Number Grid Challenge 4: Target 100 - A math puzzle worksheet where students find routes from start to finish with a total of 100.

To solve this puzzle, we need to find paths from the START (top-left) to the FINISH (bottom-right).
The rules are:
1. You can only move Right or Down.
2. The total of all numbers on your path must equal exactly 100.
3. There are 3 correct routes.

Let's break down the grid into coordinates to make it easier to track. Let's say Row 1 is the top row and Column 1 is the left column.
- Start is at (Row 1, Col 1). Note: The "START" square itself doesn't have a number, but usually in these mazes, you start counting from the first number you enter. However, looking at the grid, the green square says "START". Let's assume the value of the start square is 0 as per the instruction "Start the maze with zero." So we add the numbers in the squares we land on.
- Finish is at (Row 5, Col 5).

Let's list the numbers in the grid:
Row 1: [Start/0], 14, 7, 13, 15
Row 2: 12, 9, 11, 26, 21
Row 3: 16, 5, 12, 19, 10
Row 4: 13, 10, 17, 24, 11
Row 5: 8, 20, 4, 15, Finish

Wait, let's re-read carefully: "Start the maze with zero." This implies the starting square contributes 0 to the sum. We need to reach a total of 100 by the time we hit the "FINISH" square. The finish square is yellow and labeled "FINISH". Does it have a value? Usually, the end square is just the endpoint. Let's assume we sum the numbers in the cells we pass through *excluding* the start (which is 0) and *including* the last number before "Finish"? Or does the "Finish" cell contain a number? Looking at the image, the cell labeled "FINISH" is in the bottom right corner. The cell to its left is 15. The cell above it is 11. The cell labeled "FINISH" does not have a number inside it, just the word "FINISH". So we likely sum the numbers in the path leading up to and including the final step into the finish zone? Or perhaps the path includes the number in the square *before* the finish label?

Let's look at the standard interpretation for these "Math Salamanders" worksheets. Usually, you sum every number you land on. The "START" square is 0. The "FINISH" square is just the destination. So we need the sum of the numbers in the intermediate squares to be 100.

Let's trace possible paths. A path from top-left (1,1) to bottom-right (5,5) requires exactly 4 moves Right and 4 moves Down. Total 8 moves. We will land on 8 numbered squares (since the start is 0). Wait, if we start at (1,1) which is 0, and move to (1,2), that's one number. To get to (5,5), we make 8 steps. So we will collect 8 numbers.

Let's verify the number of steps.
From (1,1) to (5,5):
Rows: 1 -> 2 -> 3 -> 4 -> 5 (4 downs)
Cols: 1 -> 2 -> 3 -> 4 -> 5 (4 rights)
Total 8 steps.
So we will sum 8 numbers.

Let's try to find the 3 routes.

Route 1 Attempt:
Let's try going mostly right first, then down.
Path: R, R, R, R, D, D, D, D
Squares:
(1,2)=14
(1,3)=7
(1,4)=13
(1,5)=15
(2,5)=21
(3,5)=10
(4,5)=11
(5,5)=Finish (no number?)
If the finish square has no number, the sum is $14+7+13+15+21+10+11 = 91$. This is not 100.
Maybe the "Finish" square implies we stop there. What if the last square *is* included? But it has no number.
Let's look at another possibility. Maybe the start square isn't 0? "Start the maze with zero." Okay, so current sum = 0.
Move to 14. Sum = 14.
...
Sum = 91. Too low.

Let's try a different path. We need higher numbers. The big numbers are 26, 24, 21, 20, 19, 17, 16, 15, 15, 14, 13, 13, 12, 12, 11, 11, 10, 10, 9, 8, 7, 5, 4.
Target: 100.
Average needed per step: $100 / 8 = 12.5$. So we need a mix of high and average numbers.

Let's work backward from the finish or forward with pruning.

