Is there a strategy that allows the customer to gain on average more than (M1 +M2)/2?

(It seems like there wouldn’t be, since the customer has no information about M1 and M2 other than that both values are positive and different from one another. Nonetheless...)

Try to solve this problem yourself before moving on to the solution below. Use hint if needed

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Hint

Start with two houses, then three houses, then four houses, and so on by numbering the houses in base 2.

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Solution

A natural strategy is to fix a threshold above which one is happy, for example $100, and to cash the first check if it is greater than $100, but otherwise to cash the other check.

If both sums M1 and M2 chosen by the banker are below $100, this strategy leads the client to always change envelopes. He will thus gain on average, (M1+M2)/2 . If both sums chosen by the banker are above $100, this strategy leads the client to never change envelopes. Thus, he wins, on average, (M1+M2)/2 . Finally, if M1 < 100 < M2 or the other way around, this strategy allows the client to always choose the envelope with the highest amount.

In the end, the strategy consisting of thresholding the amount one is happy with is only a good strategy if the threshold is between the two sums the banker chose. If we fix it arbitrarily, it is possible that it doesn’t land between the two checks and that the strategy is ineffective. To avoid this issue, we must choose the threshold at random, with a random variable X such that

In other words, X is a random variable with a nonzero probability of landing in any interval of R+, such as an exponential law. In this manner, the client chooses the threshold at random with the random variable X, and he knows that sometimes, the chosen threshold will be in between the two sums chosen by the banker. In any case, he performs better than (M1+M2)/2 . His average gain is thus strictly better than (M1+M2)/2. We can compute the average gain G as a function of M1 and M2 and of

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(For example, with six houses, the mailman distributes the mail to house numbers 2, 4, and 6, then to number 3 (he skips number 1 and ignores number 2), number 1 (he ignores number 4 and skips number 5), and finally f inishes with number 5)

Try to solve this problem yourself before moving on to the solution below. Use hint if needed

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Hint

Start with two houses, then three houses, then four houses, and so on by numbering the houses in base 2.

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Solution

It is house 73 that will get its mail last. Let us compute generally for n houses (n >=2) which one gets its mail last. Let mn be the number of this house. By observing the small values of n (m2 = 1, m3 = 3, m4 = 1, m5 = 3,...) we can notice that the sequence will increase by taking only odd values until n before falling back to 1. It is not easy to write this as an explicit formula. This is because we aren’t working in the right base. Let us consider base 2 (see reminder 7); n is written as 1bp−1...b0 in base 2, with p>=1 and bi taking value either 0 or 1. We will show, by induction on n>=2, that in base 2, mn is written as bp−1...b01.

Induction Assume the property holds for some rank n>=3 and show it holds at rank n+1. The postman delivers mail to house number 2. He must then deliver the mail to the n remaining houses. We can use the induction hypothesis if we renumber the houses beforehand as follows:

House number 3 becomes house number 1, number 4 becomes number 2, …, n + 1 becomes n - 1, and 1 becomes n. If mn = n, given the renumbering performed, mn+1 = 1. By uniqueness of the base 2 representation b0 = 1, b1 = b0, …, bp−1 = bp−2, and thus n = 1 . . . 12. In this case, n+1 = 10 . . . 02, and the induction property holds: mn+1 = 1.

If not, mn 6= n, and, given the numbering, mn+1 = mn + 2.

n+1 has no more digits in base 2 than n, otherwise n = 1...12 = mn, which is forbidden. Therefore,

with the Bi taking values in {0, 1}.

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We wish to cover it with tiles in the shape of an elongated domino:

Is it possible?

Try to solve this problem yourself before moving on to the solution below. Use hint if needed

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Hint

One must color the tiles of the bathroom with three colors.

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Solution

It is impossible. Let us assume for contradiction that we have successfully paved the bathroom. We then paint the tiles as follows

We notice that there are 22 white tiles, 21 gray tiles, and 20 black tiles. On the other hand, each domino-shaped tile must cover a white square, a gray square, and a black square. Thus, the numbers of white tiles, gray tiles, and black tiles must all be equal to each other. This result is a contradiction.

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We wish to cover it with tiles in the shape of an L, which we can also invert:

Is it possible?

Try to solve this problem yourself before moving on to the solution below. Use hint if needed

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Hint

One must color the tiles of the bathroom column-wise.

