Dice and Cube

This is the most common dice question of the exam. The same die is shown in two positions, and exactly one face appears in both. Let us see why a neat trick works here, because the trick is only safe when you know the reason.

Call the common face N. The four faces around N form a ring: A, B, C, D, going round in order. In this ring, A is opposite C and B is opposite D (they are across from each other). The sixth face, which you never see, is opposite N.

View 1 stands at one corner of N and shows N with two neighbours in the ring, say A and B. View 2 stands at the diagonal corner of N and shows N with the other two, C and D. Read both views clockwise starting from N, and the lists run in the same direction round the ring. So the first of view 1 faces the first of view 2, and the second faces the second.

[[fig:0]]

This gives the stacking method:
1. Write N at the left of two rows.
2. Row 1: the other two faces of view 1, clockwise after N.
3. Row 2: the other two faces of view 2, clockwise after N.
4. Faces in the same column are opposite. The number that appears in neither view is opposite N.

In the figure, N = 5. View 1 gives 2, 6 and view 2 gives 4, 1. The columns are 2 ↕ 4 and 6 ↕ 1, and the unseen 3 is opposite 5.

Both lists must be read in the same sense, and both must start right after the common face. Reading one view clockwise and the other anticlockwise gives a wrong answer.

Two quick checks before you stack. (1) If the question asks for the opposite of the common face N itself, there is nothing to stack: the answer is the one face that appears in neither view. (2) Faces seen together are neighbours, never opposites. So every face drawn beside the asked face, in the view where the asked face appears, can be struck out of the options at once. Usually two options survive, and the stack decides between them.

The number of faces the two views share decides the method. There are only three cases. One common face: stack (Type 1). Two common faces: the leftover method of Part A. None: only a given pair, or a standard die, can give a definite answer (Part B).

Part A: two common faces

Now the two pictures share two faces. Two faces that appear together in a picture are neighbours, so the two common faces meet along an edge. The two corners sit at the two ends of that edge. Each corner shows one more face, and these two extra faces lie at opposite ends of the edge. So they are opposite each other.

Take View 1 = Top 3, Left 4, Right 1 and View 2 = Top 1, Left 2, Right 3. The common faces are 3 and 1. Cross them out. What is left is 4 in view 1 and 2 in view 2.

[[fig:0]]

The leftover face of view 1 is opposite the leftover face of view 2.

Here is the limit of this trick, and exam setters know it. It fixes only the two leftover faces. The opposite of a common face (3 or 1) is one of the two faces you have not seen at all (5 or 6), and two views cannot say which. If the question asks for the opposite of a common face, you need a third view, or the answer is "cannot be determined".

Do not use the stacking method here. Stacking was built for two corners that sit diagonally on a single common face, which is not the case now.

Part B: no common face

If two views have no face in common, between them they show all six faces. That means they are the two opposite corners of the die, and every face of view 1 is opposite one face of view 2. But the pictures do not say which one. In fact there are three different ways to pair them, and all three fit the pictures (we checked every case). So, with no extra information, the honest answer is cannot be determined.

There are two ways to get a definite answer.

1. The die is standard. Then the opposite of n is 7 − n, and you do not need the pictures.

2. The question gives one pair, for example "A is opposite D", with A in view 1 and D in view 2. Write A over D. Read view 1 clockwise from A and view 2 anticlockwise from D. Faces in the same column are opposite.

[[fig:1]]

Why is the second view read the other way round? The two views are at opposite ends of the die. The same ring of four faces that turns clockwise when seen from one end turns anticlockwise when seen from the other end. So the two lists run round the ring in opposite senses.

No common face: read view 1 clockwise from the given face and view 2 anticlockwise from its partner.

The geometry does not change when the faces carry letters, symbols or colours. Every face still has exactly one opposite, and a face is never seen together with its opposite. So all the earlier tricks work. In addition, letters allow a quick elimination method.

