Fill in the letters in the correct order to show the correct path that Emile must take through the equivalent fractions.

To solve this puzzle, Emile must start at the "Start" point and move through a path of equivalent fractions, eventually reaching "Home". The goal is to find a sequence of letters where each fraction is equivalent to the previous one, forming a continuous path from Start to Home.

We are told that he must visit all equivalent fractions on his way home. This suggests that the correct path will be a chain of fractions that are all equal in value, moving from one cell to an adjacent one (likely horizontally or vertically).

Let’s begin by simplifying each fraction and identifying which ones are equivalent.

---

Step 1: Simplify all fractions

| Letter | Fraction | Simplified |
|--------|----------|------------|
| A | 2/3 | 2/3 |
| B | 12/18 | 2/3 |
| C | 8/12 | 2/3 |
| D | 6/9 | 2/3 |
| E | 14/21 | 2/3 |
| F | 48/60 | 4/5 |
| G | 400/500 | 4/5 |
| H | 12/15 | 4/5 |
| I | 4/5 | 4/5 |
| J | 24/30 | 4/5 |
| K | 36/45 | 4/5 |
| L | 16/21 | 16/21 |
| M | 50/40 | 5/4 |
| N | 32/40 | 4/5 |
| O | 12/16 | 3/4 |
| P | 20/25 | 4/5 |
| Q | 6/9 | 2/3 |
| R | 23/40 | 23/40 |
| S | 12/15 | 4/5 |
| T | 80/100 | 4/5 |
| U | 28/35 | 4/5 |
| V | 40/50 | 4/5 |
| W | 3/4 | 3/4 |
| X | 20/29 | 20/29 |

Now group them by simplified value:

---

Group 1: Equivalent to 2/3

- A: 2/3
- B: 12/18 = 2/3
- C: 8/12 = 2/3
- D: 6/9 = 2/3
- E: 14/21 = 2/3
- Q: 6/9 = 2/3

So: A, B, C, D, E, Q → all equal to 2/3

---

Group 2: Equivalent to 4/5

- F: 48/60 = 4/5
- G: 400/500 = 4/5
- H: 12/15 = 4/5
- I: 4/5
- J: 24/30 = 4/5
- K: 36/45 = 4/5
- N: 32/40 = 4/5
- P: 20/25 = 4/5
- S: 12/15 = 4/5
- T: 80/100 = 4/5
- U: 28/35 = 4/5
- V: 40/50 = 4/5

So: F, G, H, I, J, K, N, P, S, T, U, V → all equal to 4/5

---

Group 3: Equivalent to 3/4

- O: 12/16 = 3/4
- W: 3/4

So: O, W

---

Group 4: Others (not equivalent)

- L: 16/21 → not reducible
- M: 50/40 = 5/4 → not matching others
- R: 23/40 → unique
- X: 20/29 → unique

So only 2/3, 4/5, and 3/4 appear multiple times.

But notice: Emile starts at the left side, and the Home is at the far right column (H column), so likely he moves across the grid from left to right.

Also, the instruction says: “visiting all the equivalent fractions on his way.” That may mean he must visit all cells with equivalent fractions — but only one group can form a connected path from Start to Home.

Let’s look at the layout:

```
Row 1: A(2/3) B(2/3) C(2/3) D(2/3) E(2/3) F(4/5) G(4/5) H(4/5) → Home
Row 2: I(4/5) J(4/5) K(4/5) L(16/21) M(5/4) N(4/5) O(3/4) P(4/5)
Row 3: Q(2/3) R(23/40) S(4/5) T(4/5) U(4/5) V(4/5) W(3/4) X(20/29)
```

He starts at A (2/3). So the first fraction is 2/3.

From A, possible adjacent cells:
- Right: B (2/3) ✔
- Down: I (4/5) ✘
- Down-right: J? No, only adjacent.

So he can go from A → B → C → D → E → ... but then what?

After E (2/3), next is F (4/5), not equivalent.

So unless there's a jump, he cannot continue.

But wait — maybe the path doesn’t have to go strictly right? Maybe it can go down, up, etc., as long as adjacent and equivalent.

But if he stays in 2/3 group, he can only go through:
- A, B, C, D, E, Q

Q is below A.

But Q is in bottom-left.

Then from Q, can he go to R? R is 23/40 → no.

So the 2/3 group is isolated in the top-left and bottom-left.

Similarly, 4/5 group is large — spans most of the grid.

But he starts at A (2/3), so unless there's a transition, he can't switch to 4/5.

Unless... maybe the starting point is not part of the path?

Wait — the image shows:

> Start → [I] → ...

Wait! Look carefully.

The arrow points from "Start" to I, not A.

Let me re-read.

It says:

> Start → I

And I is 4/5.

So Emile starts at I, which is 4/5.

That changes everything!

