# Quantitative Interview questions and answers

#### Duane Rich

This is my crack at number 4. I think it is 3/4.

Place the numbers one at a time. The first one can go anywhere. The second one needs to be within a half circle from that point, so it can also go anywhere.

The placement of the last point is the real question. Well if the first two points landed right on top of each other, then the last point can also go anywhere on the circle and be within one half circle of the other two points (100% chance). However, if the first two points are on opposite sides of the circle, then the last point must be placed on a defined half circle, so it has a 50% chance of meeting that criteria.

This is my leap of faith; As I consider the extreme cases, I suspect the actual probability to be right in the middle of 100% and 50% --> 75%.

#### Duane Rich

Number 6 is 2.

Let a = sqrt{2+sqrt{2+sqrt{ ...
Then a^2=2+a, which has 2 as the only positive solution.

#### Amir Yousefi

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 ?

--> I think there is a slight discreepancy here .... the solutions to question 4 and question 5 here must be related as ans Q5 = 1 - ans Q4

the way i think about it is as .... we know that 3 randomly choosen line segments will form a triangle IFF the sum of any 2 sides is greater than the 3rd side....

Think of choosing points on a circle as breaking up a line segment --- choosing 3 points on a semi circle gives you 3 line segments..... if 3 of them lie on the same side of a semi circle ==> the sum of 2 of them is smaller than the third one => they cannot make a triangle....

I liked your answer and I think you are of those natural genius people in danger of extinction by nerds!

You are right. both probablities are 50%

#### Amir Yousefi

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)(x = \sqrt{2})
2)2
3) 60/11 minutes after 1

4) You can assume that you are choosing an inscribed angle on the circle. the angle range from 0 to 180 degrees. all the angles smaller that 90 degrees represent 3 points on the same semi-cirle ( the end point and other two point that the angle hit the circle represent those 3 points)
the probability of choosing an angle smaller that 90 out of 180 is 50%
5) assume that the longest piece is called x. If the x is longer than 1/2 it is not possible to make the triangle. other wise you have made a triangle with 3 pieces ( Note that x is the longest so if it is less than 1/2 others would be so)
The probability that x not to be larger than 1/2 is 50%, so in 50% we have our triangle!
6) x=2 ( x^2=x+2 => x=2 , x=-1(unacceptable))

#### Juan Sebastian Ovalle

For #6 is 2 right?

#### yatici

Two fun ones that are both pretty doable:

1. An airplane has N seats, and N passengers are waiting to board it, not in any particular order. Miraculously, everyone is assigned to a different seat on the airplane; however, the first passenger to board is a jerk and selects a seat at random. Thereafter, passengers board one at a time according to the following rule: If his or her assigned seat is vacant, the passenger sits there; otherwise, the passenger selects a vacant seat at random.
What's the probability that the last passenger to board gets his or her assigned seat?

2. We have two concentric circles. A chord of the larger circle is tangent to the smaller circle and has length 8. What's the area of the annulus--the region between the two circles?

I didn't thoroughly solve it but a quick look resembles a simple solution if you realize recursion. If first passenger sits at a random seat given it is someone else's we will move to the point where we initially started. So last step you have 2 seats. The 1st person sit randomly so the last passenger has a 50% chance of sitting on his seat as the first guy is equally likely to sit on either. Let me know if I am wrong and I will actually try to calculate it.

#### yatici

2. .5

5. .5

2 can be proved by taking the differential. ( dN(W_t)=\frac{1}{2 \pi} e^ {\frac{-w^2}{2}} dW_t + \frac{-2w}{2 \pi} e^ {\frac{-w^2}{2}} dt )
( N(W_t)=.5+Ito + \int_0 ^t \frac{-2w}{2 \pi} e^ {\frac{-w^2}{2}} dt )
( \mathbb{E} [N(W_t)]= .5+ \int_0 ^t \mathbb{E}[\frac{-2w}{2 \pi} e^ {\frac{-w^2}{2}}] dt )
( = .5+ \int_0 ^t \int_{-\infty} ^ {\infty} \frac{-2w}{2 \pi} e^ {\frac{-w^2}{2}} e^{\frac{-w^2}{2}} dw dt )
(=.5 )

For 5, the price of the option is the solution to ( \mathbb{E} [e^{-r \tau}] ) where ( \tau ) is the first time the stock hits 40. This is simply the laplace transform of ( \tau ), which is ( e^{-(\sqrt{2r+a^2}-a)m} ) (if I assume B.M. with drift). However, since I assume the stock follows R.N. GBM with constant coefficients, I have to rewrite as follows:
( S_t = 20 e^{\sigma W_t +(r-\frac{1}{2}\sigma^2 )t} )
( \frac{S_t}{20} = e^{\sigma W_t +(r-\frac{1}{2}\sigma^2 )t} )
( log\large(\frac{S_t}{20}\right) = \sigma W_t +(r-\frac{1}{2}\sigma^2 )t )
( \frac{1}{\sigma}log\large(\frac{S_t}{20}\right) = W_t +(\frac{r}{\sigma}-\frac{1}{2}\sigma)t )

