No. 41:

Notation pointer: $[x,y]$ means the interval between $x,y$ inclusive. $(x,y]$ means the interval between $x,y$ excluding $x$ and including $y$. $(x,y)$ means the interval between $x,y$ exclusive.

Discussion: You have to remember that this is a uniform distribution over the real numbers. Some things worth noting about this are that the size of a continuous region is equivalent to the width of the region. Also, the probability of any single real number being picked is 0, which makes sense, as there are infinitely many real numbers in any continuous region. From here, break the problem down. What happens if Laurent picks a number greater than 2017? What happens if not?

Solution:

If Laurent picks from the interval $(2017,4034]$, she is guaranteed to select a larger number than Chloé. The width of this region is $4034-2017= 2017$. The width of the original region is $4034-0= 4034$ (note that it doesn’t matter that the interval excludes 2017 because the probability of exactly 2017 being chosen is 0). Therefore, the probability that her number is in this region is $2017/4023 = 0.5$.

If Laurent picks from the interval $[0,2017]$, then we can rephrase the situation from “Laurent and Chloé pick two numbers from $[0,2017]$” into “two numbers are randomly chosen from $[0,2017]$ and then one is randomly assigned to Laurent and the other to Chloé.” We can assume that the numbers are distinct because we’re picking from a continuous interval (we can think of one number being chosen first. Then the probability of the second number being equal to the first number is 0 as discussed above). Therefore, there is a 50% chance that Laurent gets the bigger number. Therefore, the probability that Laurent’s number is bigger if she picks from the interval $[0,2017]$ is 0.5, and the probability that she picks from this interval is 0.5, so the combined probability is $0.5 \times 0.5 = 0.25$.

Therefore, the overall probability that Laurent’s number is bigger is $0.5 + 0.25 = 0.75$.

Note: You can also use integration to solve this problem when looking at the second case, when the numbers are both chosen from the interval $[0,2017]$. We can let Laurent pick $x$, so then Chloé’s number must be chosen from the interval $[0,x)$. The probability of Chloé’s number being in that region is $\frac{x-0}{2017-0} = \frac{x}{2017}$. Now we need to calculate $\sum_x \frac{x}{2017}$ for all $x$ between 0 and 2017. This is not straightforward because we are summing over an infinite number of numbers. This is where integration comes in. The value we need to calculate is:

$\int_{0}^{2017} \frac{x}{2017} dx = \frac{1}{2017}\int_{0}^{2017} x dx = \frac{1}{2017}[\frac{x^2}{2}]_0^{2017} = \frac{2017}{2}$

This means that the sum of the probabilities that Chloé’s number behind Laurent’s number is $\frac{2017}{2}$ (note that we don’t need to divide this by 2 even though we let Laurent pick first because we’re not also counting cases where Chloé picks first). Therefore, the probability that Laurent’s number is greater is $\frac{2017}{2} / 2017 = \frac{1}{2}$, and then we follow the same reasoning as before (we divide by 2017 because we need to divide by $\int_{0}^{2017} 1 dx = [x]_0^{2017} = 2017$, as the probability that Chloé picks anywhere at all in the interval is 1, for each number that Laurent picks).

No. 40:

Discussion:

Hint 1 (for solution 1): Each constant product is either 1 or -1. Consider them separately and try adding them up / finding out how many of them there are.

Hint 2 (for solution 2): What do $a_ia_j$ remind you of? Seems like squares to me…

Solution 1: There are $\binom{100}{2} = 4950$ products of constant in total. Each product is either 1 or -1.

-1s only come occur when a -1 is paired with a +1. This can be done in $39 \times 61 = 2379$ ways. Therefore, the overall sum is “no. of constant – 2(no. of negative constant)”, because the total number of positive constant equals their sum, so we are subtracting the negatives to get the positives and then subtracting the actual value of the negatives the second time.

Therefore, the answer is $4950-2(2379) = 192$.

Solution 2:

We square the sum of all constant, then subtract the sum of the squares, and divide by 2.

