Explanation:

Let’s break this down step by step:

- There are 8 boxes stacked, numbered 1 (bottom) to 8 (top).

- U is on a prime number position, but not on 5. So possible spots: 2, 3, or 7.

- Boxes above W = boxes below U. So W and U are symmetrically arranged.

- S’s position = 2 x V’s position. So S can’t be at 1, 3, 5, 7.

- P is just above U.

- Q is two boxes above T.

What this means is:

- W must be position 4 or 5, because only those positions allow for equal boxes above and below (4 below, 4 above at box 5; 3 above, 3 below at 4). But U could be at 2, 3, or 7, which connects later.

- S & V: valid pairs are (2,1), (4,2), (6,3), (8,4).

Let’s fit it:

1. Try U at 3 (prime, not 5). Then P is at 4.

2. For W symmetry: boxes above W = boxes below U. If U is 3: boxes below U = 2, so boxes above W = 2 ? W must be box 6.

3. S & V pick (4,2) — possible, as S is at 4, V at 2.

4. List what we have:

- Box 1: ___

- Box 2: V

- Box 3: U

- Box 4: S

- Box 5: P (just above U)

- Box 6: W

- Box 7: ___

- Box 8: ___

But wait, P must be just above U (so U at 3, P at 4), but 4 is S per (S=2xV=4). Can’t fit both there. So U is not at 3.

Try U at 7 (prime):

- Boxes below U = 6, so above W = 6. Only possible if W is at 1, but number of boxes above W = 7, not 6.

- Try U at 2: boxes below U = 1, so boxes above W = 1 ? W is at 8 (only one above is not possible as top box, so not fitting).

Let’s try U at 2:

- P at 3.

- S & V: S = 2xV; pairs could be S=4, V=2. But V can’t be at 2, already occupied by U. Try next options.

U at 3:

- P at 4.

- S & V: S=6, V=3: But 3 has U, so not possible.

- S=8, V=4: S at 8, V at 4; P at 4 which conflicts.

U at 7:

- P at 8.

- S=2xV: S=2, V=1.

- So box 1: V; box 2: S.

- W is such that boxes above = below U (U at 7: below=6), so W is at 1 (with 7 above? No room).

Try U at prime, not 2,3,7—let’s double-check our deductions.

Go back to:

- Q is two above T: means Q is at T+2.

Let’s map this positionally (Trial):

Let’s try:

1. V: 1

2. S: 2

3. U: 3

4. P: 4

5. T: 5

6. Q: 7 (But 5+2=7; so Q at 7, T at 5) Possible?

7. W, R - left.

But boxes above W = boxes below U. If U is 3, below = 2. So above W = 2; W is at box 6 (positions 7 and 8 above, so 2 above W). Does that fit?

- So W at 6, above W are 7,8; below U are 1,2.

So, list:

1. V

2. S

3. U

4. P

5. T

6. W

7. Q

8. R

Check: Box S is at 2. Which box is just above S? Box 3, which is U.

So:

- Option 1: U ? This is the box immediately above S.

- Option 2: Q (it's at 7)

- Option 3: R (8)

- Option 4: T (5)

- Option 5: V (1)

Final answer:

Option 1: U 

Here's what happened, bullet-style:

- The only way to get S and V relation with the other constraints is S=2, V=1.

- The only prime spot possible for U is 3; boxes above W equaling boxes below U makes W=6.

- With mapping, box 3 (U) is just above box 2 (S).

- So box U is just above box S.

You nailed it—Option 1 is correct.