Counting, Probability & Logic

Let’s call the four-digit number \(N\). The last clue tells us that none of the digits 7, 6, 4 or 2 is a digit in \(N\). Then the fourth clue shows that 1 is a digit in \(N\), but it is not the fourth digit. The first clue now tells us that \(N\) involves a 3 but not as its third digit. It also shows that 1 is not the second digit. The second clue now tells us that either 8 is the second digit of \(N\) and 9 is not one of its digits or else 9 is the first digit of \(N\) and 8 is not one of its digits. Suppose that 8 were the correct second digit. Then the third clue would tell us that both 0 and 5 were correct digits. But this would mean that all of 1, 3, 8, 0 and 5 were digits of the four-digit number \(N\). Therefore 8 is incorrect and so 9 is correct and is the first digit. Knowing this, the third clue shows us that exactly one of 5 and 0 is correct and, moreover, it is in the right place. It can’t be 5 because the first digit of \(N\) is 9. So 0 is the correct second digit. We already know 3 is correct, but is not the third digit; so the last digit of \(N\) is 3 and \(N\) is 9013.

Since each game involved two girls, the number of games played is (10 + 15 + 17) ÷ 2 = 21. Alice played 10 of these, and rested for 11 of them. The maximum amount of resting possible is obtained by alternately losing and resting. To rest 11 times, Alice must have rested the odd-numbered games (1st, 3rd, etc) and lost the even numbered games (2nd, 4th, etc). Hence Alice lost the second game.

Consider first how to join each individual layer. In the 4 × 4 layer there are 4 × 3 × 2 = 24 points of contact. Similarly, in the 3 × 3 layer there are 3 × 2 × 2 = 12 points of contact and in the 2 × 2 layer there are 2 × 1 × 2 = 4 points of contact. Now consider how to join the layers together. To join the 4 × 4 layer to the 3 × 3 layer, each sphere in the 3 × 3 layer has 4 points of contact with the lower layer, making 3 × 3 × 4 = 36 points of contact. Similarly to join the 3 × 3 layer to the 2 × 2 layer, each sphere in the 2 × 2 layer has 4 points of contact with the lower layer making 2 × 2 × 4 = 16 points of contact and the single sphere in the top layer has 4 points of contact with the 2 × 2 layer. Hence the total number of points of contact is 24 + 12 + 4 + 36 + 16 + 4 = 96 and therefore 96 blobs of glue are used.

The large cube is 4 × 4 × 4. There are only two possible configurations of the three painted faces: either they all share a common vertex, or they don’t.

When the three faces share a common vertex, the edges that they share will have more than one face painted, leaving 9 cubes on each of the three faces with exactly one face painted, as shown in the diagram on the left. This is 27 cubes altogether.

When the three faces don’t share a common vertex, then each of the cubes on the common edges will have more than one face painted. This leaves two of the large faces having 12 cubes with one face painted, and the middle large face having 8 cubes, giving a total of 32 cubes, shown in the diagram on the right.

The table can be partially completed as follows. Berta scored 5 points in total. Therefore her score from judge III is 3. Clara’s total is 3. Therefore she cannot have been given 4 by any of the judges. David’s total is 4. He cannot have been given a score of 4 by any judge, since, if so, both the other two judges must have given him 0. This is impossible, as judges I and II give 0 to Berta and Clara, respectively. Therefore judge I gives 4 to Emil, and judge II gives 4 to either Adam or Emil. Emil’s total is 11. So if he gets 4 from judges I and II, he gets 3 from judge III which is not possible as judge III gave Berta 3. Hence judge II gives 4 to Adam. Because Adam’s total is 7, it now follows that Adam gets a score of 1 from judge III.
(Note: the final four scores are not uniquely determined.)

From statement 1 it follows that if a shape is not square, then it is not large; hence all triangles are small. Using statement 2 we see that blue figures are triangles and so are small. This shows that E is true. Suppose we had a set of shapes which consisted of one small blue triangle, one small red triangle, one small red square and one large red square. Then this satisfies statements 1 and 2 but it does not satisfy A, B, C or D. So they are not true in general.

To be divisible by 18, the number must be divisible by 2 and by 9. The digit sum for any number formed is 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45, and hence every number is a multiple of 9. Therefore any even number formed will be divisible by 18. Of the 9 possible last digits, there are four even digits (2, 4, 6, 8), so the probability of the number being even is \(\tfrac{4}{9}\).

We are told there is a red cube at one end and that the three red cubes are all together within the row. Therefore there is a block of three red cubes at one end. If they were at the right-hand end, the tenth cube from the left would be red. But the tenth cube from the left is blue. Hence the red cubes are at the left-hand end and a yellow cube is at the right-hand end, with a blue cube in the tenth place from the left, as shown in the diagram.

The four green cubes are all together within the row and hence are somewhere between the 4th and 9th positions from the left. Whether they start at position 4 or 5 or 6 from the left, the cube in the 6th position from the left is green.

Since no small cubes have three faces painted red, it is impossible to build a large cube with three faces that meet at a vertex painted red. Therefore no more than four of the faces can be red and, without loss of generality, let us consider that they are the front, back and sides of the cube. We can construct a large cube with four faces completely red by arranging that the 12 small cubes along the four vertical edges of the large cube have two red faces showing and the 12 small cubes down the centre of the four vertical faces have one red face showing. Hence it is possible to build a large cube with four completely red faces.

There is exactly one route from "1" to "2", but two routes from "1" to "3" - one route directly from "1" to "3" and one route which moves from "1" to "2" to "3".
To move from "1" to "4", it is necessary to move from "2" directly to "4" or to move from "3" to "4". So the number of routes from "1" to "4" is 1 + 2 = 3.
By a similar method, the number of routes from "1" to "5" is 2 + 3 = 5 and from "1" to "6" is 3 + 5 = 8. Finally, the number of routes from "1" to "7" is 5 + 8 = 13.
(Note that the numbers of routes to each of the squares is a term in the Fibonacci sequence.)

Note that the numbers 123 to 213 are in both option A and option C. Also, numbers 213 to 231 are in both option B and option C, and numbers 231 to 312 are in both C and D. However, numbers 313 to 321 are in option E only. So that must be where Adam’s house is.

We label the five squares, from left to right, \(P, Q, R, S, T\) respectively.
In order to paint two adjacent squares, Lucy could paint \(P\) and \(Q\), or \(Q\) and \(R\), or \(R\) and \(S\), or \(S\) and \(T\). So in four of the finished grids, Lucy’s red squares are adjacent to each other.

Each team plays exactly three matches. Hence the maximum number of points any team can obtain is 3 × 3 = 9. A draw only gets 1 point. Hence the next highest total number of points possible, from two wins and a draw, is 2 × 3 + 1 = 7. Therefore it is impossible to obtain 8 points.
(Note: totals of 4, 5 and 6 points can be obtained by one win, one draw and one loss, one win and two draws and two wins and a loss respectively.)

