ↑ (2,A)
- Right → (3,B) via ×100
- Down → (4,A) via ×1000

From (3,B):
- Left ← (3,A)
- Right → (3,C) via ×10
- Down → (4,B) via ÷100

From (3,C):
- Left ← (3,B)
- Up ↑ (2,C)
- Down → (4,C) via ×10

From (4,A):
- Up ↑ (3,A)
- Right → (4,B) via ÷10

From (4,B):
- Left ← (4,A)
- Right → (4,C) via ÷100
- Up ↑ (3,B)

From (4,C):
- Left ← (4,B)
- Up ↑ (3,C)
- Right → (4,D) [END] via ÷100? Wait, no — from (4,C) to END is via ÷100? Actually, looking at image:

In row 4:

(4,A) --÷10--> (4,B) --÷100--> (4,C) --???--> END?

Wait, actually, from the image description:

Bottom row:

Leftmost box → ÷10 → middle box → ÷100 → END box.

And above END box is ×10 from (3,C).

Also, from (3,C) down is ×10 to (4,C), then from (4,C) left is ÷100 to (4,B), etc.

But to reach END, you must come from (4,C) via ÷100? Or is END connected directly?

Looking back at original problem statement: “Find your way through the maze to give the biggest value at the end.”

And the END box is in bottom right.

From the diagram logic, likely:

To enter END, you come from the left (from (4,C)) via ÷100? Or from above?

Actually, in the image, the arrow into END is from the left: (4,C) → END via ÷100? But that would reduce the number.

Wait — perhaps I misread.

Let me re-express the correct path that maximizes the result.

After trying several paths, let’s try this one:

Start at 5

→ Go DOWN: ×100 → 500

→ Go DOWN again: ÷10 → 50

→ Go RIGHT: ×100 → 5000

→ Go DOWN: ÷100 → 50

→ Go LEFT: ÷10 → 5 → too small.

Not good.

Another try:

Start → RIGHT: ×10 → 50

→ DOWN: ÷10 → 5 → stuck.

No.

What if:

Start → RIGHT: ×10 → 50

→ RIGHT: ÷100 → 0.5

→ DOWN: ×1000 → 500

→ LEFT: ×100 → 50,000

→ DOWN: ÷100 → 500

→ LEFT: ×10 → 5000

→ DOWN: ÷100 → 50

→ LEFT: ÷10 → 5

Still ends at 5.

This isn't working. Maybe we need to go all the way down early?

Try:

Start → DOWN: ×100 → 500

→ DOWN: ÷10 → 50

→ RIGHT: ×100 → 5000

→ RIGHT: ×10 → 50,000

→ DOWN: ×10 → 500,000

→ LEFT: ÷100 → 5,000

→ LEFT: ÷10 → 500

Ends at 500 — better!

Let’s write that path clearly:

Path:

1. Start: 5
2. Down (×100) → 5 × 100 = 500
3. Down (÷10) → 500 ÷ 10 = 50
4. Right (×100) → 50 × 100 = 5,000
5. Right (×10) → 5,000 × 10 = 50,000
6. Down (×10) → 50,000 × 10 = 500,000
7. Left (÷100) → 500,000 ÷ 100 = 5,000
8. Left (÷10) → 5,000 ÷ 10 = 500

Ends at 500

Is there a better path?

What if after step 5 (50,000), instead of going down, we go... but from (3,C), down is ×10 to (4,C), which we did.

Alternatively, from (3,B) [which is 5,000], could we go down to (4,B) via ÷100? That would be 5,000 ÷ 100 = 50, then right to (4,C) via ÷100? 50 ÷ 100 = 0.5, then to END? Worse.

Another idea: Can we hit ×1000 somewhere?

Earlier we had a path that went to 0.5 then ×1000 to 500, but then we divided a lot.

What if:

Start → RIGHT: ×10 → 50
→ RIGHT: ÷100 → 0.5
→ DOWN: ×1000 → 500
→ LEFT: ×100 → 50,000
→ DOWN: ÷100 → 500
→ LEFT: ×10 → 5,000
→ DOWN: ÷100 → 50
→ LEFT: ÷10 → 5

Ends at 5 — no.

Wait — what if from 50,000 (after ×100 from 500), we go down to (3,B) via ×100? No, from (2,C) down is ÷100.

Let’s list all possible paths systematically? Too long.

Perhaps the maximum is 500,000 before final divisions.

But we have to end at END, which requires two more steps: from (4,C) to END via ÷100? Or is there a direct connection?

Looking back at the original image description:

In the bottom row:

Box (4,A) --÷10--> (4,B) --÷100--> (4,C) --???--> END

Actually, in the image, the END box is to the right of (4,C), and the arrow from (4,C) to END is labeled ÷100? Or is it from (4,B) to END?

I think I made a mistake.

Let me reinterpret the maze based on standard such mazes.

Typically, the END is reached from the left or above.

In the given diagram:

- The last operation before END is from the box to its left, via ÷100.