Let's try Path A:
Go Down early to get bigger numbers in the lower rows?
Row 2 has 26. Row 4 has 24. Row 5 has 20.
Let's try to include 26.
To get to 26 at (2,4), we could come from (1,4)=13 or (2,3)=11.
Let's try reaching (2,4)=26.
Path to (2,4):
Option 1: R, R, R, D. Squares: 14, 7, 13, 26. Sum = $14+7+13+26 = 60$.
Remaining moves to (5,5): From (2,4) to (5,5) needs 3 Downs and 1 Right.
We need $100 - 60 = 40$ more from 4 numbers. Average 10.
Possible next steps from 26:
- Down to 19 (3,4). Sum = $60+19=79$. Need 21 from 3 numbers.
- From 19, can go D to 24 (4,4) or R to 10 (3,5).
- If D to 24: Sum = $79+24=103$. Already over 100. And we still have 2 moves left. So this branch is dead.
- If R to 10: Sum = $79+10=89$. Need 11 from 2 numbers.
- From 10 (3,5), must go D to 11 (4,5). Sum = $89+11=100$.
- From 11 (4,5), must go D to Finish (5,5).
- Wait, do we count the number in the Finish square? The square at (5,5) is labeled FINISH. It has no number.
- If the path ends *at* the finish line, do we include the last number entered?
- Let's check the count.
- Path so far: (1,2)14, (1,3)7, (1,4)13, (2,4)26, (3,4)19, (3,5)10, (4,5)11.
- Moves: R, R, R, D, D, R, D.
- Current position: (4,5). Next move is D to (5,5) Finish.
- Numbers summed: $14+7+13+26+19+10+11 = 100$.
- This sums to exactly 100 with 7 numbers.
- But we made 7 moves to get to (4,5). To get to (5,5) we need one more move.
- Total moves from Start(1,1) to Finish(5,5) is 8 moves.
- So we visit 8 numbered squares?
- Let's re-evaluate the coordinates.
- Start at (1,1). Value 0.
- Move 1: to (1,2) Val 14.
- Move 2: to (1,3) Val 7.
- Move 3: to (1,4) Val 13.
- Move 4: to (2,4) Val 26.
- Move 5: to (3,4) Val 19.
- Move 6: to (3,5) Val 10.
- Move 7: to (4,5) Val 11.
- Move 8: to (5,5) Finish.
- If the Finish square has no value, the sum is of the 7 numbers visited before entering the finish square? Or does the 8th square have a value?
- In the grid, the square at (5,5) is the Finish square. It is colored yellow. It does not contain a digit.
- The square at (4,5) contains 11.
- The square at (5,4) contains 15.
- If the sum of the numbers on the path is 100, and we found a path where the first 7 numbers sum to 100, that might be a valid route if the rule implies "sum the numbers you pass through". Since the last step lands on "Finish" (no number), the sum remains 100.
- Let's assume this is Route 1:
14 → 7 → 13 → 26 → 19 → 10 → 11
Sum: $14+7=21$; $21+13=34$; $34+26=60$; $60+19=79$; $79+10=89$; $89+11=100$.
Moves: Right, Right, Right, Down, Down, Right, Down.
Let's check if this is a valid path on the grid.
Start (1,1) -> R (1,2) -> R (1,3) -> R (1,4) -> D (2,4) -> D (3,4) -> R (3,5) -> D (4,5) -> D (5,5 Finish).
Yes, this is a valid path.

Route 2 Attempt:
Let's look for other combinations.
We need 8 numbers if the finish square had a number, but since it doesn't, we are summing the numbers in the 7 squares preceding the finish? Or maybe 8 squares if we count the start as 0? No, start is 0.
Actually, let's look at the path length again.
Start (1,1). End (5,5).
Steps: 8.
Cells visited excluding start: 8 cells.
Cell 1: (r,c)
...
Cell 8: (5,5).
If Cell 8 (Finish) has no number, we sum Cells 1 through 7? That seems arbitrary.
Usually, in these puzzles, every square you land on has a number. Here, the Finish square is special.
However, look at the square (5,4) which is 15. And (4,5) which is 11.
If I enter Finish from (5,4), the last number added is 15.
If I enter Finish from (4,5), the last number added is 11.

Let's test if there are paths where the sum of 8 numbers is 100. This would imply the Finish square *does* have a hidden value or I'm miscounting. But "Start with zero" suggests the start is 0. If the finish was also 0, we'd sum 8 numbers. But Finish is not 0.
Let's stick with the hypothesis that we sum the numbers in the squares traversed *before* stepping onto the Finish tile, OR the Finish tile is just the end marker and the last number collected is the one in the square immediately preceding it.
In Route 1, we collected 7 numbers.
Is it possible to collect 8 numbers?
If we go through (5,4)=15 then to Finish, the last number is 15.
Let's try to find a path ending with 15 that sums to 100.
So, sum of first 6 numbers + 15 = 100 => Sum of first 6 = 85.
Or sum of 7 numbers + 15? No, total 8 steps.
Let's assume the standard rule: Sum all numbers in the path. The path consists of the squares you occupy.
Square 1 to Square 8.
If Square 8 is "Finish" and has no number, maybe it counts as 0?
If Finish = 0, then Sum of 7 numbers + 0 = 100.
This matches Route 1 perfectly.
So, Rule: Sum the numbers in the 7 squares before the Finish square (which acts as 0).
Or simply: The path consists of 8 moves. We sum the values of the 8 squares landed on. If the last square is "Finish", its value is effectively 0 for the sum? Or maybe the problem implies the path *ends* when you reach 100? No, "finish the maze with a total of 100".

