1. Find \(x\) if \(x^{x^{x^{\ldots}}}=2\)
2. Find all real and complex root of \(x^6=64\)
3. The hour and minute hands of a clock meet at 12'oclock. When will be the first time they meet again ?
4. 3 points are randomly drawn on a circle. What is the probability of them being on the same semi-circle ?
5. A unit length is broken off into 3 pieces. What is the probability of them forming a triangle ?
6. Calculate \(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\ldots}}}} \)
7. There are 14 identical-looking balls. 13 of them have the same weight while one of them is heavier than the rest. What is the minimum times you can weight to identify the heaviest ball ? How do you generalize for n balls ?
1. Recurrence relations problem:
Let \(s_n = x^{s_{n-1}}\)
with \( s_1 = x. \)
Clearly, \(s_n \to 2\) as \(n \to \infty\). Thus \( 2 = x^2 \), which is based on the fact that \(a^y\) is a continuous function on \(y > 0 \) for any positive real \( a\) (most people ignore that we have used the implicit substitution property). Hence \( x = \sqrt 2 \).
2. In fact, we can geometrically see the points lying on a circle of radius 2, forming a 6-gon with vertices \( 2 e^{i\pi/3}, 2 e^{2i\pi/3}, 2 e^{4i\pi/3}, 2 e^{5i\pi/3}, 2, -2 \), the last two of which are real.
3. It is more like a physics problem in dealing with angular velocities and angular positions. Define \(x(t)\) and \(y(t)\) as the angular positions of the hour hand and the minute hand respectively, and clockwise movement as positive. Then
\( x(t) =t \frac{5}{60} \times 2\pi \times \frac{1}{3600} rad/s \)
and
\( y(t) =t \frac{60}{60} \times 2\pi \times \frac{1}{3600} rad/s \).
Now for the first time they meet together, \( y(t) - 2\pi = x(t) \), which implies \( t = 3600 s \times \large(1 + 1/11\right) \), which means they meet together at 13:05:27 or 1:05:27.
4. + 5. I don't think I can explain better than Bob.
6. Define \( x_n = \sqrt{2 + x_{n-1}} \) with \( x_1 = \sqrt{2}\). Then we see that \( x_1 < 2; x_2 = \sqrt{2+\sqrt 2} < \sqrt{ 2 + 2} = 2 \); by induction, we see that \( x_n < 2\) for all \( n =1,2,\cdots \). Also, \( \frac{x_n^2}{x_{n-1}^2} = \frac{1}{x_{n-1}} + \frac{2}{x^2_{n-1}} > \frac{1}{2} + \frac{2}{2^2} = 1 \) and so \( x_n > x_{n-1} \) for all \( n =1,2,\cdots \), since \( x_n > 0 \). Hence \( (x_n) \) is an increasing sequence with an upper bound 2, and thus converges. Now let the limit of such sequence be s, so that \( s = \sqrt{2+ s} \) (remember we know \(\sqrt{} \) functions are continuous on the set of non-negative real numbers and we have used the implicit substitution property), and the only solution to this is \( s = 2 \) (you can form it as a quadratic equation to get this).
7. Assume that you measure with a two-dish mass balance. We should observe that we always weigh even number of balls on the balance to compare the weight of both sides. I think it's a standard question.
But before solving this problem, we can try to see the results by induction.
Suppose \( n = 1 \). Then it doesn't need any measurement to see this is the heaviest one.
Suppose \( n = 2 \). Then it only requires one measurement to see one of them is heavier, and \( 1 \) is the smallest number because if \( 0 \) is the smallest (we don't measure at all), then we will realise that we cannot guarantee which is the heavier ball.
Suppose \( n = 3 \). By placing a ball each side on the balance, we see if both sides are of the same weight, then the remaining ball must be heavier; if both sides are not, then one ball on one side must be heavier. In conclusion, \( 1 \) is the smallest number because we cannot know which is heavier without measurement.
Suppose \( n =4 \). If we weigh only \( 2 \) balls (one ball each side) and these \( 2 \) balls are equal, we still have to determine the heavier ball by measuring the other two balls once more. So we should measure \( 4 \) balls altogether, and this eliminates two balls one side after the first measurement, and then obtain the heavier ball in the second. So in either case the minimum number of measurements is 2, and we are sure this is the minimum.
Thus we can hypothesise that when \( n =2^m \), where \( m \) is a non-negative integer, the minimum number of measurements undertaken is \(m\).
Now we consider the case for \( n = 14 \).
Take 4 balls on each side, so the 6 balls are not on the balance and 8 balls are on the balance.
1) if both sides are of the same weight, then the 6 balls contain the heavier ball. Weigh the 4 out of the 6 balls by placing 2 balls on each side:
a) if they are the same, then the other two balls would contain a heavier one, and we can just measure these two balls once more, (3 times)
b) if they are not the same, then one side of two balls contains a heavier ball, and so measuring once more determines the ball; (3 times)
2) if both sides are not of the same weight, then one side of 4 balls must be heavier, and so measuring two more times guarantees the heavier ball.
In conclusion, the minimum number is \(\max\{3,3,3\} = 3\) (why maximum is used because it is for the worst case).
In fact, we must remember how logic (falsification) allows us to cut down the number of measurements. So let's say we have \(n\) balls. In the first measurement, we weigh \(2^x\) number of balls, where \(x\) is the maximum integer such that \( 2^x \leq n \), for \(n \geq 1\) (It includes the equal sign; for example when \(n= 4\), when \( n= 2 )\).
Case 1): If these are of the same weight (this happens only when \(n - 2^x > 0)\), then the other \( n - 2^x \) balls (for such \(x)\) would have a heavier weight. Now we weigh \(2^y\) number of balls, where \(y\) is the max. integer such that \(2^y \leq n - 2^x\);
Case 2): if these are not of the same weight, we can weigh the \(2^{x-1}\) balls on the heavier side. Clearly, this only takes \((x-1)\) more times to find out the heavier ball so that \(x\) times of measurement will get the heavier ball.
In fact, for case 1), \( y + 1\leq x \); otherwise, if \( y + 1 > x \), then \( y + 1 \geq x + 1\), which implies \( y \geq x \) and so \( n \geq 2^y + 2^x \geq 2^{x+1} \), a contradiction since \( x \) is the largest integer such that \( 2^x \leq n\). So if the heavy ball lies in the \( 2^y \) number of balls are the same, then we can iterate to weigh at most \(y\) more times to guarantee the heavier ball; otherwise, the heavy ball lies in the \( n - 2^y \) number of balls. By induction and labelling \( x = x_1 , y = x_2 \leq x - 1, x_3 \leq x - 2, \cdots, x_m \leq x - (m-1) (m \leq x)\) etc., we can show that the minimum number of measurements is \(x\) where \(x\) is the maximum integer satisfying \( 2^x \leq n \), for \( n \geq 1\).