Because in the bottom row, it shows:

[box] --÷10--> [box] --÷100--> [END]

So to reach END, you must come from the immediate left box, and apply ÷100.

Similarly, the box above END has an arrow down to it? In the image, from (3,C) down is ×10 to (4,C), then from (4,C) left is ÷100 to (4,B), etc.

Actually, let's assume the following connectivity for the bottom row:

Position (4,A) -> (4,B) via ÷10

(4,B) -> (4,C) via ÷100

(4,C) -> END via ???

But in the image, the END box is separate, and the arrow into END is from the left, so likely (4,C) -> END via some operation.

Looking at the user's image description: "÷ 100" arrow pointing to END from the left.

So probably, the box immediately left of END is (4,C), and the operation to enter END is ÷100.

Therefore, whatever value is in (4,C), when you do ÷100, you get the END value.

So to maximize END, we want to maximize the value in (4,C), then divide by 100.

So our goal is to get the largest possible number in (4,C), then divide by 100 for the final answer.

How to get to (4,C)?

From the diagram:

(4,C) can be reached from:

- Above: (3,C) via ×10

- Left: (4,B) via ? But arrow from (4,B) to (4,C) is not shown; usually arrows indicate direction.

In the image, from (3,C) down is ×10 to (4,C)

From (4,B) right is ÷100 to (4,C)? Or to END?

I think there's confusion.

Let me define the grid properly.

Assume the maze has 4 rows and 4 columns of decision points, but START and END are included.

Based on common such worksheets, here's a reliable path that gives a large number:

Start at 5

Go DOWN: ×100 → 500

Go DOWN: ÷10 → 50

Go RIGHT: ×100 → 5,000

Go RIGHT: ×10 → 50,000

Go DOWN: ×10 → 500,000 [this is now in (4,C)? Let's say]

Then from (4,C), to reach END, if the operation is ÷100, then 500,000 ÷ 100 = 5,000

Is that possible?

Let's trace the positions:

- (1,A): START 5

- (2,A): 5 × 100 = 500

- (3,A): 500 ÷ 10 = 50

- (3,B): 50 × 100 = 5,000

- (3,C): 5,000 × 10 = 50,000

- (4,C): 50,000 × 10 = 500,000 [if down from (3,C) is ×10]

Then from (4,C) to END: if the arrow is ÷100, then 500,000 ÷ 100 = 5,000

Yes! And this matches the diagram: from (3,C) down is ×10 to (4,C), then from (4,C) to END is ÷100.

In the bottom row, the operations are between boxes, but END is separate.

In the image, the last part is:

... -> [box] --÷100--> END

And that box is (4,C), which comes from (3,C) via ×10.

So yes, this path works.

Let's confirm the sequence:

1. Start: 5

2. Down to (2,A): ×100 → 5 * 100 = 500

3. Down to (3,A): ÷10 → 500 / 10 = 50

4. Right to (3,B): ×100 → 50 * 100 = 5,000

5. Right to (3,C): ×10 → 5,000 * 10 = 50,000

6. Down to (4,C): ×10 → 50,000 * 10 = 500,000

7. Right to END: ÷100 → 500,000 / 100 = 5,000

Final value: 5,000

Is there a path that gives more?

What if we go to (2,C) earlier?

For example:

Start → Right: ×10 → 50

→ Right: ÷100 → 0.5

→ Down: ×1000 → 500

→ Left: ×100 → 50,000

→ Down: ÷100 → 500

→ Left: ×10 → 5,000

→ Down: ÷100 → 50

→ Left: ÷10 → 5

Worse.

Or from 50,000 at (2,C), go down to (3,C) via ÷100 → 500, then down to (4,C) via ×10 → 5,000, then to END ÷100 → 50. Worse.

Another path:

Start → Down: ×100 → 500

→ Right: ÷100 → 5

→ Down: ×100 → 500

→ Right: ×100 → 50,000

→ Down: ÷100 → 500

→ Right: ×10 → 5,000

→ Down: ×10 → 50,000

→ Then to END? From (4,C) ÷100 → 500

Less than 5,000.

So the path giving 5,000 seems best.

But wait — in the path I described, step 6 is down to (4,C) via ×10, getting 500,000, then ÷100 to END = 5,000.

Is there a way to get to (4,C) with a larger number?

What if we do:

Start → Down: ×100 → 500

→ Down: ÷10 → 50

→ Right: ×100 → 5,000

→ Down: ×100 → 500,000 [to (4,B)?]

From (3,B) down is ÷100 to (4,B), according to diagram.

In the image, from (3,B) down is ÷100 to (4,B)

Then from (4,B) right is ÷100 to (4,C)? Or to END?

Assume:

From (3,B) down: ÷100 → (4,B)

From (4,B) right: ÷100 → (4,C)

From (4,C) right: ÷100 → END? That would be three divisions.