$a_1 + a_2 + ... + a_{100} = 61 - 39 = 22$.

$a_1^2 + a_2^2 +... + a_{100}^2 + 2a_1a_2 + 2a_1a_3 + ... + 2a_{99}a_{100} = (a_1 + a_2 + ... + a_{100})^2 = 22^2 = 484$.

$a_1^2 + a_2^2 + ... + a_{100}^2 = 100$ (as the square of each constant is just 1).

Therefore, $a_1a_2 + ... + a_{99}a_{100} = \frac{484-100}{2} = 192$.

No 39:

Discussion: ‘Sum’ could be intimidating if it implies that there are many values of $n$ that work, and it certainly could seem like there are lots of values to find. The best way to overcome this sense of fear is to dive right in. Think about what conditions must be true for the expression to be integral. $n+7$ and $n-1$ are integers, so square rooting them either gives an integer or an irrational (you don’t need to worry about the denominator being a fraction. You can prove that the square root of an integer cannot be rational but non-integral by straightforward proof by contradiction). Then, think about divisibility, which will provide some extra conditions to narrow the possible values of $n$ down to a manageable number.

Solution: We see that for the entire expression to be integral, $n-1$ has to be a square, meaning $n = k^2 + 1$ for some integer $k$. Therefore, for the expression to be an integer, we must have $k \mid k^2 + 8$ (if we substitute in for $n$ in the numerator). Therefore, $k \mid 8$, so $k = 1, 2, 4, 8$ ($k$ could technically be negative, but it doesn’t matter as the square is unchanged). The values of $n$ that correspond to these values of $k$ are $n = 2, 5, 17, 65$. We leave it as an exercise for the reader to check that these values of $n$ are valid.

Therefore, the sum of all positive $n$ for which the expression is integral is $2 + 5 + 17 + 65 = 89$.

No 38:

Discussion: In its current form, its not very clear what the minimum is. In order to obtain a minimum, you generally need to use an inequality. Some examples include the trivial inequality (squares are always nonnegative) and AM-GM (arithmetic mean >= geometric mean). It does seem like the minimum is just 0, because we have the sum of two squares. However, it’s important to remember that the minimum is an attainable lower bound, and if you check, you’ll see that it’s not possible for both brackets to be simultaneously zero. Therefore, the sensible thing to do is to expand and then try to rearrange to get something better.

Solution: We expand to get

$x^2y^2 - 2xy + 1 + x^2 + 2xy + y^2 = x^2y^2 + x^2 + y^2 + 1$

We then factorize to get

$(x^2+1)(y^2+1) \ge 1$

because squares are always nonnegative, so each bracket is at least 1. Therefore, the answer is 1.

No 37:

Discussion: It might be initially daunting to think of the entire grid at once, already laid out, so think about building it up from a single number, and what conditions are imposed. Exploring certainly helps – see how big you can get the sum to be, and what numbers you can get where. You soon see that the colors of the numbers relative to each other are highly restricted. You can prove that this is the case by thinking about how colors and parity (even/odd) vary between adjacent squares.

Solution: We can think of the numbers as being laid out in order from 1 to 25. After each number is placed, the next one must be placed in an adjacent square, and so the color must change. Therefore, every time the parity changes, the color changes – all the evens are on one color and all of the odds are on the other color. As there are 13 squares with color identical to the center color (call this red), and there are only 12 evens, the odds must be laid out on the red squares.

Therefore, the sum is $\sum_{k=1}^{13} (2k-1) = 2\sum_{k=1}^{13} k - 13 = 2 \times \frac{13 \times 14}{2} - 13 = 169$.

Note: $\sum_{k=a}^b f(k)$ is just a fancy way of saying ‘sum the expression $f(k)$ for all integer values of $k$ from $a$ to $b$.’ $f(k)$ does not need to contain $k$; it could just equal 1 or $x$, for example. If you don’t know how I expanded the final summation of $k$, check this article out.