Let's look for other routes that sum to 100 using 7 numbers (entering Finish from Top or Left).

Candidate Route 2:
Try entering Finish from the left, i.e., coming from (5,4)=15.
So the last number is 15.
We need the previous 6 numbers to sum to $100 - 15 = 85$.
Path ends: ... -> (5,4)=15 -> Finish.
Preceding square must be (5,3)=4 or (4,4)=24.

Case A: Come from (4,4)=24.
Sequence ends: ..., 24, 15.
Sum so far: $24+15=39$.
Need remaining 5 numbers to sum to $100 - 39 = 61$.
Path to (4,4): Needs 3 Rights and 3 Downs from Start.
Let's try to find 5 numbers summing to 61 ending at (4,4).
Average ~12.
Let's try going through the high number 26 again?
If we use 26 at (2,4):
Path to (2,4): 14, 7, 13, 26. (Sum 60).
From (2,4) to (4,4): Down to (3,4)=19, Down to (4,4)=24.
Numbers: 19, 24.
Total Sum: $60 (first 4) + 19 + 24 + 15 (last) = 118$. Too high.

Let's try avoiding 26.
Try going down early.
(1,1)->D(2,1)=12.
(2,1)->D(3,1)=16.
(3,1)->D(4,1)=13.
(4,1)->R(4,2)=10.
(4,2)->R(4,3)=17.
(4,3)->R(4,4)=24.
(4,4)->R(4,5)... wait, we want to end at (5,4).
From (4,4)=24, go D to (5,4)=15.
Path: 12, 16, 13, 10, 17, 24, 15.
Sum: $12+16=28$; $28+13=41$; $41+10=51$; $51+17=68$; $68+24=92$; $92+15=107$. Too high.

Let's adjust. Replace some high numbers with lower ones.
Instead of 17, maybe go through 12?
Path: 12, 16, 13, 10...
From (4,2)=10, go D to (5,2)=20?
(5,2)=20 -> R (5,3)=4 -> R (5,4)=15.
Path: 12, 16, 13, 10, 20, 4, 15.
Sum: $12+16=28$; $28+13=41$; $41+10=51$; $51+20=71$; $71+4=75$; $75+15=90$. Too low.

Let's try mixing.
We need sum 85 for first 6 numbers ending at (4,4) or (5,3).
Let's try ending at (5,3)=4, then R to (5,4)=15.
Last two: 4, 15. Sum 19.
Need first 5 numbers to sum to $100 - 19 = 81$.
Ends at (5,3). Preceding is (5,2)=20 or (4,3)=17.

Subcase B1: Come from (4,3)=17.
Seq: ..., 17, 4, 15.
Sum last 3: $17+4+15=36$.
Need first 4 numbers to sum to $100 - 36 = 64$.
Ends at (4,3).
Possible paths to (4,3):
1. 14, 7, 11, 17?
Path: R(14), R(7), D(11 is at 2,3? No, 2,3 is 11. Yes.), D(12 is at 3,3? No, 3,3 is 12. Wait.
Grid:
R1: 14, 7, 13, 15
R2: 9, 11, 26, 21
R3: 5, 12, 19, 10
R4: 10, 17, 24, 11

Let's re-map carefully.
(1,2)=14, (1,3)=7, (1,4)=13, (1,5)=15
(2,1)=12, (2,2)=9, (2,3)=11, (2,4)=26, (2,5)=21
(3,1)=16, (3,2)=5, (3,3)=12, (3,4)=19, (3,5)=10
(4,1)=13, (4,2)=10, (4,3)=17, (4,4)=24, (4,5)=11
(5,1)=8, (5,2)=20, (5,3)=4, (5,4)=15, (5,5)=Fin

Back to Subcase B1: Need 4 numbers summing to 64, ending at (4,3)=17.
Try: 14, 7, 11, 17?
Path: (1,2)14 -> R(1,3)7 -> D(2,3)11 -> D(3,3)12 -> D(4,3)17.
Wait, that's 5 numbers: 14, 7, 11, 12, 17.
Sum: $14+7+11+12+17 = 61$.
Then add 4, 15.
Total: $61 + 4 + 15 = 80$. Too low.