Let the largest number of teams that can enter the tournament be \(n\). Hence there would be \(3n\) players in total. Each player in a team will play a game against every player from all other teams and hence the total number of games played is \(\tfrac{3n \times (3n - 3)}{2} = \tfrac{9n(n - 1)}{2}\). Since no more than 250 games can be played, we have \(\tfrac{9n(n - 1)}{2} \leqslant 250\). Therefore \(n(n - 1) \leqslant \tfrac{500}{9} = 55\tfrac{5}{9}\). Since \(8 \times 7 = 56 > 55\tfrac{5}{9}\), we have \(n < 8\). Also, since \(7 \times 6 = 42 < 55\tfrac{5}{9}\), \(n = 7\) is a possible solution. Therefore the largest number of teams that can enter the tournament is 7.

Since no two circles that are joined directly are painted the same colour and circles 2, 5 and 6 are joined to each other, they are all painted different colours. Similarly circles 2, 6 and 8 join to each other and hence are painted different colours. Therefore circles 5 and 8 must have been painted the same colour. It is easy to check that, given any other pair of circles in the diagram, it is possible for them to be coloured differently.

Since the integer 7 is joined by a diameter to the integer 23, we can deduce that there are 23 − 7 + 1 = 15 integers between them on each side. Therefore there are 2 × 15 + 2 = 32 integers in total round the circle. Hence the value of \(n\) is 32.

Since each box has a different number of pears, and we want as many boxes as possible, we could start by putting no pears in box 1, 1 pear in box 2, 2 pears in box 3, and so on. By the 11th box we have 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55 pears, and there are not enough pears for another box. However, we can’t use 11 boxes because we need to share the 60 apples evenly and 60 is not a multiple of 11. So, the maximum possible is 10 boxes with 6 apples in each box, and the pears distributed as stated but with 24 in the 10th box.

Since no integer is both odd and even, exactly one of the first two speakers was telling the truth. No square is also prime; so at most one of the third and fourth speakers was telling the truth. Hence the number telling the truth must be one or two. If there was just one telling the truth then the statements of speakers 1 and 4 would both be true - which is a contradiction. If there were two telling the truth, then speakers 2 and 3 would be the ones telling the truth; and this is indeed two speakers.

Since Dora shook hands four times, she shook hands with all the other four people. Hence, since Alan only shook hands once, it was with Dora. Since Claire shook hands three times and did not shake hands with Alan, she shook hands with Bella, Dora and Erik. Hence, since Bella only shook hands twice, it was with Dora and Claire. Therefore Erik shook hands twice (with Dora and Claire).

For each of the 5 ways in, there are 4 ways out, so there are 5 × 4 = 20 ways.

The information in the question tells us that Lotar finished before Manfred who finished before Jan who finished before Victor. Since Eddy also finished before Victor, it was Victor who finished last of these five runners.

The smallest total that can be achieved is 1 + 1 + 1 = 3, and the greatest is 6 + 6 + 6 = 18. Every integer total in between can be obtained, so there are 16 possibilities.

In order from 00:00, the times when the digits 2, 0, 1, 9 appear in some order are 01:29, 02:19, 09:12, 09:21, 10:29, 12:09, 19:02, 19:20, 20:19 and 21:09. So, in total, there are ten such times.

Kerry uses the digit 0 five times and hence the value of \(n\) is smaller than 60. She uses the digit 9 six times and hence the value of \(n\) is at least 59. Therefore the value of \(n\) is 59.

It is known that any diagram with at most two points where an odd number of lines meet can be drawn without lifting your pencil off the page and without drawing along the same line twice. Any diagram with more than two such points cannot be drawn in this way. Of the options given, only diagram D has more than two such points. Hence the diagram which cannot be drawn is D.

Each column will contain two numbers that add to a multiple of 3. The only possible pair for 3 is 6 so one of these will appear in the top row, and the other on the bottom row. The numbers in the same row as 3 must also add to a multiple of 3, so could be 1, 2 or 1, 5 or 2, 4 or 4, 5. Once this row is decided, the other row is entirely determined by this choice; e.g. if the top row is 3, 1, 2, then the bottom row must be 6, 5, 4. Hence, ignoring order in the first instance, there are just these four choices for what goes in the rows.

For each of these, the rows can be swapped, which doubles the number of options to 8.
For each of these, the three columns can be arranged in 6 ways (3 choices for first column, 2 choices for second, so 3 × 2 = 6 choices). Hence there are 8 × 6 = 48 choices.

The dominoes in the line contain three ends with four spots and three ends with six spots, as shown in diagram 1. Therefore, a correctly arranged set of these dominoes will have four spots at one end and six spots at the other, as is currently the case. Hence, Paul does not need to move either of the end dominoes.
If he swaps the third and the fifth dominoes from diagram 1, he obtains the row shown in diagram 2 which has the same number of spots in the adjacent ends of the fourth, fifth and sixth dominoes. Next, if he swaps the second and third dominoes from diagram 2 to obtain the line shown in diagram 3, he has matched the spots at the adjacent ends of the first and second dominoes. Finally, rotating the third domino in diagram 3, he obtains the correctly arranged line as shown in diagram 4. This shows that it is possible to arrange the dominos correctly in three moves.
To see that two moves is not sufficient, note that, whatever else needs to happen, the two 1s must be correctly placed next to each other. To do that requires one of the dominos with a 1 to be rotated and then one pair of dominos to be swapped so that the two 1s are now next to each other. A similar argument applies to the two dominos with a 3. However, this is not possible in only two moves. Therefore the smallest number of moves he needs to make is 3.

In between a vertex labelled \(n\) and a vertex labelled \(m\) (with \(m > n\)), there are \(m - n + 1\) vertices (including the end points). Hence between vertex 18 and vertex 1018, there are 1001 vertices; between vertex 1018 and vertex 2000 there are 983 vertices. Continuing clockwise from vertex 2000 to vertex 18, we count 37 vertices (and the polygon also has vertex 1018, making 38). Therefore the resulting polygons have 1001, 983 and 38 vertices.

The number of votes left to be cast is 130 – (24 + 29 + 37) = 40. Let the number of these votes Alan receives be \(x\). Since Katie is the closest challenger to Alan, for Alan to be certain of having the most votes, \(37 + x > 29 + 40 - x\). Therefore \(2x > 32\) and hence \(x > 16\). Therefore Alan needs at least 17 more votes to be certain to finish with the most votes.

The slogan suggests that there will be at most 15 days without sun in 2018. In this case it is possible to have a run of 31 days without having two consecutive days of sun by alternating a sunny day with a non-sunny day, starting with a sunny day. Hence Will Burn has to stay for one further day, or 32 days in total, to be certain of having two consecutive days of sun.