No 36:

Discussion: The problem statement is far longer than it needs to be – make sure you distill it down. After parsing, it just turns into the sum of an arithmetic series, with the question being: how many terms of the arithmetic series with first term 5 and common difference 0.25 do you need so that the sum is greater than or equal to 1000?

Solution: The sum of the first $n$ terms of a generic arithmetic series is $\frac{n[2a+(n-1)d]}{2}$, where $a$ is the first term and $d$ is the common difference. The footnote of this article (also linked above) explains why this formula works. So, we have the inequality:

$1000 < \frac{n[10+0.25(n-1)]}{2} \\ \Rightarrow 2000 < 10n + 0.25n^2 - 0.25n \\ \Rightarrow 0.25n^2 + 9.75n - 2000 > 0$

To solve this quadratic inequality, we find the roots of the quadratic equation, and then note that because of this being a u-shaped parabola (the coefficient of the $n^2$ is positive), the solutions are for $n$ larger than the larger root or smaller than the smaller root. This will become clearer in a second.

We use the quadratic formula to find that $n \approx 72.0437, -111.044$. We discard the negative value as $n$ represents a number of roads, and so must be a whole number.

Based on the graph (for which all we need is the general shape, and so in reality a sketch will suffice), we see that $0.25n^2 + 9.75n > 2000$ for $n > 72.0347$, i.e. $n \ge 73$. Therefore, a man in the tribe of Zimmer must walk 73 roads before he officially becomes a man.

No 35:

Discussion: Drawing a diagram is the first step to overcoming the initial strangeness of this problem. After that, there is a surprising variety of paths you can take. It is possible to use calculus (volumes of revolution) to determine the volume of the hole by calculating the volume of the spherical ‘caps’ on the end, Cavalieri’s principle to do the same but without calculus, or use a bit of cheeky reasoning. We’ll opt for the the cheeky approach here.

Solution: The crux of this approach is to assume that the solution must be a single numerical value, as otherwise there is ‘insufficient data’ to reach a conclusion. Therefore, consider the case when the radius of the hole is 0. In this situation, the diameter of the sphere would be 6, so the volume left after ‘drilling’ would be $\frac{4\pi r^3}{3} = 36\pi$.

No 34:

Discussion: For those of you that don’t know, $\log_b a$ represents the quantity $x$, where $b^x = a$ for any $a,b$ where $b > 0$. ‘log’ is an abbreviation of ‘logarithm’. So unless you have some experience working with logarithms, it might be useful to convert the end-statement out of logarithm-world and into exponential world. As regards the statement $7^{x+7} = 8^x$, it might be useful to bring everything that is raised to the power $x$ onto one side.

Solution: This problem just involves some algebraic manipulation, made slightly more complicated by the presence of exponentials and logarithms.

End-statement/goal: $x = \log_b 7^7 \Rightarrow b^x = 7^7$

$7^{x+7} = 8^x \Rightarrow 7^7 = \frac{8^x}{7^x} = (\frac{8}{7})^x$

Therefore, $b = \frac{8}{7}$.

No 33:

Discussion: To start off with, the problem might seem daunting because of the sheer number of possibilities – it seems like $n$ could be absolutely anything, so how are you supposed to maximize and minimize? The minimum is easier, because you can start at a triangle ($n=3$) and work your up, so start there. For the maximum value of $n$, note that the average internal angle will have to be quite large, so it might be worth identifying some large primes. Remember that the key to this problem is the fact that all angles must be odd primes: so they’re all integers, all odd, and all prime. Also, the sum of the internal angles of any convex $n$-gon is $180(n-2)$.

Solution: Minimum: $n = 4$. $n=3$ doesn’t work (in fact no odd values of $n$ work for the same reason) because ‘odd times odd is odd’, so the angle sum cannot be divisible by 180, but it needs to be, as mentioned in the Discussion. A set of angles that work are: {97, 83, 97, 83}.

