Box/Stack Puzzles - Intermediate-Advanced Level: stacking logic INTERMEDIATE-ADVANCED

Step-by-step Solution:
1. Initial State:
- Stack 1 (bottom to top): A(1), B(2), C(3), D(4), E(5)
- Stack 2: Empty

2. Operation 1: Move top box of Stack 1 to Stack 2
- Remove E from Stack 1 (was at position 5)
- Stack 1 after: A(1), B(2), C(3), D(4)
- Stack 2 after: E(1)

3. Operation 2: Move box at position 2 of Stack 1 to top of Stack 2
- Box at position 2 is B
- Remove B from Stack 1
- Stack 1 after: A(1), C(2), D(3) [positions renumber]
- Stack 2 after: E(1), B(2) [B placed on top]

4. Operation 3: Move top box of Stack 1 to top of Stack 2
- Top box of Stack 1 is D (position 3)
- Remove D from Stack 1
- Stack 1 after: A(1), C(2)
- Stack 2 after: E(1), B(2), D(3) [D placed on top]

5. Final Stack 2 (bottom to top): E, B, D

6. Answer: Box E is at the bottom of Stack 2

Step-by-step Solution:

1. Fixed positions from given conditions:
- T is at bottom → Position 1 = T
- S is exactly in the middle → For 6 boxes, middle positions are 3 or 4
- P is two positions above Q → P's position = Q's position + 2

2. Determine valid arrangement:
- Since P = Q + 2, possible pairs: (1,3), (2,4), (3,5), (4,6)
- Position 1 is T, so (1,3) invalid
- Try Q=2, P=4:
* Then S can be at position 3 (middle)
* R must be at even position (2,4,6) but 2 and 4 taken → R=6
* U must not be adjacent to S (pos3) → cannot be at 2 or 4 → U=5
* All positions filled: 1=T, 2=Q, 3=S, 4=P, 5=U, 6=R ✓

3. Test if each box can be at top (position 6):
- T: Fixed at bottom → Cannot be at top ❌
- Q: If Q at 6, then P would be at 8 (invalid) → Cannot be at top ❌
- P: Can be at top if Q=4, P=6 (with S=3, R=2, U=5) → Possible ✓
- R: Can be at top as shown in our arrangement → Possible ✓
- S: Can be at top if S=6 (but then middle would be 3, possible with adjustments) → Possible ✓
- U: Can be at top with different arrangement → Possible ✓

4. Only Q and T cannot be at top:
- T is explicitly fixed at bottom
- Q is mathematically impossible at top due to P=Q+2 constraint

5. The question asks "which box" (singular):
- Since T is explicitly stated to be at bottom, it's obvious
- Q is the non-obvious answer that requires deduction
- Therefore, Q is the intended answer

Answer: Box Q can never be at the top

Verification:
- In any valid arrangement, Q's maximum position is 4 (when P=6)
- Q at position 6 would require P at 8 (outside stack)
- Therefore Q is impossible at top ✓

Step-by-step Solution:

1. Apply stacking rule: Heavier below lighter
- So weights must decrease from bottom to top
- Position 1 (bottom) = heaviest
- Position 5 (top) = lightest

2. Fixed positions:
- Position 1 = Box O (given)
- Position 5 = Box K (given)
- Therefore: O is heaviest (55kg), K is lightest (15kg)

3. Box L's weight is exactly 25kg
- So L must be at position 4 (since 25kg is second lightest)
- Position 4 = L (25kg)

4. Remaining boxes: M and N for positions 2 and 3
- Remaining weights: 35kg and 45kg

5. Constraint: N > M (in weight)
- So N must be heavier than M
- Therefore: N at position 2 (45kg), M at position 3 (35kg)

6. Verify additional constraints:
- "Box N's weight is less than 50kg" → 45kg < 50kg ✓
- "Box immediately above M has weight < 30kg"
- M at position 3 → box above is position 4 = L (25kg) < 30kg ✓

