Let \(n\) be an odd number such that some power of 2 leaves remainder 1 on division by \(n\). Show, by considering the sequence of remainders of \(1, 2, 2^2, \ldots\) on division by \(n\), that there is a number \(m < n\) such that \(2^k - 1\) is divisible by \(n\) if and only if \(k\) is divisible by \(m\). If \(2^n - 1\) is divisible by \(n\), show that \(2^m - 1\) is divisible by \(m\). Deduce that for no number \(n\) greater than 1 is \(2^n - 1\) divisible by \(n\).
\(ABC\) is a triangle, whose angles are \(3\alpha, 3\beta, 3\gamma\). Points \(P, Q, R\) interior to the triangle are such that \begin{align} \angle PBC &= \beta, \quad \angle PCB = \gamma,\\ \angle PBC &= \beta, \quad \angle CRQ = \frac{1}{3}\pi + \beta,\\ \angle PCQ &= \gamma, \quad \angle BPR = \frac{1}{3}\pi + \gamma. \end{align} The points \(H\), on \(AC\), and \(K\), on \(AB\), are such that \(\angle QHC = \frac{1}{3}\pi + \beta\), \(\angle RKB = \frac{1}{3}\pi + \gamma\). Prove (i) that the triangle \(PQR\) is equilateral, (ii) that \(A, K, R, Q, H\) lie on a circle, and (iii) that \(AR, AQ\) trisect the angle \(A\).
The number of hours of sleep of a group of patients was recorded. On a subsequent night the patients were each given a sleeping pill and the number of hours of sleep was again recorded. The results were as follows:
If \(x_1, x_2, \ldots, x_n\) is a random sample from the uniform distribution with density function \(f(x) = 1/\theta\), \(0 < x < \theta\), where \(\theta\) is an unknown parameter:
Solution:
Ten different numbers are chosen at random from the integers 1 to 100. If the largest of these is divisible by 32 find an expression for the probability that the smallest is divisible by 24.
Solution: The largest can be, 32, 64, 96. \begin{array}{c|c|c} \text{largest} & \text{smallest} & \text{number in between} & \text{choices} \\ \hline 32 & 24 & 7 & \binom{7}{8} = 0 \\ 64 & 24 & 39 & \binom{39}{8} \\ 64 & 48 & 15 & \binom{15}{8} \\ 96 & 24 & 71 & \binom{71}{8} \\ 96 & 48 & 47 & \binom{47}{8} \\ 96 & 72 & 23 & \binom{23}{8} \\ \end{array} Therefore the probability that both the largest number is divisible by 32 and the smallest by 24 can be given by: The probability the largest is divisible by 32 is \begin{align*} && \mathbb{P}(\text{largest divisible by }32) &= \frac{\binom{31}{9} + \binom{64}{9} + \binom{96}{9}}{\binom{100}{100}} \\ \Rightarrow && \mathbb{P}(\text{smallest divisible by }24|\text{largest divisible by }32) &= \frac{\binom{39}{8}+\binom{15}{8}+\binom{71}{8}+\binom{47}{8}+\binom{23}{8}}{\binom{31}{9} + \binom{63}{9} + \binom{95}{9}} \end{align*}
In a game between two players both players have an equal chance of winning each point. The game continues until one player has scored \(N\) points. Find the probability \(p_r\) that the winning player has a lead of exactly \(r\) points when the game is completed. Deduce that $$(2N - r - 1)p_{r+1} = 2(N - r)p_r \quad (r = 1, 2, \ldots, N),$$ and hence find the expected value of the lead at the end of the game.
A uniform solid circular cylinder of radius \(a\) and mass \(M\) has, rigidly attached to the cylinder, a uniform rod of length \(a\) and mass \(M\). The rod lies in a plane perpendicular to the axis of the cylinder, passes through the axis and has one end on the circumference of the cylinder, as illustrated in the diagram. The cylinder rolls on a rough horizontal table with the rod just overhanging the edge of the table. The system is released from rest with the rod horizontal. By energy considerations, show that, when the rod is inclined at an angle \(\theta\) to the downward vertical, the angular velocity \(\dot{\theta} = d\theta/dt\) is given by $$a\dot{\theta}^2(19 - 8\cos\theta) = 8g\cos\theta.$$ Find the frictional force between the cylinder and the table as a function of \(\theta\) (not involving the time derivatives of \(\theta\)), and show that it vanishes at a value of \(\theta\) between \(\cos^{-1}0.8\) and \(\cos^{-1}0.4\).
A smooth wedge of mass \(M\) rests on a smooth horizontal plane. The sloping face of the wedge makes an acute angle \(\alpha\) with the horizontal. A particle of mass \(m\) is dropped from a point vertically above the centre of mass of the wedge, so that its velocity immediately before impact is \(u\). The coefficient of restitution for impacts between the particle and the wedge is \(e\). Immediately after the \(n\)th impact, the velocity of the wedge is \(U_n\), the component of the velocity of the particle parallel to the sloping face of the wedge is \(V_n\), and the component of the relative velocity of the particle and wedge perpendicular to the sloping face of the wedge is \(W_n\), all measured so as to have positive values. Find \(U_1\), \(V_1\) and \(W_1\), and show that \begin{align} (M + m\sin^2\alpha)U_n &= m(V_n\cos\alpha + W_n\sin\alpha),\\ V_{n+1} &= V_n + 2W_n\tan\alpha\\ \text{and} \quad W_{n+1} &= eW_n. \end{align}
A small ring of mass \(m\) is placed around the midpoint of a rough uniform rod \(AB\) of mass \(M\) and length \(2l\). The coefficient of friction between the ring and the rod is \(\mu\). The rod is pivoted at its end \(A\) so as to be free to swing in a vertical plane, but the ring is such that this plane is constrained to rotate about the vertical through \(A\), pivoted at constant angular velocity \(\omega\). If \(\mu\) is sufficiently large that the ring does not slide along the rod, then a position of equilibrium will exist in which the rod makes an angle \(\theta > 0\) with the downward vertical through \(A\). Assuming the validity of the use of `centrifugal forces', find the moment about \(A\) of the centrifugal forces acting on the rod in this equilibrium situation. Hence show that $$\omega^2 = \frac{3(M + m)g}{4Ml + 3ml}\cos\theta.$$ Show also that, for the ring not to slip in this position, one must have $$\mu\sin 2\theta - \cos 2\theta \geq 7 + (6m/M).$$ Find the minimum value of \(\mu\) for which this has a solution for \(\theta\). What is the value of \(\theta\) in this case? [It may be assumed that the rod has circular cross-section and that the ring sits loosely around it. There is thus only one point of contact between the rod and the ring, and the direction of the reaction at that point lies in the vertical plane through the rod.]