Each of the trains leaves one town and arrives at another town so there are a total of 80 start/finish points. Forty are already mentioned, so forty must begin/end at Jena.

Consider the three possible positions of the lion in turn. If the lion were in room 1, the notes on both room 1 and room 2 would be true. Hence the lion is not in room 1. If the lion were in room 2, the notes on all three rooms would be false. Hence the lion is not in room 2. However, if the lion were in room 3, the notes on room 1 and room 3 would be false but the note on room 2 would be true. Hence the lion is in room 3.

The number of terms in the sequence up to and including the \(n\) occurrences of \(n\) is \(1 + 2 + 3 + \ldots + n = \tfrac{1}{2}n(n + 1)\), which equals 105 when \(n = 14\). Hence the numbers divisible by 3 are 3 (three times), 6 (six times), 9 (nine times) and 12 (twelve times), giving a total of \(3 + 6 + 9 + 12 = 30\) numbers.

The total mass is 621g so any three masses with total mass over 310.5g could be in the heavier pan. There are eight of these triples that include the 106g mass: (106, 105, 104), (106, 105, 103), (106, 105, 102), (106, 105, 101), (106, 104, 103), (106, 104, 102), (106, 104, 101), and (106, 103, 102).
Without 106, there are 2 ways to make a set over 310.5g: (105, 104, 103) and (105, 104, 102).
Hence the probabililty that the 106g mass is included in the heavier pan is \(\tfrac{8}{8 + 2} = \tfrac{8}{10}\) or 80%.

Let the five positions in the photograph be numbered 1, 2, 3, 4, 5. Then the boys may occupy a total of six positions: 1 and 3; 1 and 4; 1 and 5; 2 and 4; 2 and 5; 3 and 5. For each of these positions, the boys may be arranged in two ways as they can interchange places. So there are 12 ways of positioning the boys. For each of these, the girls must be placed in three positions. In each case, the first girl may choose any one of three positions, the second girl may choose either of two positions and then there is just one place remaining for the third girl. So for each arrangement of the two boys there are 3 × 2 × 1 different ways of arranging the three girls. Therefore the total number of line-ups is 12 × 6 = 72.

Each of the nine tunnels in Stan's cube is five cubes long. However, the three tunnels starting nearest to the top front vertex of the cube all intersect one cube in. Similarly, the three tunnels starting at the centres of the faces all intersect at the centre of the large cube and the final three tunnels all intersect one cube in from the other end to that shown. Hence the number of small cubes not used is 9 × 5 − 3 × 2 = 45 − 6 = 39.

Each ‘triple’ consisting of a cell and the two cells immediately below can have at most two odds (for if the bottom two are both odd, the one above is even, so they cannot be all odd). The whole diagram can be dissected into six of these (shaded) triples as shown in the top diagram, with three other (white) cells left over. These six triples have at most \(6 \times 2 = 12\) odds between them. Moreover, the three remaining white cells cannot all be odd; if we assign the values \(A\) and \(C\) to the lowest of these white cells, and \(B\) to the cell between them, then the cells above have values \(A + B\) and \(B + C\). The top white cell then contains \(A + 2B + C\), which is even when \(A\) and \(C\) are both odd. Hence the three white cells have at most two odds, giving the whole diagram at most 12 + 2 = 14 odds.
The second diagram shows one possible way of achieving this maximum of 14 odds.

The password has length 7 so the different digits making it up must add to 7. The possibilities are: {7}, {6, 1}, {5, 2}, {4, 3}, {4, 2, 1}. Using only the digit 7 produces just one password, 7777777. Using two digits gives two possibilities, depending on which digit goes first, so the three pairs give 2 × 3 = 6 phone numbers. Three different digits can be arranged in six ways. This gives 1 + 6 + 6 = 13 possibilities.

We can get a negative product if the first die is negative and the second positive, with probability \(\tfrac{3}{6} \times \tfrac{2}{6} = \tfrac{6}{36}\), or if the first die is positive and the second is negative, with probability \(\tfrac{2}{6} \times \tfrac{3}{6} = \tfrac{6}{36}\). Together this gives a probability of \(\tfrac{12}{36} = \tfrac{1}{3}\).

Label the kangaroos facing right as K1, K2, K3, K4, K5 and K6 as shown in the diagram.

No further exchanges will be possible only when the kangaroos facing right have moved past all the kangaroos facing left. Kangaroos K1, K2 and K3 each have four left-facing kangaroos to move past while kangaroos K4, K5 and K6 each have two left-facing kangaroos to move past. Hence there will be (3 × 4 + 3 × 2) = 18 exchanges made before no further exchanges are possible.

Since 5 and 2 differ by 3, Tom can obtain the same integer from two different integers in the first list that also differ by 3 by adding 5 to the smaller integer and adding 2 to the larger integer. In the first list there are six pairs of integers that differ by 3, namely 1 and 4, 2 and 5, 3 and 6, 4 and 7, 5 and 8 and 6 and 9. However, the integers 4, 5 and 6 appear in two of these pairs and hence the same integer in the second list can be obtained from only three pairs of integers from the first list leaving three integers in the first list unpaired. Therefore, the smallest number of different integers Tom can obtain in the second list is six.

Whichever day of the week Margot chooses for her first jogging day, there are four other days she can choose for her second day since she does not want to jog on either the day before or the day after her first chosen day. Therefore there are 7 × 4 = 28 ordered choices of days. However, the order of days does not matter when forming the timetable, only the two days chosen. Hence Margot can prepare 28 ÷ 2 = 14 different timetables.

Let \(m\) be the number of days with sunny mornings and wet afternoons. Let \(n\) be the number of days with sunny mornings and sunny afternoons. There were 5 sunny mornings so \(m + n = 5 \ldots\) (1). Since there are seven wet days, the number of days with wet mornings and sunny afternoons must be \(7 - m\). There are 6 sunny afternoons so \(n + (7 - m) = 6\), which rearranges to \(m = n + 1 \ldots\) (2).
Equations (1) and (2) together give \(m = 3\), \(n = 2\), so Johannes had 3 days with sunny mornings and wet afternoons, 2 days sunny all day, and 4 days with wet mornings and sunny afternoons, a total of 9 days (not counting any cloudy days he may have had!).

Rohan wants to obtain 9 equal pieces and so makes eight marks. Jai wants to obtain 8 equal pieces and so makes seven marks. Since 9 and 8 have no common factors (9 and 8 are co-prime), none of the marks made by either boy coincide. Therefore Yuvraj will cut at 15 marked points and hence will obtain 16 pieces of string.

Any day when Tycho jogs is immediately followed by a day without a jog. Therefore any period of seven days has three pairs of ‘jog, no-jog’ days and one extra no-jog day. There are seven possibilities for this extra non-jog day, so seven distinct schedules.

