APR Teasers

The answer is 32.

The best player will always make the final (and then win the tournament). Their opponent will be the strongest player in the opposite half of the draw. At one extreme, one half of the draw contains the best 32 players, and the other half contains the worst 32 players. In this case the best and 33^rd best players meet in the final. So, the 33^rd best player is the weakest opponent the best player can have in the final, as any weaker player can’t be the strongest player in their half of the draw, and so will be eliminated before the final.

By having the 34^th to 64^th best players in one half of the draw along with one “good” player (from the 2^nd best to 33^rd best) the best player can meet any “good” player in the final. As there are 32 choices for the “good” player, there are 32 possible finals. Note that if you consider “Player A vs Player B” as a different final from “Player B vs Player A”, we will also accept 64 as a valid answer.

Split each number into columns, so 123,456 has a 6 in the units column, a 5 in the tens column, a 4 in the hundreds column etc.

Every tenth number has a nine in the units column, so you write 1,000,000/10 = 100,000 nines in the units column. Ten numbers out of every hundred (xxx90, xxx91, … xxx99) have nines in the tens column, so you write 10*(1,000,000/100) = 100,000 nines in the tens column. One hundred out of every thousand numbers (xxx900, xxx901, …, xxx999) have nines in the hundreds column so you write 100,000 nines in the hundreds column. Similarly, you write 100,000 nines in the thousands column and the same number in the tens of thousands and hundreds of thousands column.

So you write down 600,000 nines altogether.

The answer is 161/36 or roughly 4.472.

Below is a diagram of the sample space, the green entries indicating the outcome of each throw.

Each of the green cells occurs with equal probability of 1/36.

So, the expected value we are after is:

1 * 1/36 + 2 * 3/36 + 3 * 5/36 + 4 * 7/36 + 5 * 9/36 + 6 * 11/36 = 161/36  4.472

The answer is 11/36.

See below a plot of the possible time, in minutes, after 10am that A and B can both arrive at the station in order to meet each other.

The x axis here represents times that A could arrive, the y axis the times that B could arrive. The unshaded region is the area in which they can both arrive, such that they'll meet each other at the station.

Since A and B turn up uniformly randomly, the desired probability is simply the probability of the unshaded region.

P(both meet)  = 1 - P(triangle above region) - P(triangle below region)

= 1 - 2*(1/2 * 5/6 * 5/6)

= 11/36

This gives a probability of 11/36.

The answer is (1 - d)^2

Consider any square on the chessboard. We just need to consider the area of that square that the centre of the hoop must land in, in order for the hoop to surround an area of just one colour. That area we have calculated will then be the probability we want, since the area of a given square is 1.

The centre must land between d/2 and 1 - (d/2) along in each direction, creating a square of possible places the centre could land, the side length of which is 1 - d/2 - d/2 = 1 - d

So, this area and the probability we want is just: (1 - d)^2

The answer is 53.
This can be done in a number of ways, the least satisfying of which is probably using the 'mod' function in Excel and counting how many of the 4 conditions are satisfied.

A more insightful solution is as follows:

Firstly, note that if x is a solution to the above, so is x + 210 since 210 is 2*3*5*7, i.e. we try and find a solution modulo 210.

Now, we look for a solution x of the form:
x = (3*5*7)*a + (2*5*7)*b + (2*3*7)*c + (2*3*5)*d
For a, b, c, d integers.

Note that by doing this e.g. modulo 5, to work out the remainder, we need only consider the term (2*3*7)*c since all the other terms are multiples of 5. So, we just need to pick the c that makes that term's product be 3 (mod 5).
3*5*7 = 105 is 1 (mod 2) so we can pick a = 1.
2*5*7 = 70 is 1 (mod 3) so we can pick b = 2.
2*3*7 = 42 is 2 (mod 5) so we can pick c = 4.
2*3*5 = 30 is 2 (mod 7)so we can pick d = 2.

This gives one solution of x = 3*5*7*1 + 2*5*7*2 + 2*3*7*4 + 2*3*5*2 = 473.

Noting the fact this is a solution modulo 210, we have that 473 - 2*(210) = 53is a positive integer solution to the 4 conditions. It is easy to check (or by using the Chinese Remainder Theorem) that this is indeed the smallest such solution.
(For more information about these types of problems, search for the Chinese Remainder Theorem).

The answer is 6.

A multiple of 12 is both a multiple of 3 and a multiple of 4, and so any permutation of the digits of a jiggly number must satisfy both of the following two points:

(i). For any multiple of 4, the number formed from the last two digits must also be a multiple of 4 (since 100 is a multiple of 4).

(ii). Any of multiple of 3 is such that the sum of its digits is also a multiple of 3. (this is a cool trick that can be proved, for example, using modular arithmetic).

From (i), we deduce that a jiggly number's digits must firstly be even.

Then, 2 cannot be a jiggly number's digit because of (i) and the fact that 22, 42, 62 and 82 are not multiples of 4.

Then 6 can also not be a digit because 46, 66 and 86 are all not multiples of 4.

So from (i), in summary, we get that the digits of a jiggly number must be either 4 or 8.

Noting then that 4 is one more than a multiple of 3 and 8 is one less than a multiple of 3, the only way 4 of these digits can sum to a multiple of 3 are permutations of: 4488.

So deduce from (ii), that there are 4C2 = 6 jiggly numbers (4488, 4848, 4884, 8448, 8484 and 8844).

The answer is 14.

Let C_n be the number of ways of 2n people can each shake one hand with no hands crossing. So we want to find C_4.

Clearly C_1 = 1, there is only one way 2 people can shake hands.

C_2 = 2, for 4 people in a circle, only adjacent pairs can shake hands without crossing.

Consider C_3, i.e. 6 people in a circle, call them P1, P2,…, P6. Consider who P1 shakes hands with in turn. If they shake with P2, the rest can shake in C_2 ways. If with P4, the rest in C_1*C_1 ways. If with P6, the rest in C_2 ways. So C_3 = C_2 + C_1*C_1 + C_2 = 2 + 1 + 2 = 5.

Consider now C_4, P1,…, P8 in a circle. Considering who P1 shakes hands with, and get a similar relation that: C_4 = C_3 + C_2*C_1 + C_1*C_2 + C_3 = 5 + 2 + 2 + 5 = 14.

So there are 14 ways that 8 people can shake hands without any pairs crossing.

If you want to find out more about this pattern - this is one characterisation of the Catalan numbers.

x is 5

The answer was 31.

The nth door is open at the end only if an odd number people have visited it i.e. n has an odd number of factors.

But numbers with an odd number of factors are exactly the square numbers as factors always come in pairs unless they are repeated (e.g. for 4: 1 and 4, and 2).

There are 31 square numbers below 1000 (sqrt(1000) = 31.62) so exactly 31 doors are left open.

The answer is approximately 1-1/e where e is the mathematical constant e~2.718. This result holds no matter how large the group of people.

For a group of n people, there are n! different orders in which the names can be pulled out of the hat.

We are only interested in orderings where no element appears in its original position. These are the orderings where each person doesn't pull their own name out of the hat.

In mathematics these are known as derangements and the number of derangements of n elements is given by n!/e.

Therefore, the probability that at least one person does draw their own name is then 1-1/e.

The profit from buying x lottery tickets is 1,000,000 * x/(x+10) - x.

Differentiating this with respect to x (using the product rule) and setting to zero to find the maximum:

1,000,000/(x+10) - 1,000,000x/(x+10)^2 - 1 = 0

10,000,000 = (x+10)^2

Therefore the answer is 3,152.

Answer = 40/401 = 9.975%

There were 200 episodes and each champion won five episodes, so there were 40 champions. The first episode had three new contestants and each of the subsequent 199 episodes introduced two new contestants, so the total number of contestants was 3 + 199 x 2 = 401.

There are finitely many squares. Therefore there must be a square that attains the lowest numerical value out of all the squares. Let's say that square is called X and it has some value x (which is the lowest value amongst all the values).

Let Y be the set of squares on the board from which a knight could jump, in one move, to square X (so Y cannot have more than 8 elements). Let Z be the set of numerical values of the squares in the set Y. Well, x is the mean of Z, but x is, by definition, less than or equal to all of the elements of Z.

The only way that this can happen is if all of the elements of Z are equal to x. Then apply an inductive argument to show that every square on the board must have value x. Therefore the answer is 21 (i.e. whatever arbitrary number you state in posing the question).

Answer = 1,240 kg

There were 781 gold medals, 770 silver medals and 850 bronze medals awarded.
Source: https://olympics.com/tokyo-2020/olympic-games/en/results/all-sports/medalists.htm

There were significantly more bronze medals awarded than golds and silvers because of sports like boxing awarding two bronze medals for each event.

The number of golds and silvers were not the same, partly because of the men’s high jump (where the gold medal was shared and there was no silver medal) and partly because of team sport events (football, rugby sevens, hockey etc) where the eventual gold medal winners used more players during the tournament, each of whom received a medal, than the eventual silver medal winners.

Let x, y and z be the number of marbles of each colour. We know that there are 100 marbles, so x + y + z = 100.

There are six different combinations of marbles which give one of each colour when drawn in that order:
x, y, z
x, z, y
y, x, z
y, z, x
z, x, y
z, y, x

The chance of the first one is x/100 (the probability the first marble is x-coloured) * y/99 (the probability the second marble is y-coloured, with only 99 balls left in the bag) * z/98 (the probability the third marble is z-coloured, with only 98 marbles left in the bag).

In fact, the probability will be the same for all six combinations: x*y*z / (100*99*98), with just the order of factors changing, which doesn’t matter.

So the total probability is 6*x*y*z / (100*99*98), and since we know this is 20%, we can simplify and get x*y*z = 32340.

So we need x+y+z=100 and x*y*z=32340, with x, y and z all whole numbers, and there are lots of routes to finding solutions here - whether you just check all combinations, or think about prime factors to limit the options you need to check. But however you get there, the only combination that fits is 21/35/44.

Anna = 12,

Danielle = 14

Ben = 17,

Chris = 22!

No, it's impossible!

To maximise your expected winnings, you should guess 50.

Suppose you make a guess of X. Then the probability that the host’s number is higher than your guess is (100-X)/100. The product X*(100-X)/100 gives your expected earnings and this is maximised when X=50.

Option D, 4 points & 12 points.

Ash

The total number of games played is (9 + 13 + 16) / 2 = 19.

If a player loses every game they play, then they play alternate games. So the minimum number of games it is possible to have played is 9: games 2, 4, 6, … , 18. As Ash only played 9 games, they must have played in each even numbered game and lost them all.

Option D, 4 points & 12 points.

Scenario/reasoning for a) = 4:

One team wins all six of its matches. The three other teams all draw with each other in all of their matches, thereby all finishing on 4 points. One of these teams will finish second, based on head-to-head, away goals etc.

Scenario/reasoning for b) = 12:

One team loses all six of its matches. The three other teams all beat each other once and lose to each other once, thereby all finishing on 12 points. One of these teams will finish third, based on head-to-head goal difference etc.

D

Cari by 4.  As there are no ties we know the six times Cari plays scissors must be against non scissor plays from Morgan. So Cari plays scissors against rock twice and paper four times putting her two games ahead. All remaining 4 games Morgan must now play scissors against Cari who will rock three times and paper once giving Cari another three wins and one loss. This leads to a total net win of 4 games by Cari.

D. 10

For each nth person the size of slice they receive can be shown as:

\(slice_{n} = \frac{n}{100}\ left_{n-1}\)

Where left_n is the amount of pizza left after the nth person takes their slice and can be shown as:

\(left_{n} = left_{n-1}\frac{100-n}{100}\ \)

Now consider the following ratio of the slice size received by the nth and (n-1)th person denoted ratio_n:

\(ratio_{n} = \frac{\frac{n}{100}\ left_{n}}{\frac{n-1}{100}\ left_{n-1}}\ \)

Substituting for left_n leaves:

\(ratio_{n} = \frac{\frac{n}{100}\ left_{n-1} \frac{100-n}{100}\ }{\frac{n-1}{100}\ left_{n-1}}\ \)

\(ratio_{n} = \frac{n(100-n)}{(n-1)100}\ \)

To maximise for n we now set ratio_n = 1 and solve:

\(100(n-1) = 100n - n^{2} \)

\(n^{2} = 100 \)

\(n = 10 \)

B. £14

The sign maker charges £1 for consonants, £2 for a vowel and applies a double charge for capital letters.

This can be solved by constructing simultaneous equations using the information known about the 3 partner name plates.

E. 17 mins

The key is that each pair can only cross at the speed of the slowest individual, so finding a way for the two slowest members to only have to cross once each and at the same time will improve efficiency. Hence the most efficient way to get all individuals across is as follows (all individuals are denoted by the time it takes them to cross):

1. Send 2 + 1 across first. Total time = 2 mins
2. 2 returns with the torch. Total time = 2+2 = 4 mins
3. Send 10 + 5 across . Total time = 4 + 10 = 14 mins
4. 1 returns with the torch. Total time = 14 +1 = 15 mins
5. 2 +1 cross again so all individuals are now on the correct side. Total time = 15 + 2 = 17 mins.

B. 10

Each numbered switch will be flipped a number of times equal to its number of factors. As all switches begin in the off position only those that are flipped an odd number of times will be in an on position. The only numbers that have an odd number of integer factors are square numbers so only square numbered switches will be on at the end. From 1-100 there are 10 square numbers so 10 switches will be in the on position once everyone has passed through.

A. Apples and Oranges

The key is that all boxes are incorrectly labelled.

This means that the box labelled apples and oranges cannot contain mixed fruit. Hence if you remove an apple from this box you know this box should be labelled apples and if you remove an orange it should be labelled oranges.

Relabelling this box means now have a box without a label (that you just removed to put on the previously labelled mixed fruit) and a box with a label that you know is incorrect.  So, move the label from the box you know is incorrect (the only box with the original label applied by the broken machine) to the box without a label then place the mixed fruit label on the remaining unlabelled box and they will all be correctly labelled.

C.29

Any time the bar is broken it only creates a single extra piece hence for any chocolate bar of n squares it will always take n-1 breaks to break it down into it's individual squares.

D. 6 - as demonstrated below.

C. 312211

Each number is a description of the digits which make up the previous number.

So with a starting point of 1:

1

1 has one 1, so we get 11

11 has two 1s, so we get 21

21 has one 2 followed by one 1, so we get 1211

1211 has one 1 followed by one 2 followed by two 1s, so we get 111221

111221 has three 1s followed by two 2s followed by one 1, so we get 312211

B. Bernoulli

Ignoring mathematicians who have been omitted from this problem and skating over the fact that calling Einstein a mathematician has offended some readers' sensibilities, Einstein could be last entry in volume 3 (surnames Ce-Ei). As the book sits on the shelf this would make his page the left most page.

This would make him 'next to' the first page of volume 2 (surnames Be-CA) as this is the right-most as it sits on the shelf. Of the options given only Bernoulli could be the first entry in this volume meaning he could be 'next to' Einstein.

There are two equally valid solutions for this teaser depending on the version of Excel available to you:

Excel 2019 Solution - 58 characters

SWITCH(GCD(ROW(),15),1,ROW(),3,"Fizz",5,"Buzz","FizzBuzz")

Pre 2019 Excel Solution - 61 characters

P(win) = P(win without reaching deuce) + P(reach deuce and then win)
              = P(win without reaching deuce) + P(reach deuce)*P(win from deuce)

P(win without reaching deuce) = P(win at least 4 out of first 6 points)
                                                 = P(win 4 out of 6) + P(win 5 out of 6) + P(win 6 out of 6)
                                                 = (6*5/2!)*(0.4^2)*(0.6^4) + 6*0.4*(0.6^5) + 0.6^6
                                                 = 0.31104 + 0.186624 + 0.046656
                                                 = 0.54432

P(reach deuce) = P(win exactly 3 out of first 6 points)
                        = (6*5*4/3!)*(0.4^3)*(0.6^3)
                        = 0.27648

To calculate P(win from deuce), define the function F as follows:

F(0)  = win from deuce
F(1)  = win from advantage
F(-1) = win from opponent’s advantage

Then F(0) = 0.6*F(1) + 0.4*F(-1)
                = 0.6*(0.6 + 0.4*F(0)) + 0.4*(0.6*F(0) + 0)
                = 0.36 + 0.48*F(0)

Therefore F(0) = 0.36/(1 – 0.48) = 9/13

Hence, P(win)  = 0.54432 + 0.27648*9/13
                       = 0.7357