So the start is at I, which is 4/5.

And Home is at the rightmost column, labeled "Home", which is H.

But H is also 4/5.

So we need a path from I to H, visiting only equivalent fractions, i.e., all 4/5.

And we must visit all the 4/5 fractions?

But the instruction says: “visiting all the equivalent fractions on his way.”

Possibly meaning: visit all cells that have equivalent fractions (i.e., all 4/5s), but only those.

But let's check the layout again.

Grid:

```
Row 1: A(2/3) B(2/3) C(2/3) D(2/3) E(2/3) F(4/5) G(4/5) H(4/5) → Home
Row 2: I(4/5) J(4/5) K(4/5) L(16/21) M(5/4) N(4/5) O(3/4) P(4/5)
Row 3: Q(2/3) R(23/40) S(4/5) T(4/5) U(4/5) V(4/5) W(3/4) X(20/29)
```

Start is at I (4/5)

Home is at H (4/5)

We need to go from I to H, passing through all other 4/5 fractions, moving only to adjacent cells (up/down/left/right), and only on 4/5 values.

So let's list all 4/5 cells:

- F (top row, col 6)
- G (top row, col 7)
- H (top row, col 8) → Home
- I (row 2, col 1)
- J (row 2, col 2)
- K (row 2, col 3)
- N (row 2, col 6)
- P (row 2, col 8)
- S (row 3, col 3)
- T (row 3, col 4)
- U (row 3, col 5)
- V (row 3, col 6)
- Also: M is 50/40 = 5/4 → not 4/5
- O is 12/16 = 3/4 → no
- W is 3/4 → no

Wait — did I miss any?

F: 48/60 = 4/5 ✔
G: 400/500 = 4/5 ✔
H: 12/15 = 4/5 ✔
I: 4/5 ✔
J: 24/30 = 4/5 ✔
K: 36/45 = 4/5 ✔
N: 32/40 = 4/5 ✔
P: 20/25 = 4/5 ✔
S: 12/15 = 4/5 ✔
T: 80/100 = 4/5 ✔
U: 28/35 = 4/5 ✔
V: 40/50 = 4/5 ✔

So total 12 cells with 4/5.

Now, can we connect them in a path from I to H, visiting all of them?

Let’s map their positions:

Let’s assign coordinates:

- Row 1: A(1,1), B(1,2), C(1,3), D(1,4), E(1,5), F(1,6), G(1,7), H(1,8)
- Row 2: I(2,1), J(2,2), K(2,3), L(2,4), M(2,5), N(2,6), O(2,7), P(2,8)
- Row 3: Q(3,1), R(3,2), S(3,3), T(3,4), U(3,5), V(3,6), W(3,7), X(3,8)

Now list positions of 4/5:

- F: (1,6)
- G: (1,7)
- H: (1,8) → Home
- I: (2,1)
- J: (2,2)
- K: (2,3)
- N: (2,6)
- P: (2,8)
- S: (3,3)
- T: (3,4)
- U: (3,5)
- V: (3,6)

So now, build a path from I (2,1) to H (1,8), visiting all these 12 cells, only moving to adjacent cells (up/down/left/right), and only on 4/5 cells.

Let’s try to trace a path.

Start: I (2,1)

From I (2,1), adjacent cells:
- Up: A (1,1) → 2/3 → ✘
- Right: J (2,2) → 4/5 ✔
- Down: Q (3,1) → 2/3 → ✘

So only move to J (2,2)

From J (2,2):
- Left: I → already visited
- Right: K (2,3) → 4/5 ✔
- Up: B (1,2) → 2/3 → ✘
- Down: R (3,2) → 23/40 → ✘

→ Go to K (2,3)

From K (2,3):
- Left: J → visited
- Right: L (2,4) → 16/21 → ✘
- Down: S (3,3) → 4/5 ✔
- Up: C (1,3) → 2/3 → ✘

→ Go to S (3,3)

From S (3,3):
- Up: K → visited
- Right: T (3,4) → 4/5 ✔
- Down: ? → none
- Left: R (3,2) → 23/40 → ✘

→ Go to T (3,4)

From T (3,4):
- Left: S → visited
- Right: U (3,5) → 4/5 ✔
- Up: L (2,4) → 16/21 → ✘
- Down: none

→ Go to U (3,5)

From U (3,5):
- Left: T → visited
- Right: V (3,6) → 4/5 ✔
- Up: M (2,5) → 50/40 = 5/4 → ✘
- Down: none

→ Go to V (3,6)

From V (3,6):
- Left: U → visited
- Right: W (3,7) → 3/4 → ✘
- Up: N (2,6) → 4/5 ✔
- Down: none

→ Go to N (2,6)

From N (2,6):
- Down: V → visited
- Up: F (1,6) → 4/5 ✔
- Left: M (2,5) → 5/4 → ✘
- Right: O (2,7) → 3/4 → ✘

→ Go to F (1,6)

From F (1,6):
- Left: E (1,5) → 2/3 → ✘
- Right: G (1,7) → 4/5 ✔
- Down: N → visited
- Up: none

→ Go to G (1,7)

From G (1,7):
- Left: F → visited
- Right: H (1,8) → 4/5 ✔
- Down: O (2,7) → 3/4 → ✘

→ Go to H (1,8) → Home

Now, what about P (2,8)? We haven’t visited it.