If exactly one of \(N\) and \(N + 20\) has four digits, then the other has either 3 or 5 digits. If \(N\) has 3 digits and \(N + 20\) has 4 digits, then \(980 \leqslant N \leqslant 999\), giving 20 possibilities. If \(N\) has 4 digits and \(N + 20\) has 5 digits, then \(9980 \leqslant N \leqslant 9999\), giving 20 possibilities. Overall there are 40 possibilities for \(N\).

The ages of the four cousins are 3, 8, 12 and 14. When these are added in pairs, we obtain 3 + 8 = 11, 3 + 12 = 15, 3 + 14 = 17, 8 + 12 = 20, 8 + 14 = 22 and 12 + 14 = 26. Only two of these, 15 and 20, are divisible by 5. However, we are told that the sum of the ages of Alan and Dan and the sum of the ages of Carl and Dan are both divisible by 5. Hence, since Dan's age appears in both sums that are divisible by 5, his age is 12. Since Alan is younger than Carl, Alan's age is 3 and Carl's age is 8. Hence Bob's age is 14. Therefore the sum of the ages of Alan and Bob is 3 + 14 = 17.

If the student has taken six buttons, he may already have three of the same colour, but it is possible that he has exactly two of each. However, if he takes a seventh button, he is guaranteed to have three of the same colour.

Antonia is fifth to the left of Bianca, so there are four girls in between. Similarly there are seven between them to the right. Hence there are 4 + 7 + 1 + 1 = 13 girls.

Since Florence is the fourth on the left from Jess, there are three girls between them going left round the circle. Similarly, since Florence is the seventh on the right from Jess, there are six girls between them going right. Therefore there are nine other girls in the circle apart from Florence and Jess. Hence there are 11 girls in total.

P2015 shakes hands with each of the other 2015 candidates (including P2016). In particular he shakes hands with P1, leaving P1 with no more handshakes to perform. P2014 then shakes hands with each of the candidates, not including P1. But then P2 has used up his two handshakes (once with P2015 and once with P2014). Proceeding in this way, we see that P2013 uses up the third shake for P3, P2012 uses up the fourth shake for P4, and so on, until we get towards the half-way point. The handshakes of P1009 again include P2016 and use up the 1007th shake for P1007. By this point, P2016 has shaken hands with each of P1009 to P2015, and now must provide the 1008th shake for P1008, a total of 1008 shakes for P2016.

Note first that a small cube in the centre of a face of the large cube will only appear on one face while a cube appearing on the edge of a face of the large cube will appear on two faces and a cube appearing at a corner of the face of the large cube will appear on three faces. Hence, the total number of white faces on the edge of the large cube is an even number and the total number of white faces on the corners of the large cube is a multiple of 3. The five faces shown contain 1 centre white face from 1 small white cube, 12 edge white faces and 5 corner white faces. Therefore, since the total number of white faces on the corners is a multiple of 3, the missing face contains 1 or 4 white faces at its corners. None of the options contains 4 white corners so the missing face contains one white corner as in options A and E, making 6 in total. These 6 faces come from 6 ÷ 3 = 2 small white cubes. Both options A and E have two white faces on their edges, making 14 in total over the six faces from 14 ÷ 2 = 7 white cubes. Hence the number of white cubes whose positions we know is 1 + 7 + 2 = 10. The large cube is made with 12 small white cubes so there are still two more to be placed. One can be at the centre of the large cube and the only place the remaining cube can be is at the centre of the missing face. Therefore, the missing face contains one centre white face, two edge white faces and one corner white face. Hence, the missing face is A. (This proof shows that the only possible missing face for such a cube is face A. It is left to the reader to check that the five given faces and face A can indeed be fitted together consistently to form the faces of a cube.)

The smallest possible first digit in the year is 2. This eliminates December from the month, leaving only months that start with 0, or 10 (11 cannot be used because it has a repeated digit). So we will definitely use 0 for the month. Not using 0 or 2 for a day leaves us with 13 to 19, or 31, all of which use a 1. Since we are using 0 and 1 for the month and day, the earliest year we could have is 2345. The earliest month is then 06, and the earliest day that can be made is 17. This gives 17.06.2345. So the month is June.

Let the numbers of passengers in the five carriages be \(p\), \(q\), \(r\), \(s\) and \(t\) respectively with \(p\), \(q\), \(r\), \(s\) and \(t\) all at least 1. Consider the neighbours of the passengers in the first and second carriages. Since each passenger has 5 or 10 neighbours, we have \(p - 1 + q = 5\) or 10 and \(p + q - 1 + r = 5\) or 10. Therefore \(p + q = 6\) or 11 and \(p + q + r = 6\) or 11. However, we know that \(r \geqslant 1\) and hence \(p + q \leqslant 10\) and therefore \(p + q = 6\) and \(r = 5\). Similarly, considering the neighbours of the passengers in the fourth and fifth carriages, we obtain \(s + t = 6\) and (again) \(r = 5\). Therefore, the total number of passengers in the train is 6 + 5 + 6 = 17. (Note that while the total number of passengers in the train is uniquely determined, the arrangement of these passengers in all but the centre carriage is not unique.)

For brevity, let T denote a truth teller and L a liar. Clearly each T has to have an L on each side. Each L either (i) has a T on each side or (ii) has an L on one side and a T on the other side. The largest number of Ts will occur if (i) is always the case. This gives the arrangement TLTLTL… which, since 2 divides 2016, joins up correctly after going round the table. In this case the number of Ts is \(\frac{1}{2} \times 2016\). The smallest will occur if case (ii) always is the case. This gives the arrangement LLTLLTLLT… which, since 3 divides 2016, also joins up correctly. In this case the number of Ts is \(\frac{1}{3} \times 2016\). The difference is \(\frac{1}{6} \times 2016 = 336\).

Every multiple of 6 is a holiday. For \(n > 0\), the days in between \(6n\) and \(6n + 6\) will contain three consecutive non-primes \(6n + 2\) (divisible by 2), \(6n + 3\) (divisible by 3) and \(6n + 4\) (divisible by 2). Using H to represent a holiday, W a working day and ? for an unknown day, we see the numbers \(6n\) to \(6n + 6\) form the pattern H?WWW?H. We are searching for the pattern HWH but this will not fit into the pattern shown above. Hence, the only days that can possibly give HWH must occur in the first week of the month. The days 1, 2, 3, 4, 5, 6 have the pattern WHHWHH so contain one occurrence of HWH.
The first day of the month could possibly be a working day between two holidays, but a quick check shows that of course day 40 is also a working day.

The diagram below shows the day on which certain cubes turned grey.

As can be seen, at the end of the second day there are 11 + 5 + 1 = 17 grey cubes.

The pair who play in the final will have played three matches. Only Glen and Carl play three matches so they are the pair who play in the final.