Try higher numbers.
Use 26?
Path to (4,3) via 26?
(1,2)14 -> R(1,3)7 -> R(1,4)13 -> D(2,4)26 -> D(3,4)19 -> D(4,4)24... wrong target.

Let's try a different path to (4,3).
(2,1)12 -> R(2,2)9 -> R(2,3)11 -> D(3,3)12 -> D(4,3)17.
Sum: $12+9+11+12+17 = 61$. Same.

Try: (1,2)14 -> D(2,2)9 -> R(2,3)11 -> D(3,3)12 -> D(4,3)17.
Sum: $14+9+11+12+17 = 63$.
Total: $63 + 4 + 15 = 82$.

Try: (1,2)14 -> D(2,2)9 -> D(3,2)5 -> R(3,3)12 -> D(4,3)17.
Sum: $14+9+5+12+17 = 57$.
Total: $57 + 4 + 15 = 76$.

Try entering (4,3) from Left: (4,2)=10.
Path to (4,2).
Try: 14, 9, 10...
(1,2)14 -> D(2,2)9 -> D(3,2)5 -> D(4,2)10 -> R(4,3)17.
Sum: $14+9+5+10+17 = 55$.
Total: $55 + 4 + 15 = 74$.

It seems hard to get 85 with the first 5 numbers ending at (4,3). Max possible?
14, 13, 26... can't reach (4,3) easily with high numbers.
(1,2)14 -> R(1,3)7 -> D(2,3)11 -> D(3,3)12 -> D(4,3)17. Sum 61.
(1,2)14 -> D(2,2)9 -> R(2,3)11 -> D(3,3)12 -> D(4,3)17. Sum 63.
(2,1)12 -> R(2,2)9 -> R(2,3)11 -> D(3,3)12 -> D(4,3)17. Sum 61.
(2,1)12 -> D(3,1)16 -> R(3,2)5 -> R(3,3)12 -> D(4,3)17. Sum $12+16+5+12+17=62$.
(2,1)12 -> D(3,1)16 -> D(4,1)13 -> R(4,2)10 -> R(4,3)17. Sum $12+16+13+10+17=68$.
Total: $68 + 4 + 15 = 87$. Close!

Let's tweak this last one.
Path: 12, 16, 13, 10, 17, 4, 15. Sum 87.
Can we swap 10 for something bigger?
To get to (4,2), we came from (4,1)=13.
Before that (3,1)=16, (2,1)=12.
Alternative to (4,2): From (3,2)=5? No, 5 is smaller.
From (4,1) we went R to 10.
What if we went D to (5,1)=8? No, we need to end at (4,3).

Let's try a different route to (4,3).
We had sum 68 for first 5.
What if we use 14 instead of 12?
(1,2)14 -> D(2,2)9 -> D(3,2)5 -> D(4,2)10 -> R(4,3)17. Sum 55.
(1,2)14 -> D(2,2)9 -> R(2,3)11 -> D(3,3)12 -> D(4,3)17. Sum 63.

How about:
(1,2)14 -> R(1,3)7 -> D(2,3)11 -> D(3,3)12 -> D(4,3)17. Sum 61.

It seems getting to 85 with 5 numbers ending at (4,3) is difficult. The max I found is 68.
$68 + 4 + 15 = 87$.

Let's try entering (5,4) from (5,3)=4.
We established this is tough.

What if we enter Finish from Top? i.e., from (4,5)=11.
Last number 11.
Need previous 6 numbers to sum to $100 - 11 = 89$.

Let's look at Route 1 again:
14, 7, 13, 26, 19, 10, 11.
Sum: $14+7+13+26+19+10+11 = 100$.
This works. This is Route 1.

Let's find Route 2.
We need another path summing to 100.
Let's try varying Route 1 slightly.
Route 1 used: 14, 7, 13, 26, 19, 10, 11.
Moves: R, R, R, D, D, R, D.