Take the face X. Look at every view in which X appears and write down the faces seen beside it. None of these is opposite X. If X has been seen beside four different faces, the sixth face, the one never seen with X, is its opposite.

Suppose a die has the letters T, M, B, R, H, K. View 1: Top T, Left M, Right B. View 2: Top R, Left T, Right H. The letter T is seen beside M and B in view 1, and beside R and H in view 2. Four neighbours. The only letter never seen beside T is K, so K is opposite T.

[[fig:0]]

The absent letter. Recent papers also ask for the opposite of a letter that appears in neither view. When the two views share exactly one letter, together they show five different letters, and the sixth letter is absent. The common letter has been seen with all four others, so the absent letter is opposite the common letter. This is safe only when exactly one letter is common. With two common letters it is not guaranteed.

Be careful with rotated letters such as N, Z, H, S: in a turned picture they look different. Compare faces, not shapes on the page. And read the last line of the question: "opposite" is not the same as "adjacent".

Part A: three or more views

A die has only one set of opposite pairs. So when three or more views of the same die are given, they all obey that one pairing, and you may use any two of them. The whole skill is choosing the right two.

The best pair of views has exactly one common face and also contains the face you are asked about. Then you can use the stacking method from the common face. A pair with two common faces gives only the leftover result of Type 2, which usually does not help. A pair with no common face is of no use at all.

Take three views of one die: View 1 = Top 3, Left 2, Right 6. View 2 = Top 5, Left 1, Right 4. View 3 = Top 5, Left 4, Right 6. You are asked for the opposite of 4.

Views 2 and 3 share two faces (5 and 4), which would only tell you that 1 and 6 are opposite, nothing about 4. Views 1 and 3 share exactly one face, 6. Clockwise from 6: view 1 gives 6, 2, 3 and view 3 gives 6, 4, 5. The columns are 2 ↕ 4 and 3 ↕ 5. So 4 is opposite 2.

[[fig:0]]

Now use the remaining view as a check. View 2 shows 4 and 1 together, so 1 cannot be opposite 4, which agrees. Choose the pair with one common face that carries the asked face, stack it, then confirm with the other views.

Collect the sure pairs. Every two views that share two faces give one sure pair (the leftover faces). Every two views that share one face give two pairs by stacking. A die has only three opposite pairs, so once you have two sure pairs, the last two faces are opposite without any more work.

Part B: the "?" face

In this type one picture has a "?" or an empty face, and you must find which number or letter belongs there. The idea is simple: treat "?" as just another symbol, say X, and find what X is opposite to.

First find all the opposite pairs you can from the complete views. The "?" face must be adjacent to the two faces drawn beside it, so it cannot be opposite either of them. Now pick a complete view that shares exactly one face with the incomplete view and stack the two. The column of "?" tells you which face is opposite it, and that is enough to name it.

Take View 1: Top 2, Left 1, Right 5. View 2: Top 3, Left 2, Right 4. View 3: Top ?, Left 6, Right 1. Views 1 and 2 share 2. Clockwise from 2: view 1 gives 2, 5, 1 and view 2 gives 2, 3, 4. So 5 ↕ 3, 1 ↕ 4, and the unseen 6 is opposite 2.

Now view 3 shares only 1 with view 1. Clockwise from 1: view 1 gives 1, 2, 5 and view 3 gives 1, 6, ?. Columns: 2 ↕ 6 (agrees) and 5 ↕ ?. So "?" is opposite 5. But 5 is opposite 3, so ? = 3.

[[fig:1]]

Sometimes elimination leaves two candidates. Then use substitution: put each candidate in, run the stacking with another view, and reject the one that contradicts a known pair. Do not test candidates with a pair of views that share two faces, because it can look consistent when it is not.

Open dice (nets) give two kinds of question. Part A: which face is opposite which when the net is folded? Part B: which closed die can, or cannot, be made from the net? Part A needs no folding in the head, only three facts. Part B uses Part A first and then adds one more idea, the clockwise order.