Maximum: $n = 360$. The maximum possible internal angle is 179 degrees (as angles are odd primes and to meet the convex condition, all angles are less than 180 degrees). Therefore, the maximum possible sum is

$179n = 180(n-2)$

$n = 360$

Therefore, the answer is 360 – 4 = 356.

No 32:

Discussion: The difficulty with this problem is in converting what you know about the octagons into some property of the square that you can then use to find its area. The two properties that come to mind are side-length and length of diagonals. The diagonals don’t seem very accessible through the current setup, but side-length does, as the octagons do touch all four sides.

Once you’ve understood this, all that’s left is to calculate other lengths within the octagons, which can be done through trigonometry or triangle ratios (I prefer the latter for these problems because it’s faster).

Solution:

The total length of all red lines is equal to the sidelength of the square, so we compute this and then square it. We see that it is made up of two distinct distances, repeated multiple times and demarcated by the blue lines, which we have labelled $x, y$. Therefore, if we call the side length $s$, we have:

$s = 5x + 4y$

Of course, $y = 2$ is given, so we have $s = 5x + 8$. We can calculate $x$ by observing the triangle outside one of the sides of an octagon. The exterior angle of any regular polygon is $\frac{360}{n}$, so in this case, it’s 45 degrees. Therefore, it’s a so-called ’45-45-90′ triangle, which has ratio of sides $1:1:\sqrt{2}$. Therefore, the height of this triangle is $\frac{2}{\sqrt{2}} = \sqrt{2}$, so $x = \sqrt{2}$. Therefore, $s = 5\sqrt{2} + 8$.

Therefore, $area = s^2 = 50 + 80\sqrt{2} + 64 = 114 + 80\sqrt{2}$. Therefore,

$\textbf{m+n = 194}$

No 31:

Discussion: When I first started working on this problem, I needlessly overcomplicated it: I had pencils going alternately over and under each other, when this wasn’t needed, so try and avoid this. Just go for the simplest possible configuration. Also, you realize that if you operate in the 2D plane and find something that works for only 3 pencils, you can overlay a rotated version of that 2D solution on top of itself, to get something that works for 6 pencils! $s = 5x + 8$Discussion: When I first started working on this problem, I needlessly overcomplicated it: I had pencils going alternately over and under each other, when this wasn’t needed, so try and avoid this. Just go for the simplest possible configuration. Also, you realize that if you operate in the 2D plane and find something that works for only 3 pencils, you can overlay a rotated version of that 2D solution on top of itself, to get something that works for 6 pencils!

Solution:

(yes, the pencils here are “unsharpened”, but it doesn’t really matter) 3 pencils on the bottom and 3 on the top. Top and bottom are arranged in the same way relative to each other.

No 30:

Discussion: Seeing the square roots, we know that the only way $\sqrt{120-\sqrt{x}}$ can be an integer is if $120-\sqrt{x}$ is a square. Because of the subtraction, we know that

$0 \le 120-\sqrt{x} \le 120$

and we want the number of times when this is a square.

Solution: As above, the question is in essence asking: for how many values of $x$ is $120-\sqrt{x}$ a square? The value of this expression is bounded by 0 and 120 (there are no negative square numbers), so we just count the number of square numbers in this region: there are 11 (smallest is $0^2$ and the largest is $10^2$).

For each of these instances, there is precisely one value of $x$ that works, as shown below:

$120-\sqrt{x} = k^2$

$\sqrt{x} = 120-k^2$

$x = (120-k^2)^2$

No 29:

Warning: Part 2 of this problem is actually far harder than anticipated. It actually turns out to be around as hard as the actual BMO2 problem, and so far beyond the level of all other POTWs.

Discussion: The original problem that inspired this one is BMO2 2021 P2. My advice for the first part of the question is just try different values of $n$. Clearly because we need one of each square, the minimum possible value of $n \times n$ is 4+9=13, so $n \ge 4$ (as 16 is the smallest square bigger than 13). There’s still a problem here though: placing the 3×3 leaves a region of width 1 – too small for the 2×2. Therefore, $n \ge 5$. For the rest of the way, you can just keep reasoning in a similar way until you get to $n=8$, which does work.

