Problems

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Solution: There are \(1\,000\,000\) numbers between 1 and a million (inclusive). \(500\,000\) are divisible by \(2\), \(200\,000\) are divisible by \(5\) and \(100\,000\) are divisible by both. Therefore there are: \(1\,000\,000 - 500\,000-200\,000+100\,000 = 400\,000\). (Alternatively, the only numbers are those which are \(1,3,7,9 \pmod{10}\) so there are \(4\) every \(10\), or \(4 \cdot 100\,000\)). We can sum all these values similarly, \begin{align*} S &= \underbrace{\sum_{i=1}^{10^6} i}_{\text{all numbers}}-\underbrace{\sum_{i=1}^{5 \cdot 10^5} 2i}_{\text{all multiples of } 2}-\underbrace{\sum_{i=1}^{2 \cdot 10^5} 5i}_{\text{all multiples of } 5}+\underbrace{\sum_{i=1}^{10^5} 10i}_{\text{all multiples of } 5} \\ &= \frac{10^6 \cdot (10^6 + 1)}{2} - \frac{10^6 \cdot (5\cdot 10^5+1)}{2} - \frac{10^6 \cdot (2\cdot 10^5+1)}{2} + \frac{10^6 \cdot (10^5+1)}{2} \\ &= \frac{10^6 (10^5 \cdot (10-5-2+1))))}{2} \\ &= \frac{10^6 \cdot 10^5 \cdot 4}{2} \\ &= 2\cdot 10^{11} \end{align*} So the average value is \(\frac{2 \cdot 10^{11}}{4 \cdot 10^5} = \frac{10^6}{2} = 500\,000\). (Alternatively, each value can be paired off eg \(999\,999\) with \(1\) and so on, leaving averages of \(500\,000\)). Note that \(4197\) is divisible by \(3\) and \(7\). Using the same long we have: \(4179 - \frac{4179}{3} - \frac{4179}{7} + \frac{4179}{21} = 4179 - 1393 - 597 + 199 = 2388\). The sum will be: \begin{align*} S &= \underbrace{\sum_{i=1}^{4179}i }_{\text{all numbers}}- \underbrace{\sum_{i=1}^{1393}3i }_{\text{multiples of }3}- \underbrace{\sum_{i=1}^{597}7i }_{\text{multiples of }7}+ \underbrace{\sum_{i=1}^{199}21i }_{\text{mulitples of }21} \\ &= \frac{4179 \cdot 4180}{2} - \frac{4179 \cdot 1394}{2} - \frac{4179 \cdot 598}{2} +\frac{4179 \cdot 200}{2} \\ &= \frac{4179 \cdot 2388}{2} \end{align*} So the average value is \(\frac{4179}{2}\).

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Solution: \begin{align*} && a^2 &= d_1^2 + d_2^2 + d_3^2 \\ &&&= (x-x_1)^2+(y-y_1)^2 + (x-x_2)^2+(y-y_2)^2 + (x-x_3)^2+(y-y_3)^2 \\ &&&= \sum (x-\bar{x}+\bar{x}-x_i)^2 + \sum (y-\bar{y}+\bar{y}-y_i)^2 \\ &&&= \sum \left ( (x-\bar{x})^2+(\bar{x}-x_i)^2 + 2(x-\bar{x})(\bar{x}-x_i) \right)+ \sum \left ( (y-\bar{y})^2+(\bar{y}-y_i)^2 + 2(y-\bar{y})(\bar{y}-y_i) \right)\\ &&&= 3(x-\bar{x})^2 + \sum (\bar{x}-x_i)^2 + 6x\bar{x} -6\bar{x}^2-2x\sum x_i+2\bar{x}\sum x_i + \\ &&&\quad\quad\quad 3(y-\bar{y})^2 + \sum (\bar{y}-y_i)^2 + 6y\bar{y} -6\bar{y}^2-2y\sum y_i+2\bar{y}\sum y_i \\ &&&= 3(x-\bar{x})^2 + \sum (\bar{x}-x_i)^2+3(y-\bar{y})^2 + \sum (\bar{y}-y_i)^2 \\ \\ \Rightarrow && (x-\bar{x})^2+(y-\bar{y})^2 &= \frac13\left ( a^2- \sum \left((\bar{x}-x_i)^2+(\bar{y}-y_i)^2 \right) \right) \end{align*} Therefore the locus is a circle, centre \((\bar{x}, \bar{y})\). radius \(\sqrt{\frac13(a^2 - \text{sum of squares distances of centroid to vertices}})\). \(a^2\) cannot be less than this distance, because clearly the right hand side is always bigger than it!

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Solution:

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Solution:

1. Case \(b > 0\). Then \(bx+a\) is increasing and the greatest value is \(10b+a\), and the least value \(a-10b\) Case \(b=0\), then \(a\) is constant and the greatest and least value is \(a\) Case \(b < 0\), then \(bx+a\) is decreasing and the greatest value is \(-10b+a\) and the least value is \(10b+a\)
2. If \(c = 0\) we have the same cases as above. If \( c > 0\) the consider \(2cx+b\). if \(b-20c > 0\) then our function is increasing on our interval and the greatest value is \(100c+10b+a\) and the least value is \(100c-10b+a\) If \(20c+b < 0\) then our function is decreasing and that calculation is reversed. If neither of these are true, then the minimum will be when \(x = - \frac{b}{2c}\) and the max at one end point.