Part A: opposite faces from a net

An open die, or net, is a die cut along some edges and laid flat: six squares joined edge to edge that fold up into a cube. The question asks which faces end up opposite each other. You do not have to fold it in your head if you use three facts.

1. Gap pairs. In a straight row (or column), two squares with exactly one square between them become opposite faces. A straight strip wraps round the cube like a belt, so squares two apart face each other. In a row of four, the 1st faces the 3rd and the 2nd faces the 4th.

2. S or Z pairs. If four squares make an S or Z shape (a step in one direction, a sideways step, then the same first step again), the two end squares are opposite.

3. Leftovers. Whatever two squares remain unpaired are opposite each other.

[[fig:0]]

In the figure the row Q R S T has gap pairs Q ↕ S and R ↕ T. The squares P, R, S, U make an S shape (down, across, down), so the ends P and U are opposite. The leftovers, P and U, agree.

Squares that share an edge in the net are always adjacent, never opposite. An L-bend of three squares gives no opposite pair.

Two cautions. First, the gap method is for a straight run only. In a staircase, alternate squares are not opposite; use the S/Z method instead. Second, if one square seems to get two different partners, the figure cannot fold into a cube at all.

Part B: net to cube

Now you are given a net and four small pictures of closed dice, and you must find which die can be made, or cannot be made, from the net. From the net you know two things: all three opposite pairs (use Part A), and the clockwise order of faces round a corner.

An option is possible only if it passes two tests:
1. No opposite pair together. A closed die never shows opposite faces in one view.
2. The clockwise order matches the net.

Test 1 is the quick strike-out. Test 2 is what separates a real die from its mirror image, which has the same three faces in the wrong order.

How to read the order from the net. Find a corner of the net where three squares meet at one point with the fourth square missing. The printed side of the net is the outside of the cube, so the clockwise order of the three squares on the paper is the clockwise order of the three faces on the cube. In the figure, round the top-left corner of R the squares read P, R, Q clockwise. On the cube this is Top P, Right R, Left Q.

[[fig:1]]

An option picture is read the same way: Top → Right → Left is clockwise. So the option (Top P, Left Q, Right R) is correct, and (Top P, Left R, Right Q) is its mirror and is wrong.

If the three faces of an option do not meet at a single point of the net, compare it with a corner that does: treat that corner as a second view and use the stacking method of Type 1.

A standard die is a promise made in the question: every opposite pair adds up to 7. Once that promise is there, three kinds of question open up. Which view is possible or impossible (Part A)? Which face is opposite which, even with no picture (Part B)? What do the faces add up to when one or more dice are placed together (Part C)?

Part A: standard or not, possible or impossible views

A standard die is the die of a Ludo board: opposite faces add up to 7. So 1 sits opposite 6, 2 opposite 5 and 3 opposite 4. But the exam does not give you this fact for free. Most dice in the exam are ordinary, where any numbers, letters or symbols can face each other. Use 7 − n only when the question says "standard", "usual" or "ludo" die, or lists the pairs.

Why does the sum 7 matter so much? A picture of a die shows three faces that meet at one corner, and these three are neighbours of each other. So opposite faces are never visible together. Now imagine a view that shows 1, 4 and 6. On a standard die 1 and 6 are opposite faces. They cannot be neighbours. So this view cannot be a picture of a standard die.

[[fig:0]]

If two faces seen together add up to 7, the die is not standard.

The reverse does not work. Suppose a view shows 1, 2 and 3. No pair adds to 7, so the view is possible on a standard die. But that does not prove that the die is standard, because we have not seen the other three pairs. Never say "standard" because no pair gave 7.

To decide whether a view is possible on a standard die, it is enough to check that its three numbers come from three different pairs of 1–6, 2–5 and 3–4. That is the same thing as saying no two of them add to 7.

Part B: pairs given in words

Sometimes there is no picture at all. The stem lists two pairs in words, for example "1 is opposite 6 and 3 is opposite 4", and asks for the opposite of 2. A die has exactly three pairs of opposite faces, and every number belongs to exactly one pair. Two pairs are known, so the two numbers left over, 2 and 5, must form the third pair.

Given two pairs, the two leftover faces are opposite each other.

This needs no 7-pair check. It is pure counting of pairs, so it also works on an ordinary die with any numbers or letters. If the stem gives only one pair, the four remaining faces cannot be paired from that alone.

Part C: sums on standard dice

On a standard die every opposite pair adds to 7, so the six faces add to 1 + 2 + 3 + 4 + 5 + 6 = 21. Each pair also has one odd and one even number. Two simple facts then solve almost every sum question.

Top and bottom. For one die, top + bottom = 7. For k dice, (sum of the tops) + (sum of the bottoms) = 7k. So bottom total = 7k − top total.

The four lateral faces. The four side faces of one die are two opposite pairs, so they add to 14 whatever the top is. For k dice, the lateral total is 14k. Another way to see it: 21 − 7 = 14.

[[fig:1]]

In the figure four standard dice show 4, 3, 1, 5 on top. Their bottoms are 7 − 4, 7 − 3, 7 − 1, 7 − 5, that is 3, 4, 6, 2. The tops add to 13, so the bottoms add to 28 − 13 = 15.

These shortcuts need the dice to be standard. Also be careful about what counts as visible. When dice are placed on a table, the face on the table is hidden. When one die rests on another, the touching faces are hidden too. Read which faces the question actually asks about.

These questions have no stacking at all. They test plain logic about visible and hidden faces, opposite and adjacent. Because the answer comes from reasoning, not from a drawing method, paper-setters use them as a new pattern. Read the type word of the question first, then work with sets.

Part A: single view

One picture shows three faces. The other three are hidden. Here is the key idea: every hidden face is the opposite of exactly one visible face. So a single view already tells you a lot, even though it does not name the pairs.

Take Radhika's die numbered 1 to 6, and suppose the view shows 2, 3 and 6. Call the visible set V = {2, 3, 6}. The hidden set is H = {1, 4, 5}.

[[fig:0]]

Now ask: what can be opposite 1? The face 1 is hidden, so its opposite must be one of the visible faces: 2, 3 or 6. It cannot be 4 or 5, because those are hidden as well, and two hidden faces can never be opposite each other. In the same way, two visible faces are never opposite each other.

Visible faces are opposite to hidden faces, never to each other.

What a single view does not give is the exact pairing. With 2, 3 and 6 all still possible, the face 1 may be opposite any one of them. If the options leave several survivors and the question asks "what is opposite", the honest answer is cannot be determined.

The method is short: write V and H; find the asked face in V or H; its opposite lies in the other set; strike out everything else; count the survivors. One survivor is your answer. Several survivors mean "can be" or "cannot be determined", depending on the question.

Part B: statements about the die

Newer papers ask dice questions in words: "Which statement is definitely true?", "Which is possible?", "Which die cannot be made?". These test the logic of opposite and adjacent, not drawing skill.

First fix three words. A statement is definitely true if it holds on every die that fits the given facts. It is possible if it holds on at least one such die. It is definitely false if it holds on none.

Now the working points.

1. A face has exactly one opposite and four neighbours. So a face is adjacent to every face except its one opposite.
2. "X is adjacent to Y" is definitely true only when Y is certainly not the opposite of X. If the opposite of X is already known, every other face is definitely adjacent to X.
3. In a view, any two visible faces are adjacent, and any two hidden faces are adjacent too (the hidden three meet at the far corner). A visible face and a hidden face may or may not be opposite.
4. A pair given in the question is true only for that question. Do not carry it forward.

[[fig:1]]

In the figure, 2 is opposite 3. So 2 is adjacent to 1, 4, 5 and 6, and "2 is adjacent to 3" is definitely false.

Read the type word first, and underline "not" or "cannot". A missed "not" turns a correct working into a wrong answer.

This is pure formula, the cheapest marks of the chapter. The whole type rests on one number, n, the count of small cubes along one edge. First we cut (Part A), then we paint and count by painted faces (Part B), and last we count painted small faces instead of cubes (Part C).

Part A: cutting a cube

A big cube of edge L is cut into small cubes of edge s. Along one edge you get n = L ÷ s small cubes. The cube then has n layers, each layer is an n × n square of small cubes, so the total number of small cubes is n × n × n = n³.

[[fig:0]]

Number of cuts. To cut a bar into n pieces you need n − 1 cuts. A cube has three directions, so, if the pieces are not rearranged, the number of cuts is 3(n − 1). For n = 4 that is 3 × 3 = 9 cuts, as in the picture. For a cuboid, a cuts, b cuts and c cuts in the three directions give (a + 1)(b + 1)(c + 1) pieces.

If pieces may be stacked. Some questions let you pile the pieces and cut them together. Then each direction needs the smallest number k with 2^k ≥ n, because every cut can at most double the number of pieces along that direction. The minimum is 3k: for n = 4 it is 6 cuts, and for n = 5 it is 9 cuts. Read the stem to see whether rearranging is allowed.

Surface area. Cutting does not change the volume, but it increases the surface. Each small cube has surface 6s², so the total is n³ × 6s² = 6L² × n. After cutting into n³ cubes the total surface area becomes n times the original.

Do not confuse the number of pieces (n³) with the number of cuts (3(n − 1)).

Part B: painted cube, counting by painted faces

A big cube is painted on all six faces and then cut into n³ small cubes. Think about where each small cube sat in the big cube, because its position decides how many painted faces it has.

Why these counts? Each of the 12 edges has n cubes, and the two at the ends are corners, so (n − 2) cubes remain, with two painted faces each. Each face is an n × n square. Remove its border of one cube all round, and an (n − 2) × (n − 2) square is left, with one painted face each. The cubes inside form an (n − 2) cube and touch no paint.

[[fig:1]]

For n = 4 the counts are 8, 24, 24 and 8, and 8 + 24 + 24 + 8 = 64 = 4³. Always add the four counts: they must make n³. This is your free check.

At least one painted face = n³ − (n − 2)³. No small cube ever has four or more painted faces. These formulas need all six faces painted and n ≥ 2 (for n = 2 only the 8 corner cubes exist). First find n from the two edges: n = big edge ÷ small edge.

Part C: total painted faces

Some questions do not ask for the number of painted cubes. They ask for the total number of painted small faces. This is much easier than it sounds, because every painted small face lies on the surface of the big cube, and each small square on the big surface is one small face.

A big cube with edge n small cubes has 6 faces, each an n × n grid of small squares. So when all six faces are painted, total painted small faces = 6n².

[[fig:2]]

In the figure n = 3. The three visible faces show 27 painted squares, and the other three faces have 27 more: 6 × 9 = 54.

You can cross-check with the cube counts of Part B. Corners have 3 painted faces each, edges have 2, face-centres have 1:

8 × 3 + 12(n − 2) × 2 + 6(n − 2)² × 1 = 24 + 24(n − 2) + 6(n − 2)² = 6n².

For k painted big faces (not all six) the total is k × n². The question may also ask for the number of unpainted small faces: the total faces of all small cubes are 6n³, so unpainted = 6n³ − 6n² when all faces are painted.

Be careful to count faces, not cubes. The number of painted cubes is n³ − (n − 2)³, a different thing altogether.

When the solid is a cuboid, or when only some faces are painted, the cube formulas of Type 8 do not apply directly. The method stays the same: ask where each small cube sat in the big solid, and count position by position. Part A gives the formulas for a fully painted cuboid. Part B handles partly painted and two-coloured cubes by cases.

Part A: painted cuboid

A cuboid has three different edges, so it is cut into a × b × c small cubes: a along the length, b along the breadth and c along the height. To find a, b, c, divide each edge by the edge of the small cube. If the cubes are the largest possible, their edge is the HCF of the three edges.

Total number of small cubes = a × b × c.

Now paint all six faces. The position logic is the same as for the cube, only each edge has its own length.

• 3 painted faces: the 8 corners, always 8.
• 2 painted faces: cubes along the edges, not corners. Four edges run along each of the three directions, and an edge of length a has a − 2 such cubes, so the count is 4[(a − 2) + (b − 2) + (c − 2)].
• 1 painted face: the middle of each face. The six faces come in three equal pairs, with (a − 2)(b − 2), (b − 2)(c − 2) and (a − 2)(c − 2) cubes each, so the count is 2[(a − 2)(b − 2) + (b − 2)(c − 2) + (a − 2)(c − 2)].
• 0 painted faces: the inside, (a − 2)(b − 2)(c − 2).

[[fig:0]]

For 5 × 4 × 3: 8, 4 × 6 = 24, 2(6 + 2 + 3) = 22 and 3 × 2 × 1 = 6. Check: 8 + 24 + 22 + 6 = 60 = 5 × 4 × 3. Always add the four counts and compare with a × b × c.

These formulas were checked for every a, b, c from 2 to 7. If any dimension is just 1 piece, they fail, and you must count directly.

Part B: partly painted or two-coloured cube

When only some faces are painted, or different faces have different colours, the cube formulas of Type 8 no longer apply. Do not start from them and subtract a guess. Instead count by position: a small cube's painted faces are the painted faces among the big faces it touches.

All faces painted except one. The unpainted face changes the count in a layer. For a cube of edge 4 with the bottom unpainted:

• Corners: the 4 top corners keep 3 painted faces; the 4 bottom corners fall to 2.
• 2-face cubes: 4 × 2 = 8 along the top edges, and 4 × 3 = 12 on the vertical edges (the bottom end of each vertical edge now has 2): 20 in all.
• 1-face cubes: 4 on the top face, and on each side face 2 across × 3 rows (the bottom row now has just that side painted) = 6, so 24 on the sides: 28 in all.
• 0-face cubes: 8 inside and 4 in the middle of the unpainted bottom: 12.

[[fig:1]]

Check: 4 + 20 + 28 + 12 = 64.

Two colours. Suppose the top and bottom are red and the four sides black. A small cube with one red and one black face lies on one of the 8 edges that join the top or bottom to a side: 8(n − 2) such cubes (n = 4 gives 16). The corners have one red and two black faces, so they are not counted for "one red and one black".

Two adjacent faces painted. The shared edge cubes have both painted faces: n cubes with exactly two painted faces, including the two end cubes. This is quite different from two opposite faces, where no cube has two painted faces.

Count by cases and always check the total n³.

A stack is a floor plan of columns, and each column has a height. You count column by column.

Total cubes = the sum of all the column heights (the numbers in the top-view grid). If you wrongly count only the cubes you can see in the picture, you will miss the ones hidden behind taller columns.

Cubes needed to complete a solid. A solid cuboid has the same floor plan and the height of the tallest column. So cubes needed = (rows × columns × tallest height) − (cubes already there).

[[fig:0]]

In the figure the floor is 3 × 3 and the heights are 3, 2, 1 / 2, 2, 1 / 1, 1, 1. There are 14 cubes. A solid 3 × 3 × 3 has 27, so 13 more are needed.

A solid n × n × n block seen from one corner. The visible cubes are those on the three faces you see, which is n³ − (n − 1)³: for n = 3 that is 27 − 8 = 19, and for n = 4 it is 64 − 27 = 37. The cubes you cannot see from that corner are (n − 1)³. Do not mix this with the cubes that have no face on the outside at all, which are (n − 2)³.

First decide exactly what is asked: total, visible, hidden or needed. Visible and hidden counts depend on the direction you look from.