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Solution

It is impossible. Let us assume for contradiction that we have successfully paved the bathroom. We then paint the tiles as follows:

We notice that there are 27 white tiles and 21 gray tiles. Each L-shaped tile covers three white tiles and one gray or the other way around. Let W be the number of L-shaped tiles that cover three white tiles and one gray and G the number of L-shaped tiles that cover three gray tiles and one white. We obtain the following system:

Solving this system, the only solution is W = 15/2 and G = 9/2 . This result is a contradiction since W and G must be whole numbers.

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If someone carefully chooses the starting point of their walk, is it possible for them to walk around town by crossing each of the seven bridges exactly once?

Try to solve this problem yourself before moving on to the solution below. Use hint if needed

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Hint

Think of graph theory.

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Solution

We can convince ourselves that there are no solutions by trying different routes. This is what we shall prove. To do so, we will need to use a bit of graph theory, created especially for this problem by Euler. This approach is very clever, yet very simple. The idea is to reduce the problem to connecting dots without lifting the pen off the paper. We give each part of town a number as follows:

If there is a bridge between two regions, we connect those regions with a line or a curve; the result is the following graph:

We then notice that finding a path that crosses each bridge exactly once boils down to finding a path linking the four vertices of the graph without lifting the pen up. One is easily convinced by trying that it is impossible, but it remains to prove it. To do so, we will use the following property:

If a graph can be drawn without lifting the pen, then it has at most two vertices with an odd number of edges.

Consider as an example the following graph:

On each of the four vertices, there are three edges: we say that each vertex is of degree three. Since 4 > 2, it is impossible to draw this graph without lifting the pen.

Let us now prove the result. Consider a graph that we assume to be drawn without lifting the pen. It has a vertex from which we started drawing and a vertex where we finished drawing; the two vertices may be the same. We notice that all other vertices must be of even degree. Indeed, each time we draw through a vertex, we increase its degree by two, which makes the final degree even. Thus, there are indeed at most two vertices of odd degree.

Let us return to the bridge problem. All the vertices of the graph have an odd degree (3 for vertices 1, 2, and 4; 5 for vertex 3). It is thus impossible to draw without lifting the pen.

Do note that the converse of this result is true but it is somewhat more difficult to prove.

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How can he obtain a litre from these two pitchers?

(One could start by solving the simple case appearing in a famous blockbuster: p = 3 liters and q = 5 liters)

Try to solve this problem yourself before moving on to the solution below. Use hint if needed

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Hint

It suffices to pour the contents of one pitcher into the other and start this process over as many times as necessary. It remains to prove this result.

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Solution

Let us first solve the simple instance. Start by filling the 3 litre pitcher with wine from the barrel. Now fill the 5 litre pitcher with wine from the 3 litre pitcher. We fill up the 3 litre pitcher again and pour the wine from that into the 5 litre pitcher until it is full. The remainder in the 3 litre pitcher is exactly 1 litre.

Let us now move to the general case. The integers p and q are coprime, so by Bézout’s lemma, there are two integers u, v such that up+vq = 1. Since p and q are positive, either u or v is negative. Let us assume that v is negative (the case for u negative follows in the same manner). We can then write up = 1+|v|q. It suffices to fill the q liter pitcher with up liters obtained by filling the p liter pitcher u times and emptying it into the other. Indeed, if we note r as the number of liters in the p liter pitcher after the procedure, we have up belongs to r [q], but up belongs to 1 [q], thus r belongs to 1 [q]. Since 0 < r < q, r = 1.

For example, integers 11 and 47 are coprime. We thus have the following Bézout relation: 4x47 - 17x11 = 1; hence, 4   47 = 1 + 17x11. Thus, to obtain 1 liter, it is sufficient to fill the 47 liter pitcher four times while pouring the wine into the 11 liter pitcher while emptying the 11 liter pitcher each time it is full. The 11 liter pitcher will be filled 17 times; at this moment, 1 liter will remain in the 47 liter pitcher. When the first 47 liter pitcher is empty, there are 3 liters left in the 11 liter pitcher. Now fill the 47 liter pitcher again and repeat the process. After this, there are now 6 liters in the 11 liter pitcher. Repeating this procedure again leaves 9 liters in the 11 liter pitcher, and the fourth repetition will push it past 11 liters so that it needs to be emptied once more, and then it will contain exactly 1 liter at the end.