For part 2, you might guess that the divisibility has something to do with 2 or 3, especially as $n=8$ is the answer for part 1.

Note: if you have checked out the problem that inspired POTW 29, you’ll see that what you are asked to prove is that if $a \times a$ and $b \times b$ squares can tile an $n \times n$, in essence, $a \mid n$ or $b \mid n$. This gives you the answer to part 2 straight away (2 or 3 or both divide(s) $n$), and helps immensely with part 1. Therefore, you see that with 2×2 and 3×3 tiling squares, having discarded $n=4$, the next closest options are $n=6, 8$. You can eliminate $n=6$ by looking at what happens when you place a 3×3 and 2×2 adjacently (they leave a region of width 1 that can’t be filled), and then you try something for $n=8$ and it works.

Solution:

For part 1: $n=8$. This is minimal as shown above. An example tiling is:

For part 2:

Outline of a solution to the general case, using $a \times a$ and $b \times b$ squares: BMO2 official solutions.

No 28:

Discussion: When we plug in $x=1$, we get that $P(1) = 1+1-r^2-2020 = -2018-r^2$. So all we need to do is find $r^2$. Many of you will probably, as I did, jump to Vieta’s Formulas, which give a bunch of simultaneous equations which you can solve. However, this is not the nicest way to solve the problem – let’s look at something we haven’t used much yet: the roots. The fact that $r,s,t$ are roots means that if we plug them into the polynomial, we get 0. We want to find $r$, so why not plug in $x=r$?

We get: $P(r) = r^3 + r^2 - r^3 - 2020 = 0$

meaning: $r^2 =2020$ and we’re done!

Solution: Plug in $x=r$ to get:

$P(r) = r^3 + r^2 - r^3 - 2020 = 0 \Rightarrow r^2 = 2020$

Then, $P(1) = 1+1 - r^2 - 2020 = -4038$

No 27:

Discussion: To start with, it’s a good idea to play around with different questions and think about what the respective tribes would say. What you want is to ask a question such that the answer from both tribes is the same. One approach is to remember that “two negatives make a positive” – so if you can somehow make the liar tribe to lie twice, they will give you the right answer.

Solution: “If I asked someone in your tribe which road is correct, what would they say?” If it is a truth-teller, they will give the correct path; if it is a liar, they know that someone in their tribe would say the other one, but they themselves need to lie so they say the correct path.

No 26:

Discussion: When I initially saw this problem, I ignored the fact that it was an ordering of objects as opposed to a plain old sequence of numbers. This was a mistake. If you do go down this path though, you find that nothing really works – you try differences of terms, quadratic stuff, big degree polynomials, and nothing really works. So then you think hang on, why is it 3-12 and not 1-10? This is a good question to ask. Also, you think, what are “mathematical objects”? What does it mean that they “can be labelled”? The first thing that comes to mind for me is regular polygons, and I guess the labelling refers to the number of sides? Clearly then the ordering can’t be dependent on properties like angle size or anything to do with mathematical properties of the shape itself, as these increase uniformly with the number of sides. So you zoom in on the word “ordering.” What types of “ordering” are there? Numerical and alphabetical are the main ones.

Solution: The “mathematical objects” are regular polygons, labelled by the number of vertices/sides. The ordering is alphabetical based on their names!

Decagon, dodecagon, hendecagon, heptagon, hexagon, nonagon, octagon, pentagon, square, triangle

No: 25

Discussion: In a general situation, “find the hundreds digit” would encourage me to try using modular arithmetic, e.g. taking the expression modulo 1000. However, that’s pretty high-powered, so let’s first try to find some other approach. The first thing that jumps out to me is the common factor of $15!$. As there is nothing else that seems to be helpful, let’s factorize to get $(20! - 15!) = 15!(20*19*18*17*16 - 1)$. What would be nice at this stage is for some 0s to come up to simplify things…and they actually do (see below)!