7. Final stack (bottom to top):
- Position 1: O (55kg)
- Position 2: N (45kg)
- Position 3: M (35kg)
- Position 4: L (25kg)
- Position 5: K (15kg)

8. Answer: Box N weighs 45kg

Verification: All conditions satisfied ✓

Step-by-step Solution:

1. Apply stacking rule: Heavier below lighter
- So weights must decrease from bottom to top
- Position 1 (bottom) = heaviest
- Position 5 (top) = lightest

2. Fixed positions:
- Position 1 = Box O (given)
- Position 5 = Box K (given)
- Therefore: O is heaviest (55kg), K is lightest (15kg)

3. Box L's weight is exactly 25kg
- So L must be at position 4 (since 25kg is second lightest)
- Position 4 = L (25kg)

4. Remaining boxes: M and N for positions 2 and 3
- Remaining weights: 35kg and 45kg

5. Constraint: N > M (in weight)
- So N must be heavier than M
- Therefore: N at position 2 (45kg), M at position 3 (35kg)

6. Verify additional constraints:
- "Box N's weight is less than 50kg" → 45kg < 50kg ✓
- "Box immediately above M has weight < 30kg"
- M at position 3 → box above is position 4 = L (25kg) < 30kg ✓

7. Final stack (bottom to top):
- Position 1: O (55kg)
- Position 2: N (45kg)
- Position 3: M (35kg)
- Position 4: L (25kg)
- Position 5: K (15kg)

8. Answer: Box N weighs 45kg

Verification: All conditions satisfied ✓

Step-by-step Solution:

1. Understand generations:
- Gen 1 (oldest): G
- Gen 2: F, U
- Gen 3: S, N
- Gen 4 (youngest): GS

2. Apply "oldest at bottom of Stack A":
- Stack A, position 1 = G

3. Apply "Father (F) in Stack B":
- F is in Stack B (position TBD)

4. Apply "Son (S) immediately above Grandson (GS)":
- They must be in same stack with S directly above GS
- Possible positions: (GS at 1, S at 2) or (GS at 2, S at 3)
- They cannot be in Stack A because G is at position 1
- So they must be in Stack B
- Place GS at position 1, S at position 2 in Stack B

5. Stack B so far:
- Position 1: GS (gen4)
- Position 2: S (gen3)
- Position 3: F (gen2) - must go here

6. Remaining boxes for Stack A:
- Used: G (Stack A), F, S, GS (Stack B)
- Remaining: U (gen2) and N (gen3)

7. Fill Stack A:
- Position 1: G (gen1)
- Position 2: N (gen3) - can't put U (gen2) adjacent to same gen? No same gen restriction applies to same stack
- Position 3: U (gen2)

8. Verify no same generation in same stack:
- Stack A: gen1 (G), gen3 (N), gen2 (U) → all different ✓
- Stack B: gen4 (GS), gen3 (S), gen2 (F) → all different ✓

9. Final arrangements:
- Stack A (bottom to top): G, N, U
- Stack B (bottom to top): GS, S, F

10. Answer: Box U (Uncle) is at the top of Stack A

Verification: All conditions satisfied ✓

Step-by-step Solution:

1. Understand the layout:
- Row 1 (top): positions 1 (Col1), 2 (Col2), 3 (Col3)
- Row 2 (bottom): positions 4 (Col1), 5 (Col2), 6 (Col3)
- "Directly below" means same column, consecutive rows
- "Immediate right" means same row, adjacent columns

2. Direct assignment:
- Position 4 = D (given)

3. Place A and E:
- A in Row 1 (positions 1, 2, or 3)
- E immediately right of A
- Possible pairs: (A=1, E=2) or (A=2, E=3)
- Try A=1, E=2 first

4. Place B and C (B directly below C):
- Possible column pairs: (C at pos1, B at pos4), (C at pos2, B at pos5), (C at pos3, B at pos6)
- pos4 is already D, so (C=1, B=4) invalid
- Try C=3, B=6 (Column 3)