But in the diagram, the bottom row has:

(4,A) --÷10--> (4,B) --÷100--> (4,C) --???--> END

And from (3,C) down is ×10 to (4,C)

So (4,C) can be reached from above or from left.

If we come from left to (4,C), it might be from (4,B) via some operation.

But to maximize, coming from above with ×10 is better.

In our best path, we have 500,000 in (4,C), then ÷100 to END = 5,000.

Can we get more than 500,000 in (4,C)?

Suppose we go:

Start → Right: ×10 → 50

→ Down: ÷10 → 5

→ Right: ×100 → 500

→ Down: ×100 → 50,000

→ Right: ×10 → 500,000 [to (3,C)?]

From (3,B) right is ×10 to (3,C), so 50,000 * 10 = 500,000 at (3,C)

Then down to (4,C): ×10 → 5,000,000

Then to END: ÷100 → 50,000

Oh! This is better!

Let's write this path:

1. Start: 5

2. Right to (1,B): ×10 → 5 * 10 = 50

3. Down to (2,B): ÷10 → 50 / 10 = 5

4. Right to (2,C): ×100 → 5 * 100 = 500

5. Down to (3,C): ÷100 → 500 / 100 = 5 [wait, no!]

Mistake.

From (2,C) down is ÷100 to (3,C), so 500 / 100 = 5

Then from (3,C) right? No, (3,C) is column C, row 3.

From (3,C) down is ×10 to (4,C), so 5 * 10 = 50, then ÷100 to END = 0.5 — bad.

Not good.

From step 4: at (2,C) = 500

Instead of down, go left? But left is to (2,B), which is 5, then down to (3,B) via ×100 → 500, then right to (3,C) via ×10 → 5,000, then down to (4,C) via ×10 → 50,000, then ÷100 to END = 500 — still less than 5,000.

Back to the earlier good path.

Another try:

Start → Down: ×100 → 500

→ Right: ÷100 → 5

→ Down: ×100 → 500

→ Right: ×100 → 50,000

→ Down: ÷100 → 500

→ Right: ×10 → 5,000

→ Down: ×10 → 50,000 [to (4,C)?]

From (3,C) down is ×10 to (4,C), so if (3,C) is 5,000, then (4,C) = 50,000, then ÷100 to END = 500.

Not better.

Perhaps the first path I found is best: ending with 5,000.

But let's double-check the path that gave us 500,000 in (4,C):

- Start: 5

- Down to (2,A): ×100 → 500

- Down to (3,A): ÷10 → 50

- Right to (3,B): ×100 → 5,000

- Right to (3,C): ×10 → 50,000

- Down to (4,C): ×10 → 500,000

- To END: ÷100 → 5,000

Yes.

Is there a path to get to (4,C) with 5,000,000?

For example, if we can get 500,000 to (3,C), then ×10 to 5,000,000 at (4,C), then ÷100 to 50,000.

How to get 500,000 to (3,C)?

From (3,B) right is ×10, so if (3,B) is 50,000, then (3,C) = 500,000.

How to get 50,000 to (3,B)?

From (3,A) right is ×100, so if (3,A) is 500, then (3,B) = 50,000.

How to get 500 to (3,A)?

From (2,A) down is ÷10, so if (2,A) is 5,000, then (3,A) = 500.

How to get 5,000 to (2,A)?

From (1,A) down is ×100, so 5 * 100 = 500, not 5,000.

Unless we go right first.

Start → Right: ×10 → 50

→ Down: ÷10 → 5

→ Down: ×100 → 500 [to (3,A)?]

From (2,B) down is ×100 to (3,B), not to (3,A).

From (2,A) down is ÷10 to (3,A).

So to get to (2,A) with 5,000, impossible from start.

Maximum at (2,A) is 500.

So (3,A) max is 50 (from 500 ÷10).

Then (3,B) = 50 * 100 = 5,000

(3,C) = 5,000 * 10 = 50,000

(4,C) = 50,000 * 10 = 500,000

END = 500,000 / 100 = 5,000

Seems consistent.

Perhaps there's a path using the top-right ×1000.

Start → Right: ×10 → 50

→ Right: ÷100 → 0.5

→ Down: ×1000 → 500

→ Left: ×100 → 50,000

→ Down: ÷100 → 500

→ Left: ×10 → 5,000

→ Down: ÷100 → 50

→ Left: ÷10 → 5

No.

Or from 50,000 at (2,C), go down to (3,C) via ÷100 → 500, then down to (4,C) via ×10 → 5,000, then to END ÷100 → 50.

Worse.

So I think 5,000 is the maximum.

But let's confirm with another path.

What if:

Start → Down: ×100 → 500

→ Right: ÷100 → 5

→ Down: ×100 → 500

→ Right: ×100 → 50,000

→ Down: ÷100 → 500

→ Right: ×10 → 5,000

→ Down: ×10 → 50,000 [at (4,C)]

Then to END: ÷100 → 500

Less than 5,000.

So yes, the path giving 5,000 is best.

Final Answer: 5000