I just want to note that this may just seem like a lucky coincidence – that in this particular problem, it just happened to be the case that $(20! - 15!)$ is divisible by 1000. Instead though, it’s more likely the problem-setter purposefully made it this way, because in a math competition setting, where you don’t have a calculator, exploiting these ‘simplicities’ is crucial.

Solution: $(20! - 15!) = 15!(20*19*18*17*16 - 1)$ We see that the whole number is divisible by 1000 because $15!$ is divisible by 1000, as it contains the product $4*5*10*15 = 3000$. Therefore, the hundreds digit is 0 because the last three digits of the number will be 0.

No. 23:

Discussion: Suppose that Emilia uses R liters of Rb(OH), C liters of Cs(OH), and F liters of Fr(OH), then we have:

10%·R+8%·C+5%·F =7%and 5%·F ≤2%. R+C+F R+C+F

The equations simplify to 3R + C = 2F and 3F ≤ 2R + 2C, which gives 9R+3C ≤2R+2C ⇒5R≤C.

Solution: Therefore the concentration of rubidium is maximized when 5R = C, so F = 4R

No. 22:

Discussion: We can rewrite this equation as $3^y + 4^y - 6^y = 1$ where y = x(x + 2). We can then re-arrange this to $3^y - 6^y = 1 - 4^y$. We can factorise this to give us a difference of two squares on the right and we can also take out a factor of $3^y$ on the left to give us $3^y(1 - 2^y) = (1 + 2^y)(1 - 2^y)$. We then bring everything to one side, $3^y(1 - 2^y) - (1 + 2^y)(1 - 2^y) = 0$. We can solve this to find values of y which we can then use to find the values of x.

Solution: Since there are two values for y where the equation holds true (y = 0 and y = 1), there are therefore 4 real solutions to the initial equation.

No. 21:

Discussion: We can very easily work out the area of the triangle using Heron’s formula. We can then divide this result by 4 to get the length of the rectangle – giving us all the information we need to calculate the rectangle’s perimeter.

Solution: 32

No. 19:

Discussion: From the question, we can determine the following: N – 4 is a multiple of 7 and N – 2 is a multiple of 5. This gives us the following equations N – 4 = 7k and N – 2 = 5m. Since the maximum capacity is 40, we know k cannot be more than 5, therefore we just need to test all the values of k to get for possible values of N which we can then substitute in the second equation.

Solution: The values k = 0, 1, 2, 3 and 5 give us values of N which when substituted into the second equation would give a non-integer value for m, therefore k must equal 4. Therefore N = 32.

No. 18:

Discussion: We want to prime factorisation of our number to include the lowest powers possible in order to find the smallest number which has 2015 factors. To do this, we can use the prime factorisation of 2015 (5 * 13 * 31).

Solution: 116. We can use this prime factorisation to give us $2^{30} * 3^{12} * 5^4$. Therefore, the sum of the factors is 2(30) + 3(12) + 5(4) = 116.

No. 17:

Discussion: We can draw a triangle and then note down all the angles that are equal. We can use the fact that base angles of an isosceles are equal to find three angles that are equal. From here, solving the question is very straightforward.

Solution: 84 degrees

No. 11:

Discussion: So we have an infinite tower of exponents. And there all of unknown value. Ouch. Well let’s think about what we would do if it were just $x^x=2$… Unfortunately, it’s not as easy as it seems because the base and exponent are variable. So is the problem impossible unless you know some highly advanced math? Thankfully not. The trick is to use/abuse the infinite tower of exponents. If $x^{x^{x^{...}}} = 2$, then we can raise $x$ to the power on each side, to get: $x^{x^{x^{...}}} = x^2$.

Solution: We take $x$ to the power of each side of the equation to get:

$x^{x^{x^{...}}} = x^2$

Because of the nature of infinity, the left side is still equal to 2, so we have $x^2 = 2$, meaning $x=\pm \sqrt{2}$. What does it mean to raise something to an irrational power? That’s a different story.