P is at (2,8): 20/25 = 4/5 ✔

But from H (1,8), can we go down to P (2,8)? Yes — adjacent.

But we are already at H, and H is Home. So do we need to visit P?

But the path ends at H.

Can we include P?

But if we go to P after H, we’re going away from Home.

Alternatively, could we go to P earlier?

Let’s see: P is at (2,8)

Adjacent cells:
- Up: H (1,8) → 4/5 ✔
- Left: O (2,7) → 3/4 → ✘
- Down: X (3,8) → 20/29 → ✘

So only accessible via H.

So we must go to P before H, or after.

But if we go to P before H, we can go from P → H.

But currently, our path ends at H.

Can we modify the path to go through P?

Currently, we went:

I → J → K → S → T → U → V → N → F → G → H

That’s 11 cells.

Missing: P

Is there another way?

Wait — we missed P.

But P is only reachable from H.

So if we want to include P, we must go:

... → P → H

But then we must reach P before H.

But P is at (2,8), H at (1,8), so yes, we can go from P to H.

But how to get to P?

Only from H or from O, but O is 3/4 → invalid.

So only way to reach P is from H.

So unless we go to H first, then to P, but that would be going back.

But the path must end at Home (H).

So if we go to P, we must go to P before H.

But P can only be reached from H.

So impossible?

Wait — unless we go from P to H, meaning we go to P before H.

But to go to P, we must come from H.

So contradiction.

Therefore, P cannot be reached unless we go through H, but we can’t go to H first.

So P is unreachable unless we go to H first, but then we can't go to P later.

So unless there's a mistake.

Wait — is there another connection?

What about from N (2,6) to P (2,8)? Not adjacent — separated by O (2,7), which is 3/4 → not allowed.

So no direct path.

So P is isolated from the main 4/5 cluster?

But P is at (2,8), adjacent to O (2,7) and H (1,8)

O is 12/16 = 3/4 → not 4/5

H is 4/5 → so only adjacent to H.

So P can only be reached from H.

So to visit P, we must go from H to P, but then we’re not at Home anymore.

But the destination is Home, so we must end at H.

So if we go to P, we’d have to go from H to P, then back to H? But that’s not efficient.

But the problem says: “visiting all the equivalent fractions on his way.”

So perhaps we must visit all 4/5 cells.

But P is one of them.

So we must visit P.

But only way to reach P is from H.

So we must go:

... → H → P → H? But that would require going back.

But then we’re not ending at H — we’d have to end at H.

So unless the path goes:

... → P → H

But to go to P, we need to come from H.

So impossible.

Wait — unless H is not the only way to reach P.

But no — P is only adjacent to O and H.

O is 3/4 → not 4/5.

So P is only reachable from H.

So to visit P, we must go from H to P, but then we're not at Home.

But Home is H.

So unless the path ends at H, we can't visit P.

But we can go:

... → H → P → ??? → H? But no other connections.

So impossible.

Therefore, P cannot be included in a valid path that ends at H.

But P is a 4/5 fraction.

So either:
- The path does not need to visit all equivalent fractions, or
- There’s a mistake in the assumption.

Wait — maybe "visiting all the equivalent fractions" means all fractions equivalent to the starting one, but not necessarily every single one in the grid.

But the instruction says: “visiting all the equivalent fractions on his way.”

Possibly meaning: visit all that are equivalent, but maybe not all 4/5s — maybe only those along the path.

But the wording is ambiguous.

Alternatively, maybe P is not required.

But let’s count how many 4/5s we have:

List: F, G, H, I, J, K, N, P, S, T, U, V → 12 cells.

In our current path: I, J, K, S, T, U, V, N, F, G, H → 11 cells

Missing: P

So unless P is not required, we’re missing one.

But P is at (2,8), and H is at (1,8), so they are adjacent.

But to go from P to H, we must go from P to H.

So if we go to P before H, we must come from H — impossible.

So the only possibility is that P is not part of the path, or the path includes a detour.

But there’s no other way.

Wait — what if we go to P before H, but from a different direction?

No — P has only two neighbors: O (3/4) and H (4/5)

So only H is valid.

So P cannot be reached without going through H.

Therefore, the only way to visit P is to go from H to P, but then you’re not at Home.

But the path must end at Home.

So unless the path goes: ..., H, P, H — but that would require returning, which is not typical for such puzzles.

Alternatively, maybe P is not required.

But let’s double-check if P is really 4/5.

P: 20/25 = 4/5 → yes.

So it is.

Perhaps the path doesn’t need to visit all 4/5s — just the ones on the route.

But the instruction says: “visiting all the equivalent fractions on his way.”

“On his way” — so only the ones along the path.

So perhaps not all 4/5s need to be visited — only those that are on the path.

But then why list so many?

Alternatively, maybe the equivalent fractions are not all 4/5 — maybe the path is based on a specific value.

But Emile starts at I (4/5), so the path must consist of 4/5.

But then how to include P?

Wait — maybe the path doesn't have to be continuous in terms of adjacency, but the puzzle implies it's a maze-like path.

Another idea: maybe the path is not limited to 4/5 — maybe it switches between groups.

But he starts at I (4/5), so first is 4/5.

Then must go to adjacent 4/5.

So only 4/5.

So the entire path must be 4/5.

So we must include all 4/5 cells that are reachable from I to H.

But P is only reachable from H.

So if we go to P, we must go from H to P, but then we're not at Home.

So unless the final step is from P to H, but that would mean we end at H, which is fine.

But to go from P to H, we must be at P first.

But to get to P, we must come from H.

So only way is to go from H to P, then back to H.

But that would require two steps: H → P → H.

But then we're at H, but we've visited P.

But is that allowed?

Yes, if we allow revisiting.

But typically in such puzzles, you don't revisit.

Moreover, the answer requires 11 blanks.

Let’s count the number of cells in the path.

Start at I, end at H.

How many cells?

If we go:

I → J → K → S → T → U → V → N → F → G → H

That’s 11 cells.

But we skipped P.

Is there a way to include P?

Only if we go:

... → H → P → H

But then we'd have to go from H to P and back.

But that would make the path longer.

And we'd have to write P twice.

But the blank has 11 spaces.

So likely, the intended path is the one we have, and P is not required.

But let’s check if there’s another path.

Alternative: from K (2,3), instead of going down to S, can we go elsewhere?

No — only S is valid.

From V (3,6), we go to N (2,6)

But is there a way to go from V to P? No — not adjacent.

Another idea: can we go from G to P? G is (1,7), P is (2,8) — not adjacent.

Only if diagonal, but usually only orthogonal.

So no.

So the only way to reach P is from H.

So unless the path is:

I → J → K → S → T → U → V → N → F → G → H → P → H

But then we end at H, but we've visited P.

But we've used 13 cells, and the answer has only 11 blanks.

So likely not.

Moreover, we would have to repeat H.

So probably not.

Therefore, the intended path is the one we have, and P is not required.

But why is P there?

Perhaps it's a distractor.

Maybe the path is not required to visit all 4/5s, but only the ones that are on the way.

So our path:

I → J → K → S → T → U → V → N → F → G → H

Letters: I, J, K, S, T, U, V, N, F, G, H

That’s 11 letters.

Now, check if all are 4/5:

- I: 4/5 ✔
- J: 24/30 = 4/5 ✔
- K: 36/45 = 4/5 ✔
- S: 12/15 = 4/5 ✔
- T: 80/100 = 4/5 ✔
- U: 28/35 = 4/5 ✔
- V: 40/50 = 4/5 ✔
- N: 32/40 = 4/5 ✔
- F: 48/60 = 4/5 ✔
- G: 400/500 = 4/5 ✔
- H: 12/15 = 4/5 ✔

All good.

And all are adjacent:

- I to J: right
- J to K: right
- K to S: down
- S to T: right
- T to U: right
- U to V: right
- V to N: up
- N to F: up
- F to G: right
- G to H: right

Yes, all valid moves.

And we start at I, end at H.

Now, is there a shorter path? No, because we need to visit all the 4/5s in the lower part.

But we missed P, but maybe it's not required.

Perhaps the "equivalent fractions" refers to a specific value, and not all 4/5s.

But since we start at 4/5, and end at 4/5, and all are equivalent, it makes sense.

Moreover, the path visits 11 cells, and there are 11 blanks.

So likely this is the intended path.

So Emile's path is:

I, J, K, S, T, U, V, N, F, G, H

Let’s write it:

Emile's path is: I, J, K, S, T, U, V, N, F, G, H

Now, check if any other cell could be included.

For example, is there a way to go from K to something else?

No.

Or from V to P? No.

So this seems optimal.

Therefore, the answer is:

I, J, K, S, T, U, V, N, F, G, H

✔ All equivalent to 4/5
✔ Adjacent moves
✔ Starts at I
✔ Ends at H (Home)
✔ Uses 11 cells, matches number of blanks

So the final answer is:

I, J, K, S, T, U, V, N, F, G, H