5. Place remaining box F:
- Available position: pos5
- F in Row 2 (pos5) and not below A
- A at pos1 (Col1), F at pos5 (Col2) → not below ✓

6. Final arrangement:
- Position 1: A
- Position 2: E
- Position 3: C
- Position 4: D
- Position 5: F
- Position 6: B

7. Answer: Box F is at position 5

Verification: All conditions satisfied ✓

Step-by-step Solution:

1. Start with fixed information:
- Position 1 = Empty
- Position 8 = Empty
- Position 5 = A

2. List remaining positions for boxes B, C, D, E:
Available: 2, 3, 4, 6, 7

3. Apply "Box B is three positions below Box C":
- Possible (B, C) pairs: (2,5), (3,6), (4,7), (5,8)
- Position 5 is A, position 8 is Empty → (2,5) and (5,8) invalid
- Remaining: (3,6) or (4,7)

4. Apply "Box E is at an even position":
- Even positions available: 2, 4, 6
- Position 6 is a candidate for C in option (3,6) → then E would need another even

5. Apply "Box D is immediately above an empty position":
- Empty positions: 1 and 8 (fixed), plus possibly others
- Positions above empty:
* Above pos1 (empty) → position 2
* Above pos8 (empty) → none (pos9 doesn't exist)
- So D MUST be at position 2 (above empty pos1)

6. Place D at position 2

7. Re-evaluate B and C with D placed:
- Available now: 3, 4, 6, 7
- (B,C) options: (3,6) or (4,7)

8. Place E at even position:
- Even positions available: 4, 6 (2 is taken by D)
- Try (B,C) = (4,7):
* Then B=4, C=7
* E must be at even → E=6
* Remaining position 3 becomes Empty
- Check D at 2 (above empty pos1) ✓
- Check E at 6 (even) ✓

9. Final arrangement (bottom to top):
- Position 1: Empty
- Position 2: D
- Position 3: Empty
- Position 4: B
- Position 5: A
- Position 6: E
- Position 7: C
- Position 8: Empty

10. Answer: Box C is at position 7

Verification: All conditions satisfied ✓

Step-by-step Solution:
1. Initial State:
- Stack 1 (bottom to top): A(1), B(2), C(3), D(4), E(5)
- Stack 2: Empty

2. Operation 1: Move top box of Stack 1 to Stack 2
- Remove E from Stack 1 (was at position 5)
- Stack 1 after: A(1), B(2), C(3), D(4)
- Stack 2 after: E(1)

3. Operation 2: Move box at position 2 of Stack 1 to top of Stack 2
- Box at position 2 is B
- Remove B from Stack 1
- Stack 1 after: A(1), C(2), D(3) [positions renumber]
- Stack 2 after: E(1), B(2) [B placed on top]

4. Operation 3: Move top box of Stack 1 to top of Stack 2
- Top box of Stack 1 is D (position 3)
- Remove D from Stack 1
- Stack 1 after: A(1), C(2)
- Stack 2 after: E(1), B(2), D(3) [D placed on top]

5. Final Stack 2 (bottom to top): E, B, D

6. Answer: Box E is at the bottom of Stack 2

Step-by-step Solution:

1. Establish weight relationships:
- A > B and C > A → So C > A > B
- D is lightest → D is smallest
- E > A and C > E → So C > E > A
- B > D

2. Combine all inequalities:
- From C > E > A and A > B, we get: C > E > A > B
- And B > D, so: C > E > A > B > D

3. Complete weight order (heaviest to lightest):
C > E > A > B > D

4. Stacking rule: Heaviest at bottom (position 1), lightest at top (position 5)
- Position 1 (bottom): C (heaviest)
- Position 2: E
- Position 3: A
- Position 4: B
- Position 5 (top): D (lightest)

5. Answer: Box A is at position 3

Verification:
- Heaviest (C) at bottom ✓
- A heavier than B (pos3 vs pos4) ✓
- A lighter than C (pos3 vs pos1) ✓
- D lightest at top ✓
- E heavier than A (pos2 vs pos3) ✓
- E lighter than C (pos2 vs pos1) ✓
- B heavier than D (pos4 vs pos5) ✓

Step-by-step Solution:

1. Understanding the numbering system:
- Position 1 = TOP
- Position 5 = BOTTOM

2. Fixed positions:
- Position 1 = P
- Position 5 = Q
- Position 4 = R

3. Remaining positions: 2 and 3 for boxes S and T

4. "S is immediately above T":
- They must occupy consecutive positions with S above T
- Therefore: S at position 2, T at position 3

5. Final arrangement (top to bottom):
- Position 1: P
- Position 2: S
- Position 3: T
- Position 4: R
- Position 5: Q

6. Answer: Box R is at position 4

Verification: All conditions satisfied ✓

Step-by-step Solution:

1. Create vertical stack (1=bottom, 7=top)

2. Direct assignments:
- Box T at bottom → Position 1 = T

3. Constraint 1: Box P is 3 positions above Box Q
- Possible pairs: (1,4), (2,5), (3,6), (4,7)
- Position 1 is T, so (1,4) invalid
- Try Q at 4, P at 7

4. Constraint 2: Box R immediately below Box S
- They occupy consecutive positions: (1,2), (2,3), (3,4), (4,5), (5,6), (6,7)
- Position 4 is Q, position 7 is P
- Try R at 2, S at 3

5. Constraint 3: Box U between R and P
- R at 2, P at 7 → positions between: 3,4,5,6
- Position 3 is S, position 4 is Q
- Place U at position 5
- Remaining box V goes to position 6

6. Final arrangement (bottom to top):
- Position 1: T
- Position 2: R
- Position 3: S
- Position 4: Q
- Position 5: U
- Position 6: V
- Position 7: P

7. Count boxes between S (pos3) and V (pos6):
- Positions between: 4 and 5 → Boxes Q and U
- Total: 2 boxes

Answer: 2 boxes

Verification:
- T at bottom ✓
- P at 7, Q at 4 → 3 positions above ✓
- R at 2, S at 3 → immediately below ✓
- U at 5 is between R(2) and P(7) ✓
- All boxes used exactly once ✓

Step-by-step Solution:

1. Stack to Circle mapping:
- Seat n = Box at stack position n

2. Fixed assignments:
- Stack pos1 = A → Seat1 = A
- Stack pos6 = B → Seat6 = B
- Box C at seat4 (given) → Stack pos4 = C

3. Box C immediately above Box D in stack:
- C at pos4 → D must be at pos3
- Therefore: Stack pos3 = D

4. Remaining positions in stack: pos2 and pos5 for boxes E and F
- Stack pos2 → Seat2
- Stack pos5 → Seat5

5. In circular arrangement, E opposite F:
- In a 6-seat circle, opposite pairs are: (1,4), (2,5), (3,6)
- Seat1 = A, Seat4 = C, Seat3 = D, Seat6 = B
- For E and F to be opposite, they must occupy seats (2,5)
- Therefore: Seat2 and Seat5 = E and F (in some order)

6. Determine which is which:
- Seat2 = Stack pos2 = either E or F
- Seat5 = Stack pos5 = the other one
- Both arrangements are valid

7. Complete the arrangement:
- Stack: pos1=A, pos2=E/F, pos3=D, pos4=C, pos5=F/E, pos6=B
- Circle: Seat1=A, Seat2=E/F, Seat3=D, Seat4=C, Seat5=F/E, Seat6=B

8. Find box opposite A:
- A is at Seat1
- Opposite seat to Seat1 is Seat4 (since 1+3=4)
- Seat4 = C

9. Answer: Box C sits opposite Box A

Final verification:
- A at bottom ✓
- B at top ✓
- C(pos4) immediately above D(pos3) ✓
- E and F at seats 2 and 5 (opposite) ✓
- C at seat4 ✓

Answer: Box C

Step-by-step Solution:

1. Create vertical stack (1=bottom, 7=top)

2. Direct assignments:
- Box T at bottom → Position 1 = T

3. Constraint 1: Box P is 3 positions above Box Q
- Possible pairs: (1,4), (2,5), (3,6), (4,7)
- Position 1 is T, so (1,4) invalid
- Try Q at 4, P at 7

4. Constraint 2: Box R immediately below Box S
- They occupy consecutive positions: (1,2), (2,3), (3,4), (4,5), (5,6), (6,7)
- Position 4 is Q, position 7 is P
- Try R at 2, S at 3

5. Constraint 3: Box U between R and P
- R at 2, P at 7 → positions between: 3,4,5,6
- Position 3 is S, position 4 is Q
- Place U at position 5
- Remaining box V goes to position 6

6. Final arrangement (bottom to top):
- Position 1: T
- Position 2: R
- Position 3: S
- Position 4: Q
- Position 5: U
- Position 6: V
- Position 7: P

7. Count boxes between S (pos3) and V (pos6):
- Positions between: 4 and 5 → Boxes Q and U
- Total: 2 boxes

Answer: 2 boxes

Verification:
- T at bottom ✓
- P at 7, Q at 4 → 3 positions above ✓
- R at 2, S at 3 → immediately below ✓
- U at 5 is between R(2) and P(7) ✓
- All boxes used exactly once ✓

Step-by-step Solution:

1. Establish weight relationships:
- A > B and C > A → So C > A > B
- D is lightest → D is smallest
- E > A and C > E → So C > E > A
- B > D

2. Combine all inequalities:
- From C > E > A and A > B, we get: C > E > A > B
- And B > D, so: C > E > A > B > D

3. Complete weight order (heaviest to lightest):
C > E > A > B > D

4. Stacking rule: Heaviest at bottom (position 1), lightest at top (position 5)
- Position 1 (bottom): C (heaviest)
- Position 2: E
- Position 3: A
- Position 4: B
- Position 5 (top): D (lightest)

5. Answer: Box A is at position 3

Verification:
- Heaviest (C) at bottom ✓
- A heavier than B (pos3 vs pos4) ✓
- A lighter than C (pos3 vs pos1) ✓
- D lightest at top ✓
- E heavier than A (pos2 vs pos3) ✓
- E lighter than C (pos2 vs pos1) ✓
- B heavier than D (pos4 vs pos5) ✓

Step-by-step Solution:

1. Stack to Circle mapping:
- Seat n = Box at stack position n

2. Fixed assignments:
- Stack pos1 = A → Seat1 = A
- Stack pos6 = B → Seat6 = B
- Box C at seat4 (given) → Stack pos4 = C

3. Box C immediately above Box D in stack:
- C at pos4 → D must be at pos3
- Therefore: Stack pos3 = D

4. Remaining positions in stack: pos2 and pos5 for boxes E and F
- Stack pos2 → Seat2
- Stack pos5 → Seat5

5. In circular arrangement, E opposite F:
- In a 6-seat circle, opposite pairs are: (1,4), (2,5), (3,6)
- Seat1 = A, Seat4 = C, Seat3 = D, Seat6 = B
- For E and F to be opposite, they must occupy seats (2,5)
- Therefore: Seat2 and Seat5 = E and F (in some order)

6. Determine which is which:
- Seat2 = Stack pos2 = either E or F
- Seat5 = Stack pos5 = the other one
- Both arrangements are valid

7. Complete the arrangement:
- Stack: pos1=A, pos2=E/F, pos3=D, pos4=C, pos5=F/E, pos6=B
- Circle: Seat1=A, Seat2=E/F, Seat3=D, Seat4=C, Seat5=F/E, Seat6=B

8. Find box opposite A:
- A is at Seat1
- Opposite seat to Seat1 is Seat4 (since 1+3=4)
- Seat4 = C

9. Answer: Box C sits opposite Box A

Final verification:
- A at bottom ✓
- B at top ✓
- C(pos4) immediately above D(pos3) ✓
- E and F at seats 2 and 5 (opposite) ✓
- C at seat4 ✓

Answer: Box C

Step-by-step Solution:

1. Stack to Circle mapping:
- Seat n = Box at stack position n

2. Fixed assignments:
- Stack pos1 = A → Seat1 = A
- Stack pos6 = B → Seat6 = B
- Box C at seat4 (given) → Stack pos4 = C

3. Box C immediately above Box D in stack:
- C at pos4 → D must be at pos3
- Therefore: Stack pos3 = D

4. Remaining positions in stack: pos2 and pos5 for boxes E and F
- Stack pos2 → Seat2
- Stack pos5 → Seat5

5. In circular arrangement, E opposite F:
- In a 6-seat circle, opposite pairs are: (1,4), (2,5), (3,6)
- Seat1 = A, Seat4 = C, Seat3 = D, Seat6 = B
- For E and F to be opposite, they must occupy seats (2,5)
- Therefore: Seat2 and Seat5 = E and F (in some order)

6. Determine which is which:
- Seat2 = Stack pos2 = either E or F
- Seat5 = Stack pos5 = the other one
- Both arrangements are valid

7. Complete the arrangement:
- Stack: pos1=A, pos2=E/F, pos3=D, pos4=C, pos5=F/E, pos6=B
- Circle: Seat1=A, Seat2=E/F, Seat3=D, Seat4=C, Seat5=F/E, Seat6=B

8. Find box opposite A:
- A is at Seat1
- Opposite seat to Seat1 is Seat4 (since 1+3=4)
- Seat4 = C

9. Answer: Box C sits opposite Box A

Final verification:
- A at bottom ✓
- B at top ✓
- C(pos4) immediately above D(pos3) ✓
- E and F at seats 2 and 5 (opposite) ✓
- C at seat4 ✓

Answer: Box C

Step-by-step Solution:

1. Fixed positions from given conditions:
- T is at bottom → Position 1 = T
- S is exactly in the middle → For 6 boxes, middle positions are 3 or 4
- P is two positions above Q → P's position = Q's position + 2

2. Determine valid arrangement:
- Since P = Q + 2, possible pairs: (1,3), (2,4), (3,5), (4,6)
- Position 1 is T, so (1,3) invalid
- Try Q=2, P=4:
* Then S can be at position 3 (middle)
* R must be at even position (2,4,6) but 2 and 4 taken → R=6
* U must not be adjacent to S (pos3) → cannot be at 2 or 4 → U=5
* All positions filled: 1=T, 2=Q, 3=S, 4=P, 5=U, 6=R ✓

3. Test if each box can be at top (position 6):
- T: Fixed at bottom → Cannot be at top ❌
- Q: If Q at 6, then P would be at 8 (invalid) → Cannot be at top ❌
- P: Can be at top if Q=4, P=6 (with S=3, R=2, U=5) → Possible ✓
- R: Can be at top as shown in our arrangement → Possible ✓
- S: Can be at top if S=6 (but then middle would be 3, possible with adjustments) → Possible ✓
- U: Can be at top with different arrangement → Possible ✓

4. Only Q and T cannot be at top:
- T is explicitly fixed at bottom
- Q is mathematically impossible at top due to P=Q+2 constraint

5. The question asks "which box" (singular):
- Since T is explicitly stated to be at bottom, it's obvious
- Q is the non-obvious answer that requires deduction
- Therefore, Q is the intended answer

Answer: Box Q can never be at the top

Verification:
- In any valid arrangement, Q's maximum position is 4 (when P=6)
- Q at position 6 would require P at 8 (outside stack)
- Therefore Q is impossible at top ✓

Step-by-step Solution:

1. Establish weight relationships:
- A > B and C > A → So C > A > B
- D is lightest → D is smallest
- E > A and C > E → So C > E > A
- B > D

2. Combine all inequalities:
- From C > E > A and A > B, we get: C > E > A > B
- And B > D, so: C > E > A > B > D

3. Complete weight order (heaviest to lightest):
C > E > A > B > D

4. Stacking rule: Heaviest at bottom (position 1), lightest at top (position 5)
- Position 1 (bottom): C (heaviest)
- Position 2: E
- Position 3: A
- Position 4: B
- Position 5 (top): D (lightest)

5. Answer: Box A is at position 3

Verification:
- Heaviest (C) at bottom ✓
- A heavier than B (pos3 vs pos4) ✓
- A lighter than C (pos3 vs pos1) ✓
- D lightest at top ✓
- E heavier than A (pos2 vs pos3) ✓
- E lighter than C (pos2 vs pos1) ✓
- B heavier than D (pos4 vs pos5) ✓

Step-by-step Solution:

1. Stack to Circle mapping:
- Seat n = Box at stack position n

2. Fixed assignments:
- Stack pos1 = A → Seat1 = A
- Stack pos6 = B → Seat6 = B
- Box C at seat4 (given) → Stack pos4 = C

3. Box C immediately above Box D in stack:
- C at pos4 → D must be at pos3
- Therefore: Stack pos3 = D

4. Remaining positions in stack: pos2 and pos5 for boxes E and F
- Stack pos2 → Seat2
- Stack pos5 → Seat5

5. In circular arrangement, E opposite F:
- In a 6-seat circle, opposite pairs are: (1,4), (2,5), (3,6)
- Seat1 = A, Seat4 = C, Seat3 = D, Seat6 = B
- For E and F to be opposite, they must occupy seats (2,5)
- Therefore: Seat2 and Seat5 = E and F (in some order)

6. Determine which is which:
- Seat2 = Stack pos2 = either E or F
- Seat5 = Stack pos5 = the other one
- Both arrangements are valid

7. Complete the arrangement:
- Stack: pos1=A, pos2=E/F, pos3=D, pos4=C, pos5=F/E, pos6=B
- Circle: Seat1=A, Seat2=E/F, Seat3=D, Seat4=C, Seat5=F/E, Seat6=B

8. Find box opposite A:
- A is at Seat1
- Opposite seat to Seat1 is Seat4 (since 1+3=4)
- Seat4 = C

9. Answer: Box C sits opposite Box A

Final verification:
- A at bottom ✓
- B at top ✓
- C(pos4) immediately above D(pos3) ✓
- E and F at seats 2 and 5 (opposite) ✓
- C at seat4 ✓

Answer: Box C

Step-by-step Solution:

1. Stack to Circle mapping:
- Seat n = Box at stack position n

2. Fixed assignments:
- Stack pos1 = A → Seat1 = A
- Stack pos6 = B → Seat6 = B
- Box C at seat4 (given) → Stack pos4 = C

3. Box C immediately above Box D in stack:
- C at pos4 → D must be at pos3
- Therefore: Stack pos3 = D

4. Remaining positions in stack: pos2 and pos5 for boxes E and F
- Stack pos2 → Seat2
- Stack pos5 → Seat5

5. In circular arrangement, E opposite F:
- In a 6-seat circle, opposite pairs are: (1,4), (2,5), (3,6)
- Seat1 = A, Seat4 = C, Seat3 = D, Seat6 = B
- For E and F to be opposite, they must occupy seats (2,5)
- Therefore: Seat2 and Seat5 = E and F (in some order)

6. Determine which is which:
- Seat2 = Stack pos2 = either E or F
- Seat5 = Stack pos5 = the other one
- Both arrangements are valid

7. Complete the arrangement:
- Stack: pos1=A, pos2=E/F, pos3=D, pos4=C, pos5=F/E, pos6=B
- Circle: Seat1=A, Seat2=E/F, Seat3=D, Seat4=C, Seat5=F/E, Seat6=B

8. Find box opposite A:
- A is at Seat1
- Opposite seat to Seat1 is Seat4 (since 1+3=4)
- Seat4 = C

9. Answer: Box C sits opposite Box A

Final verification:
- A at bottom ✓
- B at top ✓
- C(pos4) immediately above D(pos3) ✓
- E and F at seats 2 and 5 (opposite) ✓
- C at seat4 ✓

Answer: Box C