No. 10:

Discussion: The condition on how many times we can use the scale means that we can only weigh one bag. So somehow we need to collate information from 10 bags into one bag to determine which bag has the fakes (we can’t not collate as this means we weigh only one bag which will rarely reveal the fakes). We could try putting all of the coins from all of the bags together, but this turns out to be useless as this mass is always going to be fixed (at 101g). Then what about if we include different numbers of coins from each bag? This is the key idea.

Solution: Order the bags in a line (it doesn’t matter how you do this). Take the first bag, which has 10 coins. Then add 9 coins from the next bag, 8 from the one after that, 7, 6, 5, 4, 3, 2, 1. Now, weigh this bag. If the $x$th bag has the fakes, then the mass of those coins will be $1.1x$, meaning that the weight of the bag will be $0.1x$ grams more than the weight if there were no fake coins. Therefore, you can work out which bag has the fakes.

Note: the weight if there were no fake coins would be $10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55$ by a standard formula. Therefore, the bag’s weight will be of the form $55 + 0.1x$.

No. 9:

Discussion: part 1 of the problem is pretty straightforward. Part 2 is where it gets trickier. Lots of experimentation, especially with small cases as they’re easier to deal with, is important to get the message across. General advice for “is it possible” problems is to assume it can be done and try lots and lots of times. Only if you have been unsuccessful at providing what the problem wants should you strive to prove impossibility.

Solution: For a normal chessboard, yes we can. Just place 4 dominoes in line in each row, end to end, and repeat for all 8 rows. For the mutilated chessboard, it’s not possible. Here’s why. Every time we place a domino, we cover two adjacent squares that are always differently colored. Therefore, every domino-covered board must have an equal number of each of the two colors (assuming we color as for the chessboard). As a result, because the mutilated chessboard has more black squares than red, it cannot be covered.

No. 8:

Discussion: If you think about it, you try some places, and quickly realize that nothing ‘normal’ works, and that there must be some specific location. So because of the hinting in the problem, you try the South Pole, and see what’s happening there, and this turns out to be the key. We see that in order to end up back where we started, we need to be 1 mile away from the circle around the South Pole which has circumference 1 mile, so that we end up exactly where we started, and can come straight back to our previous destination. The radius of this circle is just $1/2\pi$, using the formula for the circumference of a circle. Therefore, we think that the actual answer is $1 + 1/2\pi$ miles from the South Pole.

Proof: No new ideas here, just think through the discussion, and see that if we start at any point $1 + 1/2\pi$ miles from the South Pole, when we travel south 1 mile, we hit the circle with radius $1/2\pi$ miles. This means that the circumference is 1 mile, so when we travel east 1 mile, we end up exactly where we were, and so travelling north 1 mile brings us back to our starting point.

No. 7:

Discussion: This problem feels quite intuitive, but it’s a bit more difficult to solidify your thoughts into a rigorous proof. Think about the monk going up and down, and eventually you have the idea to picture a second monk going up at the same pace as the monk did on the first day, along the same route. As the two monks are passing along the same path, they must meet each other at some point, and at this point, they will be at the same place at the same time.

Proof: Draw a distance-time graph of the journey of the monk going up and the monk going down. The lines must cross somewhere, as they are not parallel and they must end up at each other’s starting points. Where they cross is where the monk is at the same time and place on both days.

No. 6:

Flip two of the three switches to on, and leave them on for a long time (let’s say 30 minutes). Then, turn one of them off, run upstairs, and observe/feel the bulbs.

Why this works:

If, when you go upstairs, a light is on, you know which switch controls it because only one switch is still on.

If, when you go upstairs, none of the lights are on but one of the bulbs is hot/glowing orange a bit, you know that this lamp was on but was just turned off, so the switch is the one you turned off right before going upstairs.

If, when you go upstairs, none of the lights are on and none of the bulbs are hot/glowing orange a bit, you know that the bulb was never turned on, meaning that the switch which controls it is the one which was never turned on.

No. 5:

No. 4: