Question:

Six dice are stacked on the floor as shown (heights \(3,2,1\) like steps). On each die, opposite faces sum to \(7\). What is the maximum possible sum of the numbers on all \(21\) visible faces?

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For stacked dice, maximize visible sum by minimizing the numbers placed on all faces in contact (floor or other dice). Each hidden opposite pair contributes at least \(7\) no matter how you orient the die.
Updated On: Aug 14, 2025
  • \(88\)
  • \(89\)
  • \(96\)
  • \(147\)
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The Correct Option is B

Solution and Explanation

Step 1: Count hidden faces (faces in contact).
Arrange the dice in three columns with heights \(3,2,1\).
Hidden faces come from: bottoms on the floor, vertical contacts in a column, and side contacts between columns.
\[ \begin{aligned} &\text{Bottom faces} = 3,\\ &\text{Vertical contacts in columns } (3\rightarrow2\rightarrow1) = 2+1=3 \text{ contacts } \Rightarrow 2\cdot 3=6 \text{ hidden faces},\\ &\text{Side contacts between columns } (2+1=3 \text{ contacts}) \Rightarrow 2\cdot 3=6 \text{ hidden faces}. \end{aligned} \] Total hidden faces \(= 3+6+6=15\).
Since \(6\) dice have \(6\times 6=36\) faces in all, visible faces \(=36-15=21\) (as stated).

Step 2: Maximization principle. 
The sum on all six faces of a die is \(1+2+3+4+5+6=21\).
Hence (for each die) \(\text{visible sum} = 21 - \text{(sum on its hidden faces)}\).
So to maximize the visible total, minimize the sum on hidden faces, putting the smallest possible numbers on the faces in contact, respecting “opposites sum to \(7\)”.

Step 3: Choose the smallest numbers on hidden faces die-by-die.
Label columns left\(\to\)right with heights \(3,2,1\).
Hidden faces per die: \[ \begin{array}{l|c} \text{Die} & \text{Hidden faces (count)} \\ \hline \text{Left column: bottom (A)} & \text{bottom, top (opposites), right }(3) \\ \text{Left column: middle (B)} & \text{bottom, top (opposites), right }(3) \\ \text{Left column: top (C)} & \text{bottom }(1) \\ \text{Middle column: bottom (D)} & \text{bottom/top (opposites), left/right (opposites) }(4) \\ \text{Middle column: top (E)} & \text{bottom, left }(2) \\ \text{Right column: only (F)} & \text{bottom, left }(2) \end{array} \] - For each opposite pair hidden, the minimum sum is \(7\) (unavoidable).
Thus \(A\): \(7+\) one extra face, \(B\): \(7+\) one extra face, \(D\): \(7+7=14\).
- The remaining single hidden faces can take the smallest available numbers \(1,2,\ldots\).

Choose minimally: \[ \begin{aligned} &A:\ 7+1=8,\quad B:\ 7+1=8,\quad C:\ 1,\\ &D:\ 14,\quad E:\ 1+2=3,\quad F:\ 1+2=3. \end{aligned} \] Total minimum on hidden faces \(=8+8+1+14+3+3=37\).

Step 4: Compute the maximum visible sum.
All-face total \(=6\times 21=126\).
Maximum visible \(=126-37=89\).

 

\[ \boxed{89} \]

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