Variation: Change the beginning.
Instead of 14, 7, 13...
Try 14, 9...
(1,2)14 -> D(2,2)9.
From 9, go R to 11 (2,3).
From 11, go R to 26 (2,4).
From 26, go D to 19 (3,4).
From 19, go R to 10 (3,5).
From 10, go D to 11 (4,5).
From 11, go D to Finish.
Path: 14, 9, 11, 26, 19, 10, 11.
Sum: $14+9=23$; $23+11=34$; $34+26=60$; $60+19=79$; $79+10=89$; $89+11=100$.
Yes! This is Route 2.
Sequence: 14 → 9 → 11 → 26 → 19 → 10 → 11

Let's find Route 3.
We have two routes ending with ..., 10, 11.
Both use 26, 19, 10, 11 at the end.
Sum of tail (26,19,10,11) = 66.
Need head to sum to 34.
Route 1 Head: 14, 7, 13. Sum = 34.
Route 2 Head: 14, 9, 11. Sum = 34.

Are there other heads summing to 34 ending at (2,4)=26?
To end at 26 (2,4), we need 3 numbers before it.
Possible predecessors of 26: (1,4)=13 or (2,3)=11.

If pred is 13 (1,4):
Need 2 numbers before 13 summing to $34-13=21$.
Predecessors of 13: (1,3)=7 or (2,4)... no, (1,4) comes from (1,3) or (2,4) is below.
So from Start to (1,4):
Path must be R, R.
Squares: 14, 7. Sum $14+7=21$.
This gives Route 1: 14, 7, 13.

If pred is 11 (2,3):
Need 2 numbers before 11 summing to $34-11=23$.
Predecessors of 11 (2,3): (1,3)=7 or (2,2)=9.

Case A: Come from 7 (1,3).
Need 1 number before 7 summing to $23-7=16$.
Predecessor of 7 (1,3) is (1,2)=14.
$14 \neq 16$. So this doesn't work.
Wait, Path: Start -> 14 -> 7 -> 11.
Sum: $14+7+11 = 32$.
Total with tail 66: $32+66=98$. Close, but not 100.

Case B: Come from 9 (2,2).
Need 1 number before 9 summing to $23-9=14$.
Predecessor of 9 (2,2) is (1,2)=14 or (2,1)=12.
If 14: Path 14 -> 9 -> 11. Sum $14+9+11=34$.
This gives Route 2: 14, 9, 11.

If 12: Path 12 -> 9 -> 11. Sum $12+9+11=32$.
Total $32+66=98$.

So, are there other tails?
We assumed the tail is 26, 19, 10, 11.
What if we don't use 26?

Let's try a path through 24 (4,4).
End with ..., 24, 11?
From 24 (4,4), can go R to 11 (4,5) then D to Finish.
Tail: 24, 11. Sum 35.
Need previous 5 numbers to sum to $100 - 35 = 65$.
End at (4,4).

Let's try to get to (4,4) with sum 65 in 5 steps.
Avg 13.

Try Path:
14, 7, 11, 12, 24?
(1,2)14 -> R(1,3)7 -> D(2,3)11 -> D(3,3)12 -> D(4,4)... wait, (3,3) to (4,4) is diagonal? No.
(3,3) is 12. (4,4) is 24.
From (3,3) can go R to (3,4)=19 or D to (4,3)=17.

Let's try:
14, 7, 13, 26... we did this.

Try:
12, 9, 11, 26...
(2,1)12 -> R(2,2)9 -> R(2,3)11 -> R(2,4)26.
Sum first 4: $12+9+11+26 = 58$.
Tail from 26: 19, 10, 11. Sum 40.
Total: $58+40=98$.

Try:
12, 9, 26? No, 9 is (2,2), 26 is (2,4). Must pass 11.

Try going through 20 (5,2)?
Path ending with ..., 20, 4, 15?
Sum last 3: 39.
Need first 4 to sum to 61.
End at (5,2).
Predecessors: (5,1)=8 or (4,2)=10.

If from 10 (4,2):
Need 3 numbers before 10 summing to 51.
End at (4,2).
Try: 14, 9, 10? Sum 33. No.
Try: 14, 7, 11... no.
Try: 12, 9, 10?
(2,1)12 -> R(2,2)9 -> D(3,2)5 -> D(4,2)10.
Sum: $12+9+5+10 = 36$.
Total: $36 + 20 + 4 + 15 = 75$.

Try: 14, 9, 5, 10?
(1,2)14 -> D(2,2)9 -> D(3,2)5 -> D(4,2)10.
Sum: $14+9+5+10 = 38$.
Total: $38 + 20 + 4 + 15 = 77$.

Let's try entering Finish from Left via 15 (5,4).
We tried this and got close with 87.
Is there a way to get 100?
Need first 6 numbers to sum to 85.
We found 68 ending at (4,3).
What if we end at (5,3)=4?
Need first 5 numbers to sum to $85-4=81$? No.
Path: N1, N2, N3, N4, N5, N6(4), N7(15).
Sum N1..N6 = 85.
N6 is 4.
So N1..N5 sum to 81.
End at (5,3). Predecessor (5,2)=20 or (4,3)=17.

If from 20 (5,2):
Need N1..N4 sum to $81-20=61$.
End at (5,2).
Try: 14, 9, 5, 20?
(1,2)14 -> D(2,2)9 -> D(3,2)5 -> D(4,2)10 -> D(5,2)20.
Wait, that's 5 numbers: 14,9,5,10,20. Sum 58.
We need 4 numbers summing to 61 ending at (5,2)?
No, N1..N4 ends at predecessor of 20?
Let's restart the count for this path.
Path: N1, N2, N3, N4, N5, N6, N7.
N7=15 (5,4).
N6=4 (5,3).
N5=20 (5,2).
Need N1+N2+N3+N4 = $100 - 15 - 4 - 20 = 61$.
End at (5,2). Predecessor is (4,2)=10 or (5,1)=8.

If N4=10 (4,2):
Need N1+N2+N3 = $61 - 10 = 51$.
End at (4,2).
Try: 14, 9, ?
(1,2)14 -> D(2,2)9 -> D(3,2)5 -> D(4,2)10.
Sum: $14+9+5+10 = 38$. (4 numbers).
We need 3 numbers summing to 51 ending at (4,2)?
No, N1,N2,N3 end at predecessor of N4?
Let's trace:
Start -> N1 -> N2 -> N3 -> N4(10) -> N5(20) -> N6(4) -> N7(15) -> Fin.
Sum N1..N3 = 51.
End at N3, which is predecessor of 10 (4,2).
Preds of 10: (3,2)=5, (4,1)=13.

If N3=5 (3,2):
Need N1+N2 = $51 - 5 = 46$.
End at (3,2). Preds: (2,2)=9, (3,1)=16.
If N2=9: Need N1=37. No single square is 37.
If N2=16: Need N1=30. No.

If N3=13 (4,1):
Need N1+N2 = $51 - 13 = 38$.
End at (4,1). Preds: (3,1)=16, (4,2)... no.
Pred of 13 (4,1) is (3,1)=16.
So N2=16.
Need N1 = $38 - 16 = 22$. No.

If N4=8 (5,1):
Need N1+N2+N3 = $61 - 8 = 53$.
End at (5,1). Pred (4,1)=13.
So N3=13.
Need N1+N2 = $53 - 13 = 40$.
End at (4,1). Pred (3,1)=16.
So N2=16.
Need N1 = $40 - 16 = 24$. No.

Let's try one more variation for Route 3.
Look at Route 1 and 2. They share the tail 26, 19, 10, 11.
Is there a third head for this tail?
Head sum 34.
We found 14,7,13 and 14,9,11.
Is there another combination of 3 numbers ending at 26 summing to 34?
We checked 12,9,11 (Sum 32).
What about 12, 16... no, 16 is (3,1).

What if the tail is different?
Try tail: 26, 21...
(2,4)26 -> R(2,5)21.
From 21, go D to 10 (3,5).
From 10, go D to 11 (4,5).
From 11, go D to Finish.
Tail: 26, 21, 10, 11. Sum 68.
Need head sum 32.
End at 26.
Heads summing to 32:
14, 7, 11?
(1,2)14 -> R(1,3)7 -> D(2,3)11 -> R(2,4)26.
Sum: $14+7+11=32$.
Total: $32 + 68 = 100$.
Yes! This is Route 3.
Sequence: 14 → 7 → 11 → 26 → 21 → 10 → 11

So the 3 routes are:
1. 14, 7, 13, 26, 19, 10, 11
2. 14, 9, 11, 26, 19, 10, 11
3. 14, 7, 11, 26, 21, 10, 11

Let's double check calculations.
Route 1: $14+7+13+26+19+10+11 = 100$. Correct.
Route 2: $14+9+11+26+19+10+11 = 100$. Correct.
Route 3: $14+7+11+26+21+10+11 = 100$. Correct.

Final Answer Formulation:
List the three routes clearly.

Final Answer:
The three possible routes that sum to 100 are:

1.