Problems

Solution: There are \(87\) years between 1905 and 1992. Of those years, every 4th is a leap years, starting with 1908, and ending with 1992, so there are 22 leap years. There are 30 days between Apr 3 and May 3, 31 days between May 3 and Jun 3 and a further 23 days to the 26th. Therefore \(87 \times 365 + 22 \cdot 1 + 61 + 23\) total days. \(\pmod{7}\) this is \(3 \times 1 + 1 + 5 + 2 = 3\), therefore it is \(4\) week days before Friday, ie Monday.

Solution:

1. \((a+b+c)+(d+e+f)+(g+h+k) = 3n = 1 + 2 + \cdots + 9 = 45 \Rightarrow n = 15\).
2. Summing all rows, columns, diagonals through \(e\) we have \((a+e+k)+(b+e+h)+(c+e+g)+(d+e+f) = 45 + 3e = 60 \Rightarrow e = 5\).
3. Suppose that one of the corners is \(9\), then we need to find \(2\) ways to make \(6\) not using \(5\) and \(1\) (as \(5\) is in the middle and \(1\) diagonally opposite). Clearly this is not possible as the only remaining numbers are \(2,3,4\) and only \(2+4 = 6\). Therefore \(9\) cannot be in the corner or central squares, ie it's one of \(b,d,h,f\)
4. We must have \begin{array}{ccc} a & 9 & c\\ d & 5 & f\\ g & 1 & k \end{array} and so \(a\) or \(c = 4\). Once we place \(4\) by symmetry there will be another solution with \(a = 2\). So: \begin{array}{ccc} 4 & 9 & 2\\ d & 5 & f\\ g & 1 & k \end{array} we now see \(k\), then \(f\), then \(d\) then \(g\) must be determined, ie: \begin{array}{ccc} 4 & 9 & 2\\ 3 & 5 & 7\\ 8 & 1 & 6 \end{array} so our two solutions must be this and \begin{array}{ccc} 2 & 9 & 4\\ 7 & 5 & 3\\ 6 & 1 & 8 \end{array}
5. For each of the \(4\) possible placements of \(9\) there are two magic squares, so there are \(8\) possible magic squares, all related by reflection and rotation.

Solution:

1. \begin{align*} \int_{-\pi}^\pi |\sin x | \d x &= \int_{-\pi}^{0} - \sin x \d x + \int_0^\pi \sin x \d x \\ &= \left [\cos x \right]_{-\pi}^{0} +[-\cos x]_0^{\pi} \\ &= 1-(-1)+(1)-(-1) \\ &= 4 \end{align*}
2. \begin{align*} \int_{-\pi}^\pi \sin | x | \d x &= \int_{-\pi}^0 - \sin x \d x + \int_0^\pi \sin x \d x \\ &= 4 \end{align*}
3. \begin{align*} \int_{-\pi}^\pi x \sin x \d x &= \left [ -x \cos x \right]_{-\pi}^\pi + \int_{-\pi}^{\pi} \cos x \d x \\ &= \pi -(-\pi) + \left [\sin x \right]_{-\pi}^\pi \\ &= 2\pi \end{align*}
4. \begin{align*} \int_{-\pi}^{\pi} x^{10} \sin x \d x &\underbrace{=}_{x^{10}\sin x \text{ is odd}} 0 \end{align*}

Solution:

1. Multiplication by \(1+i\) rotates by \(45^{\circ}\) anticlockwise
   

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2. \(|z| \leq \sqrt{2}\), \(\textrm{Re}(z^2) \geq 0\) means \(\textrm{Re}{z} \geq \textrm{Im}{z}\)
   

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3. These are all points within \(1\) unit from a circle radius \(2\) units.
   

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Solution: \begin{align*} && p_0(x) &= (1-x)(1-x^2)(1-x^4) \\ &&&= (1-x)(1-x)(1+x)(1-x^2)(1+x^2) \\ &&&= (1-x)^2 (1+x)(1-x)(1+x)(1+x^2) \\ &&&= (1-x)^3 (1+x)^2 (1+x^2) \end{align*} \begin{align*} && p_0'(x) &= 3(1-x)^2(1+x)^2(1+x^2) + (1-x)^3 q(x) \\ \Rightarrow && p_0'(1) &= 3 \cdot 0 \cdots + 0 \cdots \\ &&&= 0 \\ && p_1'(x) &= p_0(x) + xp'_0(x) \\ \Rightarrow && p_1'(1) &= p_0(1) + 1\cdot p_0'(1) \\ &&&= 0 + 1 \cdot 0 \\ &&&= 0 \end{align*} Notice that \(p_0(x) = (1-x-x^2+x^3)(1-x^4) = 1-x-x^2+x^3-x^4+x^5+x^6-x^7\), so: \(p'_0(x) = -1-2x+3x^2-4x^3+5x^4+6x^5-7x^6 \Rightarrow p'_0(1) = 0 = -1 -2 -4 -7 + 3 + 5+6\). \((xp'_1(1))' = 0 = -1^2-2^2-4^2-7^2 + 3^2 + 5^2 + 6^2\). Consider \(q_0(x) = (1-x)(1-x^2)(1-x^4)(1-x^8)\), then \((1-x)^4\) is a factor, so in particular we know \(q_0(1), (xq_0(x))'|_{x=1} = 0,(x(xq_0(x))')'|_{x=1} = 0\), and so: \(q_0(x) = 1-x-x^2+x^3-x^4+x^5+x^6-x^7 - x^8+x^9+x^{10}-x^{11}+x^{12}-x^{13}-x^{14}+x^{15}\), and so: \(1^r+2^r+4^r+7^r+8^r+11^r+13^r+14^r = 3^r+5^r+6^r+9^r+10^r+12^r+15^r\) for \(r = 1,2,3\)

Solution: If \(g(0) \in \mathbb{Z} \Rightarrow b \in \mathbb{Z}\). If \(g(1) \in \mathbb{Z} \Rightarrow a+b \in \mathbb{Z} \Rightarrow a \in \mathbb{Z}\), therefore \(a \cdot n + b \in \mathbb{Z}\), in particular \(g(n) \in \mathbb{Z}\) for all integers \(n\). \(f(0) \in \mathbb{Z} \Rightarrow C \in \mathbb{Z}\), \(f(1) \in \mathbb{Z} = A+ B + C \in \mathbb{Z} \Rightarrow A+ B \in \mathbb{Z}\) \(f(-1) \in \mathbb{Z} = A- B + C \in \mathbb{Z} \Rightarrow A- B \in \mathbb{Z}\) \(\Rightarrow 2A, 2B \in \mathbb{Z}\) \begin{align*} f(n) &= An^2 + Bn + C \\ &= An^2-An + An+Bn + C \\ &= 2A \frac{n(n-1)}2 + (A+B)n + C \\ &\in \mathbb{Z} \end{align*} Consider \(g(x) = f(x + \alpha)\), therefore \(g(0), g(1), g(-1) \in \mathbb{Z} \Rightarrow g(n) \in \mathbb{Z} \Rightarrow f(n+\alpha) \in \mathbb{Z}\)

Solution: Consider \(E = \frac12 m \dot{x}^2 - \frac{Gm}{x}\), notice that \begin{align*} && \dot{E} &= m \dot{x} \ddot{x} + \frac{Gm}{x^2} \dot{x} \\ &&&= \dot{x} \underbrace{\left (m\ddot{x} + \frac{Gm}{x^2} \right)}_{=0 \text{ by N2}} \end{align*} Therefore \(E\) is conserved. Therefore if \(x \to \infty\) \(\frac12 m V^2 - \frac{Gm}{h} = \frac12 m u^2 - 0 \geq 0\) so \(V^2 \geqslant 2G/h\) Since \(E \approx 0\) we want to solve \begin{align*} && \dot{x} &= -\sqrt{\frac{2G}{x}} \\ \Rightarrow && -\int_h^0 \sqrt{x} \d x &= \int_0^T \sqrt{2G} \d t \\ \Rightarrow && \frac{2h^{3/2}}{3} &= \sqrt{2G}T \\ \Rightarrow && T &= \frac{\sqrt{2}h^{3/2}}{3\sqrt{G}} = \frac13 \sqrt{\frac{2h^3}{G}} \end{align*}

Solution:

1. We cannot say this, since we do not know how many people were flying or walking each year.
2. This is difficult to say without knowing the variance. We might expect this to have quite a skewed distribution (one big air crash causes lots of deaths infrequently) so it's impossible to know, although it is substantially lower.
3. If we have 1080 deaths annually, we should expect ~3 deaths per day. While a day with \(7\) deaths might seem unlikely, over the course of a year it is very likely to occur. (Perhaps the weather was bad). It is also probably a case of selective reporting, we are seeing this data point because it's notable and being reported rather than because it is significant).
4. This is certainly the most alarming, a ~25% increase is very unlikely without something else going on. (We'd expect it to be ~Po(1080) approximalely N(1080, 1080) but then this is many standard deviations away). However we also know that other factors could drive this (more walking, more people, change in reporting etc)

Solution: If you arrive between 12 to and 12 past, it will take 60 minutes + how many minutes you wait at the station. If you arrive between 12 past and 12 to, it will take 75 minutes plus waiting at the station. Let's say arrival time \(X \sim U(0,60)\) minutes past the hour, then travel time is. Let's say there are two random variables, \(X_{fast} \sim U(0,24)\) \(X_{slow} \sim U(0, 36)\). Then if you wait for a fast train your expected wait time is \(72\), if you wait for a slow time, your expected wait time is \(75 + 18 = 93\). There is a \(\frac{24}{60} = \frac{4}{10}\) chance of being in the first case, and \(\frac{6}{10}\) chance of the second, ie: \(\frac{4}{10} \cdot 72 + \frac{6}{10} \cdot 93 = \frac{846}{10} = 84.6\) expected wait time. Suppose the time the trains so the expected fraction of time waiting for the fast train is \(t\) and the slow train is \(1-t\). Then the expected time is: \begin{align*} t \l 30t + 60 \r + (1-t) \l 30(1-t) + 75 \r &= 60t^2 -75t + 105 \\ &= 60 \l t^2 - \frac{5}{4}t \r + 105 \\ &= 60 \l t - \frac{5}{8} \r^2 - ? + 105 \\ \end{align*} Threfore we should choose \(x\) such that \(t = \frac58\), which is \(~37.5\) minutes after the slower train, \(25.5\) minutes past.