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What strategy can the students use to win this challenge?

(Students choose their T-shirts all at once and in such a way that no student can draw any information from the color of the T-shirt of another student)

Try to solve this problem yourself before moving on to the solution below. Use hint if needed

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Hint

One must use the sign

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Solution

Let us begin by presenting a possible strategy and then show that it answers the problem. First of all, the students number themselves from 1 to 7. Let us assume the perspective of student number 1. She observes the hats of the six other students and orders them in increasing order of their hats. Let us suppose she obtains the following order 5, 2, 4, 7, 3, 6. She then adds her own number to the end of the ranking: 5, 2, 4, 7, 3, 6, 1. This is a permutation of f1, 2, 3, 4, 5, 6, 7g. She now only needs to compute the signature of this permutation and to take a black T-shirt if the signature is 􀀀1 and a white T-shirt if the signature is 1.

Let us now see why this strategy works. First of all, one must be convinced of the following property: “If the adopted strategy is such that whenever two students have on their hats neighboring real numbers, they choose different colored T-shirts, then the students will indeed have alternating colors.”

Next, one must show that the given strategy upholds this property. To see this, consider an example:

In this example, students are sorted into order 1, 5, 2, 4, 7, 3, 6. We notice that students 4 and 7 have neighboring real numbers (but they do not know it). Let us assume the perspective of student 4 and apply the previously designed strategy. Student 4 sees the six other students in the order 1, 5, 2, 7, 3, 6. He adds his number on the end, and computes the signature of the permutation mapping 1, . . . , 7 to 1, 5, 2, 7, 3, 6, 4. Now we consider student 7. She sees 1, 5, 2, 4, 3, 6 and computes the signature of the permutation mapping 1, . . . , 7 to 1, 5, 2, 4, 3, 6, 7. In fact, we notice that since students 4 and 7 are neighbors, they will see the same order for the six other remaining students except that a 4 will become a 7 and a 7 will become a 4. They will thus compute the signature of their permutation up to one transposition: going from 1, 5, 2, 7, 3, 6, 4 to 1, 5, 2, 4, 3, 6, 7 by transposition of the 4 and the 7. The signature of the first permutation will thus be the opposite of the signature of the second. Students 4 and 7 will thus choose T-shirts of different colors. Thus, the property holds, and the presented strategy works.

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How can the students win this challenge?

(It is not specified whether the lamp is on or off at the beginning. Do note that students can enter the office several times before one of them announces that they have all been in. If, for example, each student has been in the principal’s office 50 times before one of them declares that they all have been in, they win nonetheless)

Try to solve this problem yourself before moving on to the solution below. Use hint if needed

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Hint

One can begin by assuming that the light is initially off and that the students know this.

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Solution

Let us begin by solving the problem in the instance in which the students know that the lightbulb begins turned off. The students designate a leader who will be in charge of counting the calls to the office and announcing when everyone has been in. Other than the leader, each student flips the light on if it is off, but only the first time they’re called and find it turned off. In other words, if a student finds the light off when they enter, but they have already turned it on during a previous visit, they will leave it off. When the leader is called, she always turns off the light if it was turned on. Thus, when the leader has turned the light off 99 times, it will be because the 99 other students have been in the office and the leader will be able to announce it.

To handle the general case, we slightly tweak the previous strategy. Other than the leader, the students turn the light on if it is turned off, twice but no more. If the light is initially turned off, the leader will have to turn it off 99x2 = 198 times; otherwise the leader will have to turn it off once more, that is, 199 times. In any case, the leader will know everyone has been in the office at least once when she turns the light off for the 198th time.

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Can the termite finish its meal by eating the central cube?

Try to solve this problem yourself before moving on to the solution below. Use hint if needed

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Hint

Color some of the small cubes of the cube.

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Solution

Let us color the small cubes in black and white in such a way that each small cube is surrounded by adjacent cubes of the other color, as below:

The termite travels through cubes of alternating colors. Since it eats an odd number of cubes, it must begin with a white cube to finish with the central cube, which is white. But then, if the termite eats all the cubes starting with a white cube and ending with a white cube, there will be one more white cube than black cubes. But there are 13 white cubes and 14 black cubes. It is thus impossible.

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At most, how many bananas can he transport to the oasis?

(First off, we notice that if the merchant does the trip without stopping, loading up his elephant as much as possible, he will arrive at the oasis with no bananas. It is thus necessary to make stops: the idea is to drop off some merchandise on the way, come back to pick up more bananas, and then pick up the bananas left on the way. Note, we ask not only for an optimal strategy, but also the proof of its optimality)

Try to solve this problem yourself before moving on to the solution below. Use hint if needed

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Hint

The answer is 533 bananas. The hardest part is proving one can’t do better.

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Solution

The elephant can carry at most 533 bananas to the oasis. There are several possible strategies.

The elephant carries as many bananas as possible, kilometer by kilometer. It starts from kilometer 0 with 1,000 bananas on its back, and arrives at the first kilometer with 999, drops 998 off, and then returns to the start with the last banana. It does this again and drops off 998 at the first kilometer once more. Finally, it returns to pick up the last 1,000 bananas and drops 999 off at the first kilometer mark. All in all, the elephant ate five bananas to carry 998   2 + 999 = 2,995 bananas the first kilometer. It starts over now and carries 2,990 bananas to the second kilometer mark. For each kilometer, the elephant consumes five bananas to carry all the others. The elephant continues in this way until the 200th kilometer, to which it brings 2,000 bananas. It leaves the 200th kilometer with 1,000 bananas on its back, drops 998 off at the 201st kilometer, keeping one for the trip back. It takes the last 1,000 bananas and drops 999 off at the 201st kilometer. All in all, the elephant ate three bananas to carry 998 + 999 = 1,997 bananas from the 200th kilometer to the 201st kilometer. It continues and carries 1,994 bananas to the 202nd kilometer. It consumes three bananas for each kilometer to carry the rest and continues in this way until it reaches the 503rd kilometer with 1,001 bananas. It then leaves behind one banana and leaves directly for the oasis with the 1,000 bananas remaining. It arrives 467 kilometers later with 533 bananas.

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(For example, with only three weights, one weighing 1kg, one weighing 2kg and one 5kg, it is possible to weigh any object between 1kg and 8kg. Do note, we ask not only for an optimal set of weights, but also the proof of its optimality)

Try to solve this problem yourself before moving on to the solution below. Use hint if needed

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Hint

Each weighing has three possible outcomes; what is the maximum number of weights that one could discriminate between? One will be able to use the method of comparing outcomes to hypotheses, which is detailed in the reminder on the pigeonhole principle

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Solution

First of all, notice that starting with one weight of 1 kg and another weight of 3 kg, we can already determine the weight of all objects between 1 kg and 4 kg. To weigh an object of 2 kg with these weights, it suffices to place the object with the 1 kg weight on the left and the 3 kg weight on the right.

If we choose weights of mass 1 kg, 3 kg, 9 kg, 27 kg, …, 3^9 kg, we can determine the weight of any object between 1 kg and 1 + 3 +. . . . . . . + 3^9 = (3^10−1)/2 kg. Indeed, let us show by induction on n   0 that with weights of mass 1 kg, 3 kg, ..., 3^n kg, we can determine the weight of any object between 1 kg and (3^(n+1)−1)/2 kg.

Initialization is simple. Suppose the proposition holds for n>=0. We have weights of mass 1 kg, 3 kg, …, 3^(n+1) kg. If we want to weigh a mass between 1 kg and (3^(n+1)−1)/2 kg, doing so is straightforward; we don’t need the 3^(n+1) kg weight. To weigh an object whose mass is between (3^(n+1)+1)/2 kg and (3^(n+2)−1)/2 kg, we begin by subtracting 3^(n+1) kg from this mass by placing the corresponding weight on the other side. This brings the mass down to between -(3^(n+1)−1)/2 kg and (3^(n+1)−1)/2 kg, which we can weigh with the other weights, which completes the induction.

Can we do better? No. We use the comparison method for outcomes and hypotheses. Let us compute the maximum number of different weighings we can produce with 10 weights. For each weight we have three possibilities: place it on the left side, place it on the right side, or do not use it at all. This gives 310 possible weighings. However, we actually have only 3^10 - 1 weighings, since not placing anything on the scales does not count as a valid weighing. We must then divide the result by 2 since the plates are symmetric. We thus obtain at most (3^10−1)/2 possible results from weighing. This maximum is reached with the previous system of weights; hence the result is proven.