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Solution: \begin{align*} && y &= \sin(k \sin^{-1} x ) \\ &&y' &= \cos (k \sin^{-1} x) \cdot k \frac{1}{\sqrt{1-x^2}} \\ && y'' &= -\sin (k \sin^{-1} x) \cdot k^2 \frac{1}{(1-x^2)} - \cos(k \sin^{-1} x) \cdot k \frac{x}{(1-x^2)\sqrt{1-x^2}} \\ && (1-x^2)y'' &= -k^2y -xy' \\ \Rightarrow && 0 &= (1-x^2)y''+xy' + k^2y \end{align*} \begin{align*} && y &= Ax^3 + Bx^2 + Cx + D \\ && y' &= 3Ax^2 + 2Bx + C \\ && y'' &= 6Ax+2B \\ && 0 &= (1-x^2)(6Ax+2B) - x( 3Ax^2 + 2Bx + C) + 9(Ax^3 + Bx^2 + Cx + D ) \\ &&&= x^3(-6A-3A+9A) + x^2(-2B-2B+9B) + x(6A-C+9C) + (2B +9D) \\ \Rightarrow && B &= 0 \\ \Rightarrow && D &= 0 \\ \Rightarrow && C &= -\frac34 A \\ \\ x = 0, y = 0, y' = 0: && y &= 3x-4x^3 \\ \end{align*} And so \(\sin 3 x = 3 \sin x - 4\sin^3 x\)

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Solution: If \(T_1\) is the time they meet after \(\frac12\)km from the starting points and \(T_2\) is the time they meet a second time, then \begin{align*} && \frac12 &= vT_1 \\ &&&= V(T_1-\tfrac12) \\ && X - \frac54 &= vT_2 \\ && X + \frac54 &= V(T_2 - 1) \\ && \frac{T_2}{T_1} &= \frac{4X-5}{2} \\ && X + \frac54 + V &= VT_2 \\ && \frac12 + \frac12 V &= VT_1 \\ \Rightarrow && \frac{T_2}{T_1} &= \frac{4X + 5 + 4V}{2(1+V)} \\ \Rightarrow && \frac{4X-5}{2}&=\frac{4X + 5 + 4V}{2(1+V)} \\ \Rightarrow && V(4X-9) &= 10 \\ \Rightarrow && V &= \frac{10}{4X-9} \\ \\ && T_1 &= \frac{1}{2V} + \frac12 \\ &&&= \frac{4X+1}{20} \\ && v &= \frac{1}{2T_1} \\ &&&= \frac{10}{4X+1} \end{align*} \begin{align*} && 2X &= \frac12 V \\ \Rightarrow && 2X(4X-9) &= 5 \\ \Rightarrow && 0 &= 8X^2-18X-5 \\ &&&= (4X+1)(2X-5) \\ \Rightarrow && X &= -\frac14, \frac52 \end{align*} Since \(X\) is positive, we must have \(X = \frac52\)km

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Solution: The net force on the planets will always be towards the centre of mass (by symmetry or similar arguments). Therefore it suffices to check whether we can find a speed where the planets follow uniform circular motion, ie \(F = mr \omega^2\). (But clearly this is possible, we just need to find the speed)

\begin{align*} && F &= m r \omega^2 \\ && 2\frac{\gamma m^2}{a^2} \cos 30^{\circ} &= m \frac{a}{\sqrt{3}} \omega^2 \\ \Rightarrow && \frac{\sqrt{3}\gamma m^2}{a^2} &= \frac{ma \omega^2}{\sqrt{3}} \\ \Rightarrow && \omega^2 &= \frac{3\gamma m}{a^3} \\ \Rightarrow && \omega &= \left ( \frac{3\gamma m}{a^3}\right)^{1/2} \end{align*} In the second scenario, we are interested in when: \begin{align*} && F &= m r \omega^2 \\ && \underbrace{2\frac{\gamma m^2}{a^2} \cos 30^{\circ}}_{\text{to other symmetric planets}} + \underbrace{\frac{\gamma \lambda m^2}{a^2}}_{\text{central planet}} &= m \frac{a}{\sqrt{3}} \omega^2 \\ \Rightarrow && \frac{(\sqrt{3}+\lambda)\gamma m^2}{a^2} &= \frac{ma \omega^2}{\sqrt{3}} \\ \Rightarrow && \omega^2 &= \frac{(3+\sqrt{3}\lambda)\gamma m}{a^3} \\ \Rightarrow && \omega &= \left ( \frac{(3+\sqrt{3}\lambda)\gamma m}{a^3}\right)^{1/2} \end{align*}

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Solution:

1. Notice that if \(x > 0\) we must have \begin{align*} && \left ( \sqrt{x} - \frac{1}{\sqrt{x}} \right)^2 &\geq 0 \\ \Leftrightarrow && x - 2 + x^{-1} & \geq 0 \\ \Leftrightarrow && x + x^{-1} & \geq 2 \end{align*} Let \(S\) be the sum, and assume all probabilities are equal \begin{align*} && \mathbb{P}(S = 2) &= q_1 r_1 \\ && \mathbb{P}(S = 12) &= q_6 r_6 \\ && \mathbb{P}(S = 7) &= \sum_{i=1}^6 q_i r_{7-i} \\ \Rightarrow && q_1r_1 &= q_6r_6 \\ \Rightarrow && q_1r_6+q_6r_1 &\leq q_1r_1 \\ \Rightarrow && \frac{r_6}{r_1} + \frac{q_6}{q_1} &\leq 1 \\ \Rightarrow && q_1r_6+q_6r_1 &\leq q_6r_6 \\ \Rightarrow && \frac{q_1}{q_6} + \frac{r_1}{r_6} &\leq 1 \\ \Rightarrow && \frac{r_6}{r_1} + \frac{q_6}{q_1}+\frac{q_1}{q_6} + \frac{r_1}{r_6} &\leq 2\\ \text{but} && \frac{r_6}{r_1} + \frac{q_6}{q_1}+\frac{q_1}{q_6} + \frac{r_1}{r_6} &\geq 4 \end{align*} Since we have a contradiction they cannot all be equal.
2. We would like \(\displaystyle \sum q_i^2 \leq 1/36\) (subject to \(\displaystyle \sum q_i = 1\), clearly this cannot be true since: \begin{align*} && 1 &= \left ( \sum_{i=1}^6 q_i \right)^2 \\ &&&= \sum_{i=1}^6 q_i^2 + \sum_{i \neq j} 2q_i q_j \\ &&&\leq \sum_{i=1}^6 q_i^2 + 5\sum_{i=1}^6 q_i^2 \\ &&&=6 \sum_{i=1}^6 q_i^2 \\ \Rightarrow && \sum_{i=1}^6 q_i^2 &\geq 1/6 > 1/36 \end{align*} [For a weaker solution to the last part, notice that the largest value of \(q_i\) is \(\geq 1/6\) and therefore \(q_{max}^2 \geq 1/36\), but if equality holds then the other values must also be non-zero, and therefore the inequality cannot hold]

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Solution: There are \(N-1\) gaps between the magnets which are independently gaps or not gaps. Therefore the total number of gaps is \(X \sim Binomial(N-1, \frac12)\) and \begin{align*} \mathbb{E}(X) &= \frac{N-1}{2} \\ \textrm{Var}(X) &= \frac{N-1}{4} \end{align*}

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Solution:

1. \begin{align*} \log_{10}(\log_{10} a_1) &= \log_{10} (\log_{10} (x!) \\ &\approx \log_{10} (\log_{10} \sqrt{2 \pi} x^{x+\frac12} e^{-x}) \\ &= \log_{10} \l \log_{10} \sqrt{2 \pi} + (x+\frac12) \log_{10} x-x \r \\ &= \log_{10} \l \log_{10} \sqrt{2 \pi} + (100x+50)-x \r \\ &= \log_{10} \l 99x + \epsilon \r \\ &\approx \log_{10} 99 + \log_{10} x \\ &\approx 2 + 100 = 102 \end{align*}
2. \begin{align*} \log_{10}(\log_{10} a_2) &= \log_{10}(\log_{10} x^y) \\ &= \log_{10} y + \log_{10} \log_{10} x \\ &= x + 2 \end{align*} \begin{align*} \log_{10}(\log_{10} a_3) &= \log_{10}(\log_{10} y^x) \\ &= \log_{10} x + \log_{10} \log_{10} y \\ &= 100 + \log_{10} x \\ &= 200 \end{align*} \begin{align*} \log_{10}(\log_{10} a_4) &= \log_{10}(\log_{10} z^x) \\ &= \log_{10} x + \log_{10} \log_{10} z \\ &= 100 + \log_{10} y \\ &= 100+x \end{align*} \begin{align*} \log_{10}(\log_{10} a_5) &= \log_{10}(\log_{10} e^{xyz}) \\ &= \log_{10} x + \log_{10}y+\log_{10} z+ \log_{10} \log_{10} e \\ &\approx 100 + x + y \end{align*} \begin{align*} \log_{10}(\log_{10} a_6) &= \log_{10}(\log_{10} z^{1/y}) \\ &= \log_{10}(\log_{10} 10) \\ &= 0 \end{align*} \begin{align*} \log_{10}(\log_{10} a_7) &= \log_{10}(\log_{10} y^{z/x}) \\ &= \log_{10}z-\log_{10} x + \log_{10} \log_{10} y \\ &= y \end{align*} Since \(0 < 102 < 200 < x+2 < x+100 < y < y+x+100\) we must have \(a_6 < a_1 < a_3 < a_2 < a_4 < a_7 < a_5\)