The most matches that any player could play is three. Any player who wins twice will play in the final. Hence Celine and Gina must be the two finalists, and Gina beat Celine. This means Gina must have won three matches altogether, but only two are recorded. Hence the missing result must be that Gina beat someone.
All the other players must have lost once, but Emma has no loss recorded, so the missing result must be that Gina beat Emma.
Indeed, from the given information we can deduce that the pairings must have been as shown (using players' initials instead of their full names):

Sven can choose from {1, 2, 3, 4, 5, 6, 7, 8, 9}, but can choose at most one from each of the pairs that add to 10: {1, 9}, {2, 8}, {3, 7}, {4, 6}. Since this gives him a maximum of 4 integers, he must always pick the digit 5.

In order to get from TEAM to MATE, at some point M will have to pass each of T, E and A. Likewise Angelo will have to move A in front of T and E. So at least 5 swaps are required. The list: TEAM, TEMA, TMEA, MTEA, MTAE, MATE shows it can be done in five swaps.

Any route starts by going from \(S\) to \(P\) or \(S\) to \(Q\) or \(S\) to \(R\). Any route starting \(S\) to \(P\) must then go to \(Q\) and then has the choice of going clockwise or anticlockwise round triangle \(QSR\), giving two possible routes. By a similar argument, there are two routes that start \(S\) to \(R\). For those routes that start by going from \(S\) to \(Q\), there is then the choice of going clockwise or anticlockwise round quadrilateral \(QPSR\), giving two more routes. Therefore there are six possible routes in total.

Suppose the first of the two special points mentioned has \(a\) points to its left, and \(b\) points to its right. Then the number of line segments it lies on is \(a \times b\) (\(a\) choices for the left end, \(b\) choices for the right end). Also the number of points will be \(a + b + 1\). Similarly if the second point has \(c\) points to its left, and \(d\) to the right, then the number of line segments it lies on is \(c \times d\), and the number of points is \(c + d + 1\).
Hence we need to find integers \(a\), \(b\), \(c\), \(d\) such that \(ab = 80\), \(cd = 90\), \(a + b + 1 = c + d + 1\) (or more simply \(a + b = c + d\)).
The factor pairs of 80 (and hence the possible values of \(a\), \(b\)) are \(1 \times 80\), \(2 \times 40\), \(4 \times 20\), \(5 \times 16\), \(8 \times 10\).
The factor pairs of 90 (and hence possible values of \(c\), \(d\)) are \(1 \times 90\), \(2 \times 45\), \(3 \times 30\), \(5 \times 18\), \(6 \times 15\), \(9 \times 10\).
The only pairs for which we have \(a + b = c + d\) are 5, 16 and 6, 15. Since both of these add to 21, the number of points must be \(a + b + 1 = 21 + 1 = 22\).

There are just six ways to colour the positive integers to meet the two conditions:
(a) All positive integers are coloured red.
(b) 1 is coloured red and all the rest are coloured green.
(c) 1 is coloured red, 2 is coloured green and all the rest are coloured red.
(d) All positive integers are coloured green.
(e) 1 is coloured green and all the rest are coloured red.
(f) 1 is coloured green, 2 is coloured red and all the rest are coloured green.

Note that (d), (e) and (f) can be obtained from (a), (b) and (c) respectively by swapping round the colours red and green.
It is easy to see that these colourings meet the two conditions given. We now need to show that no other colouring does so.

We first show there is no colouring different from a), b) and c) in which the number 1 is coloured red. In such a colouring, since it is different from b), there must be some other number coloured red. Let \(n\) be the smallest other number coloured red. Then, as 1 and \(n\) are red, so is \(n + 1\). Hence \((n + 1) + 1 = n + 2\) is also red, and so on. So all the integers from \(n\) onwards are red. Since such a colouring is different from a), \(n \neq 2\), and since it is different from c) \(n \neq 3\). So \(n \geqslant 4\) and therefore 2 and 3 are green. But then 2 + 3 = 5 is green, so 2 + 5 = 7 is green and so on. So all positive integers of the form \(2k + 3\) are green. In particular \(2n + 3\) is green, contradicting the fact that all integers from \(n\) onwards are red. Hence no such colouring exists.

A similar argument shows that there is no colouring meeting the given conditions other than (d), (e) and (f) in which 1 is coloured green.

It is certainly possible to underline two numbers. In the list below both 1 and −1 are equal to the product of the other nine numbers \(1, -1, -2, \tfrac{1}{2}, 3, \tfrac{1}{3}, 4, \tfrac{1}{4}, 5, \tfrac{1}{5}\).
However, it is not possible to underline three numbers. For supposing it was possible to underline the numbers \(a\), \(b\), \(c\). Let \(N\) be the product of the other seven numbers. Then we have
\[a = b \times c \times N \ldots (1) \qquad b = a \times c \times N \ldots (2) \qquad c = a \times b \times N \ldots (3)\]
Substituting (1) into (2) gives \(b = (b \times c \times N) \times c \times N = b \times c^2 \times N^2\) so \(c^2 \times N^2 = 1\). Thus \(c \times N = 1\) or \(c \times N = -1\). But if \(c \times N = 1\), then (1) becomes \(a = b\), contradicting that \(a\), \(b\) are distinct. Hence \(c \times N = -1\), and (1) becomes \(a = -b\). Substituting this into (3) gives \(c = (-b) \times b \times N\). Multiplying both sides by \(c\) gives \(c^2 = -b^2 \times c \times N = -b^2 \times (-1) = b^2\). Hence either \(c = b\) (contradicting that they are distinct) or \(c = -b = a\) (contradicting that \(a\), \(c\) are distinct).
Hence there is no way that \(a\), \(b\), \(c\) can be distinct numbers and also underlined. (If the numbers are allowed to be the same, it is possible to underline them all by choosing the numbers to be all equal to one).

Let the five points be \(P\), \(Q\), \(R\), \(S\) and \(T\) with individual distances between them of \(w\), \(x\), \(y\) and \(z\) as shown.

The maximum distance between any two points is 22 so \(PT = 22\). The next largest distance is 20 and, since no distance is 1, this is either \(PS\) or \(QT\). Assume \(QT = 20\) so that \(PQ = w = 2\). The next largest distance is 17 and, since no distance is 3 this cannot be \(QS\) or \(RT\) and so is \(PS\). Hence \(ST = z = 5\) and \(QS = 15\). The remaining distances are 6, 8, 9 and \(k\) which represent the lengths of \(PR\), \(QR\), \(RS\) and \(RT\) in some order with \(QR + RS = 15\). Since \(k > 9\), the only possible pair of distances adding to 15 is 6 and 9 and so \(QR = x = 6\) and \(RS = y = 9\) (since \(QR = 9\) and \(RS = 6\) would mean two distances are 11) leaving \(PR = 8\) and \(RT = k = 14\). Hence the value of \(k\) is 14 with the distances between the points taking the values shown below.

(Note: The same distances would result but in the reverse order if we assumed \(PS = 20\).)

The permutations which follow UKIMC in dictionary order are UKMCI, UKMIC, UMCIK, UMCKI, UMICK, UMIKC, UMKCI, UMKIC. There are eight of these, so UKIMC is 112th in the list.

Since the three digits \(a\), \(b\), \(c\) are different and \(\overline{ab} < \overline{bc} < \overline{ca}\), it is necessary that \(a < b < c\). But this condition is also sufficient. Also note that none of \(a\), \(b\), \(c\) are zero because they each represent the tens digit of a two-digit number.
There are 9 ways to pick a non-zero digit, 8 ways to pick a second (different) digit, and 7 ways to pick a third digit. These digits can be arranged in \(3 \times 2 \times 1 = 6\) ways, but only one of these will be in ascending order (we need \(a < b < c\)). Hence there are \((9 \times 8 \times 7) \div 6 = 84\) ways to choose the digits \(a\), \(b\), \(c\).

Since there is no overlap of wires, each vertex of the cube requires at least one end of a piece of wire to form it. A cube has eight vertices and each piece of wire has two ends, so the minimum number of pieces of wire required is 8 ÷ 2 = 4.
Such a solution is possible, for example with wires of length 1 cm, 2 cm, 4 cm and 5 cm, the arrangement of which is left for the reader.

There are 3 different possibilities for the first character. The second character may be any digit from 0 to 9 inclusive, so it has 10 different possibilities. The third character differs from the second digit, so has 9 different possibilities. Once the second and third characters are determined, the fourth character is also determined since it is the units digit of the sum of the second and third characters.
So, the number of different codes is 3 × 10 × 9 = 270.

Petra can arrange the dictionaries in six ways (3 choices for the first dictionary, 2 choices for the second dictionary, 1 choice for the third, giving 3 × 2 × 1 = 6 ways). The novels can be arranged in two ways. Since the novels could be on the left of the dictionaries, or on the right, we have a total of 6 × 2 × 2 = 24 ways to arrange the books.

The 12 pentagonal panels have a total of 12 × 5 = 60 edges. The 20 hexagonal panels have a total of 20 × 6 = 120 edges. So in total the panels have 180 edges. When the panels are sewn together, two edges form each join. So the number of joins is 180 ÷ 2 = 90.

All the students have given different answers to the questions so only one, at most, can be telling the truth. Suppose no student is telling the truth; but then Daniel is telling the truth, contradicting this. Hence exactly one student, Ellen, is telling the truth and so only one student had done their homework.

The first digit can be anything from 1 to 9. Where possible, we can reduce this by three or increase it by three to get the next digit, giving the following possibilities for the first two digits: 14, 25, 30, 36, 41, 47, 52, 58, 63, 69, 74, 85, 96. Repeating for the third digit, we obtain the possibilities in numerical order: 141, 147, 252, 258, 303, 363, 369, 414, 474, 525, 585, 630, 636, 696, 741, 747, 852, 858, 963, 969, which is 20 options.

Label the internal sides of the diagram \(a\), \(b\), \(c\), \(d\) and \(e\) as shown. The side labelled \(a\) is in a triangle with a green side and in a triangle with a blue side and so is to be coloured red. This is also the case for the side labelled \(e\). Hence, the side labelled \(b\) is in a triangle with a red side and a green side and so is to be coloured blue while the side labelled \(d\) is in a triangle with a red side and a blue side and so is to be coloured green. Finally, the side labelled \(c\) is in a triangle with a green side and in a triangle with a blue side and so is to be coloured red. Hence the side labelled \(x\) is in a triangle with a side that is to be coloured blue (side \(b\)) and with a side that is to be coloured red (side \(c\)). Therefore the side labelled \(x\) is to be coloured green.

From the first picture, we can see:
\[\begin{array}{lrcll} \text{From the right scale:} & Z & > & Y & \quad (1) \\ \text{From the left scale:} & X & > & W & \quad (2) \\ \text{From the large scale:} & Y + Z & > & W + X & \quad (3) \end{array}\]
It follows from (1), (2) and (3) that \(Z + Z > Y + Z > W + X > W + W\), so \(2Z > 2W\), and hence \(Z > W\) … (4).
We can show that most swaps give a contradiction of these inequalities:
Firstly, suppose that weight \(Z\) doesn’t move. Then there are three possible swaps:
\(X\) and \(Y\) swap: then in the second picture we must have \(Z < X\) and \(Y < W\), which add to give \(Y + Z < W + X\), contradicting (3).
\(Y\) and \(W\) swap: then we would have \(Z < W\), contradicting (4).
\(X\) and \(W\) swap: then we would have \(X < W\), contradicting (2).
Hence \(Z\) must swap, which can happen in three ways:
\(Z\) and \(Y\) swap: then we would have \(Z < Y\), contradicting (1).
\(Z\) and \(X\) swap: then we would have \(Z < W\), contradicting (4).
\(Z\) and \(W\) swap: This is the only possibility left, and it can work when the weights \(W\), \(X\), \(Y\), \(Z\) are 1, 2, 3, 4. The swap would give 4, 2, 3, 1 which matches the picture on the right. However, this does depend on the values of \(W\), \(X\), \(Y\), \(Z\); it would not work for 1, 5, 3, 4.

The aliens say they can see 8, 7, 5 ears respectively, but each ear has been seen by two aliens so is counted twice. Hence the total number of ears is \(\tfrac{1}{2}(8 + 7 + 5) = 10\) ears. Each alien can see all ten ears, except its own. Timi sees 5 ears, so has \(10 - 5 = 5\) ears.

The sum of the digits on the left-hand side of the equation is 24. Hence the equation formed by inserting + and − signs must be equivalent to 12 − 12 = 0. The smallest number of the given digits required to make 12 is three (2 + 5 + 5). Therefore the smallest number of asterisks that can be replaced by + is two and they would be placed in front of two of the three 5s.

We will show that the word “YES” appears exactly three times, giving the probability 3/6 or 1/2. Firstly note that “YES” appears twice on the second die. The third die also shows “YES” and this cannot be the same as either “YES” on the second die: Under the first “YES” is “MAYBE”, but on the third die the word “NO” appears below it; to the left of the second “YES” is “MAYBE”, but to the left of the “YES” on the third die is “NO”. Hence “YES” appears at least three times. However, it appears at most three times because there are two occurrences of “NO” shown in the third die, and one “MAYBE” in the second die. The first die has not been used in the above argument, but is consistent with the faces showing “YES” three times, “NO” twice, and “MAYBE” once.

Suppose another child were to join the class. The question tells us that then one of the two conditions would no longer be true. For this to happen, there must be no day of the week available for a new boy to have been born on and no month of the year available for a new girl to have been born in. Hence there must be 7 boys and 12 girls currently in the class and so there are 19 children in total in the class.

The tetrahedron has 6 edges and 4 vertices, so the required product is 6 × 4 = 24.

There are three children who like both subjects, leaving 30 children to be shared in the ratio 2:1. Hence 20 like only Computer Science, 10 like only PE, 3 like both. The total number who like Computer Science is 20 + 3 = 23.

There are 9000 four-digit integers. To calculate the number of these which have four different digits, we note that we have a choice of 9 for the thousands digit. We now have a choice of 9 for the hundreds digit (since we can choose 0 as a possible digit). After these two digits have been chosen, we have a choice of 8 for the tens digit and then 7 for the units digit. So the number of four-digit numbers in which all digits are different is 9 × 9 × 8 × 7.
Therefore the number of four-digit numbers which have at least one digit repeated is 9000 − 9 × 9 × 8 × 7 = 9(1000 − 9 × 8 × 7) =
9 × 8 × (125 − 9 × 7) = 72 × (125 − 63) = 72 × 62.

The smallest such number is 1 + 2 = 3, whilst the largest is 99 + 100 = 199. Every number between 3 and 199 may be written as 1 + \(n\) with \(n = 2, 3, \ldots, 99\) or as 100 + \(n\) with \(n = 1, \ldots, 99\). So in total there are (199 − 3) + 1 = 197 such numbers.

Precisely one of Jenny and Willie is telling the truth since the number of people is either even or odd. Similarly, precisely one of Sam and Mrs Scrubitt is telling the truth since the number of people is either a prime number or a number which is the product of two integers greater than one. So although it is not possible to deduce who is telling the truth, it is possible to deduce that exactly two of them are doing so.

The missing die, if correctly placed in the figure, would show faces 1, 3, 5 placed in a clockwise direction around the nearest corner. An examination of each of the five proposed dice shows that only C has this property.

If Alfred is telling the truth, the other three are lying (as their statements would then be false) and we know this is not the case. Hence Alfred is lying. Similarly, if Horatio is telling the truth, the other three are lying which again cannot be the case. So Horatio is lying. Hence the two who are telling the truth are Bernard and Inigo. (A case where this situation would be realised would be if the brothers in descending order of age were Alfred , Bernard, Horatio and Inigo.)

As the words ‘three’ and ‘five’ contain 5 and 4 letters respectively, their ‘sum’ will be a 9-letter word. Of the alternatives given, only ‘seventeen’ contains 9 letters.

Only two colours are needed for the upper four faces of the octahedron. If, for example, blue and red are used then these four faces may be painted alternately red and blue. Consider now the lower four faces: every face adjacent to an upper blue face may be painted red and every face adjacent to an upper red face may be painted blue. So only two colours are required for the whole octahedron.

The key to the solution is to note that the squares of the odd numbers occur on the diagonal leading downwards and to the right from the cell which contains the number 1, and the squares of the even numbers occur on the diagonal which leads upwards and to the left of the cell which contains the number 4.
The squares of the even numbers have the form \((2n)^2\), that is, \(4n^2\).We see that the number \(4n^2 + 1\) occurs to the left of the cell containing \(4n^2\). Below \(4n^2 + 1\) there occur the numbers \(4n^2 + 2, 4n^2 + 3, \ldots, 4n^2 + 2n + 1\), and then in the cells to the right of the cell containing \(4n^2 + 2n + 1\), there occur the numbers \(4n^2 + 2n + 2, 4n^2 + 2n + 3, \ldots, 4n^2 + 4n + 1 = (2n + 1)^2\).
Now \(44^2 = 1936\) and \(45^2 = 2025\). Thus 2011 is in the same row as 2025 and to the left of it, in the sequence 1981,….,2012,…,2025, and below these occur the numbers 2163,…., \(2209 = 47^2\), with 2208 below 2025 as shown below. It follows that 2195 is the number below 2012. In the diagram below the square numbers are shown in bold.

There is the possibility of using only 3s giving one possible number 333333. Let's suppose a second digit is used, say \(x\). After the initial digit 3, there are 5 positions into which we can put either 3 or \(x\). So there are 2 choices in each of these 5 positions and so 2⁵ = 32 possible choices – except that one such choice would be five 3s. So we get 31 choices. There are 9 possible values for \(x\), namely 0, 1, 2, 4, 5, 6, 7, 8, 9. So this gives 9 × 31 = 279 numbers. Together with 333333, this gives 280 numbers.

If the difference between \(\sqrt{n}\) and 7 is less than 1, then \(6 < \sqrt{n} < 8\).
Therefore \(36 < n < 64\), so there are 27 possible values of \(n\).

If the first statement is true, then the three other statements are all false. If the first statement is false, however, then the second statement is the only true statement. Either way, exactly one of the four statements is true.

Every year, the day of the week on which a particular date falls is one day later than it fell the previous year unless February 29th has occurred in the meantime, in which case it falls two days later. As January 1st returned to a Monday after 11 years, it must have ‘moved on’ 14 days during that time, so February 29th occurred three times in those 11 years.

The next such display will be 03:12, that is in 41 minutes' time.

As it is known that the fake coin is heavier than all of the others, it is possible in one comparison to identify which, if any, is the fake in a group of three coins: simply compare any two of the three coins – if they do not balance then the heavier coin is the fake, whereas if they do balance then the third coin is the fake. This means that it is possible to find the fake coin when \(N = 9\) using two comparisons: the coins are divided into three groups of three and, using the same reasoning as for three individual coins, the first comparison identifies which group of three coins contains the fake. The second comparison then identifies which of these three coins is the fake. However, it is not possible to identify the fake coin in a group of four coins in one comparison only, so it is not always possible to identify the fake coin using two comparisons when \(N = 10\). If less than four are put on each side for the first comparison and they balance, then there are more than three left and the fake coin amongst these cannot be identified in one further comparison. Alternatively, if more than three are put on each side for the first comparison and they do not balance, then the fake coin in the heavier group cannot be identified in one further comparison.

Consider the top row of four dots. One can obtain a triple of dots by eliminating any one of the four – so there are four such triples. The same is true for each of the four rows, each of the four columns and the two main diagonals, giving 40 triples. In addition there are four diagonal lines consisting of exactly three dots, so there are 44 triples in total.

There are 51 houses numbered from 100 to 150 inclusive. Of these, 17 are multiples of 3, 11 are multiples of 5 and 4 are multiples of both 3 and 5.
So the number of houses Fiona can choose from is 51 − (17 + 11 − 4) = 27.

The only digits which will appear the same when reflected in the glass table-top are 0, 1, 3 and 8. So it is necessary to find the number of times in a 24-hour period that the display on the clock is made up only of some or all of these four digits. The first of the digits, therefore, may be 0 or 1; the second digit may be 0, 1, 3 or 8; the third digit may be 0, 1 or 3; the fourth digit may be 0, 1, 3 or 8.
So the required number is 2 × 4 × 3 × 4 = 96.

Let \(A\) and \(B\) denote the ends of the first length of rope and \(C\) and \(D\) denote the ends of the second length of rope. Then Pat chooses one of 6 different possible combinations: (\(A\), \(B\)), (\(A\), \(C\)), (\(A\), \(D\)), (\(B\), \(C\)), (\(B\), \(D\)), (\(C\), \(D\)). Sam now holds one untied length of rope and one tied loop of rope if, and only if, Pat has chosen (\(A\), \(B\)) or (\(C\), \(D\)) so the required probability is \(\dfrac{2}{6} = \dfrac{1}{3}\).
(Alternatively: Irrespective of whichever end Pat chooses first, Sam will hold one untied length of rope and one tied loop of rope if, and only if, Pat now chooses a particular one of the three remaining ends, namely the other end of the same rope. As each of the three ends is equally likely to be chosen, the required probability is \(\tfrac{1}{3}\).)

October has 31 days so, in any year, in October there are three days of the week which occur five times and four days which occur four times. As there were four Tuesdays and four Fridays, there could not have been five Wednesdays or five Thursdays, so the days which occurred five times were Saturday, Sunday and Monday. Hence October 1st fell on a Saturday, which means that October 31st was a Monday.

Let shape C have width 1 unit, height 1 unit and depth 1 unit. Then the volumes, in units³, of the five shapes are: A 2, B 2\(\tfrac{1}{2}\), C \(\tfrac{1}{2}\), D 1\(\tfrac{1}{2}\), E 4. These total 10\(\tfrac{1}{2}\) units³, so we may deduce that the cube formed by the four shapes will have side 2 units and volume 8 units³. Hence B is the shape which is not required. The cube may be formed by placing C next to D to form a shape identical to A. This combination is then placed alongside A to form a shape identical to E. If shape E is now rotated through 180° about a suitable axis, it may be placed with the combination of shapes A, C and D to form a cube.

Firstly, there are 12 unit squares which contain an even number. Every 2 × 2 square in the diagram has entries which consist of two odd numbers and two even numbers and hence have an even total. There are 16 of these. Each 3 × 3 square in the diagram, however, has entries which consist of five odd numbers and four even numbers (giving an odd total), or four odd numbers and five even numbers (giving an even total). There are 4 of the latter: those with 8, 12, 14 or 18 in the centre. Every 4 × 4 square in the diagram has entries which consist of eight odd numbers and eight even numbers and hence have an even total. There are 4 of these. Finally, the full 5 × 5 square contains 13 odd numbers and 12 even numbers, giving an odd total. So the required number is 12 + 16 + 4 + 4, that is 36.

The table shows the number on the face in contact with the table at the various stages described and also the numbers on the three faces of the die visible from the viewpoint in the question.

As the ‘6’ face is now facing East, the ‘1’ face will be facing West.

There are forty dots in the 5 × 8 array and twenty of these are illuminated to form the letter ‘t’.

Note that one cannot remove two touching coins; for if you did, any third coin originally making up a triangle with them would then be slideable. Note also that it is possible to remove the centre coin and each of the coins in a corner of the frame without enabling any of the remaining coins to slide. However, as each remaining coin touched at least one of the removed coins, it is not possible to remove any further coins. If, instead, a coin other than one of these four was removed first, then it would be one of the middle pair of an edge. Four coins would have touched this coin, so that would leave five possible coins to remove. Four of the five coins would lie along an edge of the frame, with two of these forming a triangle with the fifth. Clearly, only one of the three coins which form a triangle may now be removed, together with only one of the other two coins as they are touching. So we conclude that the maximum number of coins which may be removed is four.

Statements A, B, and E may be made by the Queen of Spades, whether it is a day on which she is telling the truth or a day on which she is lying. She may also make statement C providing she is telling the truth that day. However, if she is telling the truth on a particular day then she could not make statement D since that would be a lie. Also, if she is not telling the truth on that day then she could not make statement D either, since that would then be a true statement. So she cannot make statement D.

Each pyramid has five faces, but the square base of each one is glued to the cube and therefore does not form a face of the star. So each pyramid contributes four faces to the star. As there are six such pyramids, the star has 24 faces.

(It can be shown that the height of each of the pyramids is \(1/\sqrt{2}\). If, and only if, the pyramids had been of height 1/2, however, then the angle between a triangular face of the pyramid and its square base would have been 45°. This would result in the solid having 12 faces rather than 24, as pairs of triangular faces would combine to form rhombi. The solid so formed is known as as a rhombic decahedron: a space-filling polyhedron.)
(Diagram from http://dogfeathers.com/mark/rhdodec.html ).

The configuration includes three pairs of small rings. The rings in each pair are separate from the rings in the other two pairs so at least three rings will need to be cut. If the ring in each pair which is connected to the large ring is cut, then all of the rings can be separated. So the minimum number of rings which need to be cut is three.

Figure (i) shows a net of the cube on which a possible path has been drawn, while figure (ii) shows a diagram of the cube on which the same path has been drawn. Each edge of the network which joins a corner to a face centre has length \(1/\sqrt{2}\), while each edge which joins two adjacent corners has length 1. So the length of the path shown is \(1 + 12 \times 1/\sqrt{2}\), that is \(1 + 6\sqrt{2}\). This is the length of the shortest path along the edges of the network which passes through all 14 vertices. To prove this, we first note that to connect the 14 vertices we need a minimum of 13 edges, so the length of the shortest path must be at least \(13 \times 1/\sqrt{2}\). A path of this length would move alternately between corners and face centres, but as there are 8 corners and 6 face centres this is impossible. At least one of the edges on the shortest path, therefore, must join two corners. So the length of the shortest path must be at least \(1 + 12 \times 1/\sqrt{2}\). The diagrams show that such a path does exist so we are able to conclude that \(1 + 6\sqrt{2}\) is indeed the length of the shortest path.

If a number is divisible by 6 then it must also be divisible by 2, so is even. ‘The sum of any two odd numbers is even’ is also true. However, not all multiples of 9 are odd (e.g. 18) and the sum of any two even numbers is even rather than odd, so two of the four statements are true.

Consider the cuboid to be made up of 60 unit cubes. The front and side views show that top and bottom layers consist of the same number of cubes and from the top view we see that this number is 14. The front and side views indicate that only the 4 corner cubes remain in the middle layer, so the total number of cubes remaining is 2 × 14 + 4 = 32. The required fraction, therefore, is \(\dfrac{32}{60} = \dfrac{8}{15}\).

The 49 black squares will be in a 7 × 7 formation, so the board will measure 15 × 15 squares. Hence the number of white squares = 225 − 49 = 176.

There must be 25 pupils who all have different birthdays. If the remaining 5 pupils all have the same birthday as one of these pupils, then 6 pupils will share the same birthday.