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Determine, under the same principle, a fair way to share a pizza between n people.

Try to solve this problem yourself before moving on to the solution below. Use hint if needed

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Hint

There are many ways to proceed, none are easy to find. One possibility is to build the sharing method by induction on n. One can begin by thinking of the case in which n = 3 and try to reduce to the previous case in which there are only two people.

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Solution

The case n = 2 is known to anyone who has at least one sibling. The first sibling cuts the pizza in half and the second chooses their slice. Let us solve the problem of the case n = 3. A person cuts what seems to them like one-third of the pizza. If the second person thinks this slice is more than one-third, they can cut some off. The third person may do the same (potentially cutting some off for the second time). The rule is the following: the last person to touch the slice must eat it. Thus, no one is going to cut a slice so that it seems to them like it is less than one-third in fear of ending up with the slice if no one else touches it. Everyone is happy, the last person to touch the slice gets what they think is more than one-third, and the other two don’t think that slice is more than one-third, so they think they have at least two-thirds to share. They share the remainder without issues (by the method for the n = 2 case).

This reasoning extends easily for n people. The first person cuts the slice they like, then each of the others in turn can cut this slice further, knowing that the last one to cut the slice gets it. We then reduce to sharing between n - 1 people. We thus have an induction and the problem is solved.

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How does he do it?

(Of course, the precise triangular shape of the cake is a priori arbitrary. One can’t stack parts or flip them over)

Try to solve this problem yourself before moving on to the solution below. Use hint if needed

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Hint

One can consider the moment when the player’s success rate goes over 75%.

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Solution

Let (ABC) be the triangle that will model the cake. For any triangle, there is a foot to one of its altitudes. Here we will consider H, the foot of the altitude from B. Let I and J be the midpoints of line segments [AB] and [BC]. Two strokes of the knife following lines (HI) and (HJ) solve the problem. Indeed it is simple to rearrange the pieces cut in this manner for the triangle symmetric to (ABC).s

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(Assume that the nuts all have the same mass. In this case, it is evident that, whatever the nut the squirrel sets aside, it can always separate the 10 remaining into two groups of five nuts with the same mass. Assume now that the nuts respectively weigh 1 gram, 2 grams, …, 11 grams. If the squirrel sets aside the nut that weighs 2 grams, it can separate the 10 remaining nuts into two groups of five with the same mass: on one side the nuts weighing 1 gram, 4 grams, 6 grams, 10 grams, and 11 grams, and on the other those weighing 3 grams, 5 grams, 7 grams, 8 grams, and 9 grams. However, if it sets aside the nut weighing 1 gram, we can show that it is impossible to separate the 10 remaining nuts into two groups of five with the same mass. Thus, the distribution of masses of the nuts proposed in this particular case does not work. One must show that the only distributions that work are those in which the nuts have the same weight)

Try to solve this problem yourself before moving on to the solution below. Use hint if needed

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Hint

The easiest approach is to reason on the parity of the weights.

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Solution

We consider a distribution of weights that works, which means it is such that whatever nut the squirrel sets aside, it can always separate the 10 others into two groups of five nuts with the same mass. We want to show that all the nuts have the same mass. From now on, we will assimilate the weights of the nuts to the integers representing these weights in thousandths of a gram.

First, we notice that all the weights have the same parity. Indeed, if a nut has an even weight, since we can separate the other 10 nuts into two groups of the same weight, the sum of the other weights is even. The total sum of nuts is thus even. Likewise if a nut has odd weight. If we now take an arbitrary nut, since the sum of the 10 other weights is even and the total sum is even (respectively odd), this nut has an even (respectively odd) mass.

Now we notice that if a distribution works and contains an even weight, we can divide all the weights by two and it will still work. We can thus divide all the weights of a unit. By removing the minimum mass from every nut, we have a distribution that works with at least one zero mass. Either all the masses are zero and were all equal, or there is at least one nonzero mass. In this case, since there is an even mass (zero), we can divide by two, and so on as many times as necessary to render the nonzero mass odd. We thus have a zero weight and an odd weight: a contradiction. Hence, we have proven our result.

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Each takes their turn to take a square and the upper right corner that comes with it. The one who is forced to take the poisoned square (when it is the only square left) is declared the loser!

Show that there is a winning strategy for the player who begins.

(Of course, we don’t ask for the explicit winning strategy: this is an open problem)

Try to solve this problem yourself before moving on to the solution below. Use hint if needed

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Hint

What happens if the mathematician going second plays a winning strategy?

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Solution

Given that there is no possible equality, one of the two players (either the one who starts, or the other) has a winning strategy. Indeed, either the first player has a winning strategy, or the second player can always beat the first player regardless of the first player’s first move. In the second case, the second player has a winning strategy.

Assume, for a contradiction, that the second player has a winning strategy. If the first player begins by taking the square in the top right, the second player will find a move that, despite the first player’s best efforts, will lead to the second player’s victory. But, since the first move consisted only in taking the first square, the second move could have been played from the start by the first player. But then, in this situation, the second player is blocked; regardless of what the second player does, the first player will win (if they play optimally). This contradicts the fact that he has a winning strategy! It is thus the first player who possesses a winning strategy.

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(If each rat tastes one bottle, there has to be 999 rats to determine which bottle is poisoned. But the rats can also taste mixes from different bottles. For example, if the first rat tastes a mix of the first 100 bottles, he dies the next day if and only if one of these 100 bottles was poisoned. Note, you will have to prove that the number of rats proposed is in fact minimal)

Try to solve this problem yourself before moving on to the solution below. Use hint if needed

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Hint

One must begin by asking oneself how many rats are necessary. For this, use the method of comparing outcomes to hypotheses, which is detailed in the reminder on the pigeonhole principle. Moreover, one must number the rats in base 2.

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Solution

If the first rat tastes a mix of the first 500 bottles, its death or survival the next morning allows the problem to be cut in half. To divide the problem again, the second rat must also taste half the bottles. But we must ensure all rats are useful: if the first rat tastes a mix of the first 500 bottles and the second rat tastes a mix of the last 500 bottles, the second rat is no use. To divide by two the number of bottles as many times as wanted without difficulty, the simplest is to renumber the bottles in base 2.

We have pow(2,9) <= 999 < pow(2,10), so one will need 10 digits to number all the bottles. The numbers will go from 0000000000 to the base 2 = 0 (the number of the first bottle) to 1111100111 to the base 2 = 999 (the number of the last bottle). The first rat will taste all the bottles whose last digit is a 1. It will thus taste bottles 0000000001 to the base 2, 0000000011 to the base 2, 0000000101 to the base 2, and so on. The second rat tastes all bottles whose second-to-last digit is a 1 and so forth through the 10th rat, who tastes all the bottles whose first digit is 1. The next day, the butler looks at which rats died. If the ith rat is dead, since he tasted all the bottles whose ith digit is a 1, the butler knows that the number on the poisoned bottle has a 1 in the ith position. The butler can in this manner find the number of the poisoned bottle. This method requires 10 rats. Let us show that one can do no better.

We will show that with n rats we cannot discriminate between more than pow(2,n) bottles. In other words, if we assume that the butler can determine which bottle is poisoned among k bottles with n rats, then we have k <= pow(2,n). To do so, we will compare the number of hypotheses with the number of outcomes; this method is detailed in the reminder on the pigeonhole principle. There are k bottles. Either the first bottle is poisoned, or it is the second, or it is the last. There are thus k different possible hypotheses about the solution of the problem. When the butler has the rats taste mixes from different bottles, the next day each rat will be alive or dead. Since there are n rats, there are pow(2,n) possible outcomes. By the pigeonhole principle, to identify the bottle we must have more possible outcomes than different solutions to the problem. We thus have k<= pow(2,n). With k = 1,000 bottles, we obtain 1,000<= pow(2,n); hence, 10<=n. At least 10 rats are required.

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After all these passes, which lockers are closed?

(During her first pass, the girl opens the lockers 2, 4, 6, and so on through 100 and leaves the others shut. During the second pass, she opens locker 3, closes locker 6, opens locker 9, closes locker 12, and so on)

Try to solve this problem yourself before moving on to the solution below. Use hint if needed

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Hint

Reason on the parity of the number of divisors of the number of each locker.

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Solution

We notice that for a locker to remain closed after all the passes, the child must have changed its state an even number of times. Thus, she must have passed this locker an even number of times. In other words, the locker’s number must have an even number of factors, not counting 1, so an odd number of factors in total. This is the case if and only if it is a square. Indeed, let n be a nonzero integer, and let f1, f2, . . . , fk be its factors. We can group them by pairs in the following way: we pair factor f with factor n/f . If n is not a square, n/f is always different from f, so n has an even number of factors. If n is a square, there exists a unique factor f such that n/f = f, so n has an odd number of factors. Finally, the lockers that are closed after all passes are those whose numbers are squares: numbers 1, 4, 9, and so forth.

Another method for computing the number of factors of n is to decompose it into prime factors:

The number of factors of n is then given by the following formula:

Thus, n has an odd number of factors if and only if all the αi are even. But, to say that all the αi are even is equivalent to saying that n is a square.

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At the end of this process, how many loops has he made on average?

Try to solve this problem yourself before moving on to the solution below. Use hint if needed

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Hint

One can consider the moment when the player’s success rate goes over 75%.

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Solution

Let us denote bn as the average number of loops that the young man creates with a plate of n spaghetti. Thus, b1 = 1, and we seek b100. Let us consider a plate of spaghetti with n   2. For his first knot, the young man either ties a spaghetti to itself or the two spaghetti to each other. In the first case, the spaghetti is eliminated: there will be on average bn−1 + 1 loops. In the second case, everything happens as if the plate only had n-1 spaghetti on it: there will be on average bn−1 loops. The probability that the first spaghetti tied formed a loop is 1/(2n−1): indeed the young man chose

the first end at random and then has 2n - 1 choices for the second. The average number of loops is thus

which means that for large n,

ln(n)/2 is a good approximation of bn. We say that bn is on the order of ln(n)/2 . The growth of bn is logarithmic, and thus much slower than one might have thought.

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With the best possible strategy, in how many drops can he determine the limit floor with certainty?

(Both plates have the same solidity and do not wear during the drops. Following this method, in the worst case, 100 drops are made and only one plate is used. However, if two plates are used, we need to drop only from the even-numbered floors (2, 4, 6, etc.) until the plate breaks. Once the plate breaks, we then have to go to the floor immediately below it to find the limit. This strategy requires only 51 drops in the worst case, but it is still not optimal)

Try to solve this problem yourself before moving on to the solution below. Use hint if needed

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Hint

One could ponder the question backward by asking from how high we can drop the first plate if we only have at most n trials.

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Solution

The key is to spin the problem differently: with at most n drops, how many floors can one test? We notice already that as soon as the first plate breaks, one must test all the forgotten floors one by one with the remaining plate, starting by the lowest and going up: there is no other approach when one has only one plate left.

Assume we can’t do more than n throws. We can drop the plate from the nth floor at most. Indeed, if it breaks, one will have to test in the worst case floors 1, 2, 3, …, n - 1 with the second plate. If it doesn’t break, then there are two plates left and n - 1 trials, so we can drop the plate from the n + (n - 1) = (2n - 1)th floor and if it breaks there are n - 2 trials to test n - 2 floors. Thus, with two plates and n trials, we can test at most

un = n + (n - 1) + (n - 2) + …. + 1 floors.

For 100 stories, we have u13 < 100 < u14. It will thus take 14 drops in the worst case.

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Under these conditions, what is the probability that the older brother gets strictly more tails than his little brother does?

Try to solve this problem yourself before moving on to the solution below. Use hint if needed

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Hint

No computation!

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Solution

It may seem surprising, but the older brother has, despite his extra coin, only a 50/50 chance of getting strictly more tails than his younger brother.

There are only two possible cases: either the older brother has strictly more tails than his brother, or strictly more heads. At least one of these two cases will occur: since the older brother has one more coin, he cannot have both a number of heads less than or equal to his younger brother and a number of tails less than or equal to his younger brother. Moreover, both these cases cannot occur at once: since he only has one extra coin, the older brother cannot have both strictly more heads and strictly more tails than his younger brother. We will always have one case or the other, but never both at the same time.

Both cases are symmetric; thus, they have the same probability, and since they are complementary, their probability is 1/2 . Thus, the older brother only has a 50/50 chance of getting strictly more tails than his younger brother.

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