Therefore in the laplace transform equation ( m=\frac{log\frac{40}{20}}{\sigma} ), ( a=\frac{r}{\sigma} -\frac{1}{2} \sigma ). After completing the square this becomes

( e^{-(\frac{r}{\sigma} +\frac{1}{2} \sigma -\frac{r}{\sigma} +\frac{1}{2}\sigma) \frac{log(2)}{\sigma}} )

( =e^{-log(2)} =.5 )

for the 1st one you don't actually need to integrate things. This is a simple question.

E[Phi(W_t)] = Phi(E[W_t])
since martingale Phi(0) = 0.5 Done

5. Most likely lacks information. This needs to be P(S_T >=40| S_0 = 20). if there is no timelimit, it will definitely reach this value. So you actually need to use a inverse gaussian function to find the expected time to cross this point and discount wrt to the risk free percentage. Which will be its price. I doubt people would ask this, they probably specify a time frame to see if you can calculate:

Phi(d1) with a strike 40

which is just a basic Digital option

#### Profundus

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 ?

I got:
1. (sqrt{2})
2. real root: 2, -2
complex roots: (1 + i*sqrt{3}, 1 - i*sqrt{3}, -1 + i*sqrt{3}, -1 - i*sqrt{3})
3. 1 hour 5 minutes 27.27seconds
4. 1/2
5. 1/3
6. 4
7. For 14 balls you can do it in 3 if your weighting the balls against themselves on a balance scale, for any n it will take log base 3 of n times minimum.

#### Davis Tang

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 $$n \geq 1$$.

#### Bimark

No, the answer to this is $$x=\sqrt{2}$$

$$a=x^{x^{\ldots}}=2$$

$$ln(x^{x^{\ldots}}) = ln2$$

$$x^{x^{\ldots}}lnx = ln2$$

$$a lnx = ln2$$

$$2lnx = ln2$$

$$lnx = \frac{1}{2}ln2$$

$$lnx = ln\sqrt{2}$$

$$x = \sqrt{2}$$

$$x = \sqrt{2}$$

I believe we can use such a logic only if the sequence has a limit.
In other case if I use the same logic to equation:

$$a=x^{x^{\ldots}}=4$$

I will get:

$$ln(x^{x^{\ldots}}) = ln4$$

$$x^{x^{\ldots}}lnx = ln4$$

$$a lnx = ln4$$

$$4lnx = ln4$$

$$lnx = \frac{1}{4}ln4$$

$$lnx = ln4^ \frac{1}{4}$$

$$x = 4^ \frac{1}{4}$$

$$x = \sqrt{2}$$

Thus we have:

$$x^{x^{\ldots}}=2$$ if $$x = \sqrt{2}$$
and
$$x^{x^{\ldots}}=4$$ if $$x = \sqrt{2}$$

So I believe the equation has no solution in real numbers. May be it has solution in complex numbers...

#### Bimark

and here is the solution uploaded which uses geometric probability:

#### Attachments

• solution4.JPG
33 KB · Views: 13

#### Davis Tang

$$x = \sqrt{2}$$

I believe we can use such a logic only if the sequence has a limit.
In other case if I use the same logic to equation:

$$a=x^{x^{\ldots}}=4$$

I will get:

$$ln(x^{x^{\ldots}}) = ln4$$

$$x^{x^{\ldots}}lnx = ln4$$

$$a lnx = ln4$$

$$4lnx = ln4$$

$$lnx = \frac{1}{4}ln4$$

$$lnx = ln4^ \frac{1}{4}$$

$$x = 4^ \frac{1}{4}$$

$$x = \sqrt{2}$$

Thus we have:

$$x^{x^{\ldots}}=2$$ if $$x = \sqrt{2}$$
and
$$x^{x^{\ldots}}=4$$ if $$x = \sqrt{2}$$

So I believe the equation has no solution in real numbers. May be it has solution in complex numbers...

Read the post just before your reply. $$\sqrt 2$$ is one real solution.

According to my method, if you solve the quadratic equation, you may find $$-\sqrt 2$$ being a solution as well, but, hey, we made a premise for the quadratic equation that $$a^x$$ for positive real $$a$$ is continuous, which is the fact that guarantees the (implicit) direct substitution property; most of us have forgotten, or have not learnt, how the limits of the images of sequences under a function actually work (it's based on the continuity of the function at the limit point of the sequence). Yet the continuity of $$a^x$$ for negative real $$a$$ at any $$x$$ is no way guaranteed (in fact it can be discontinuous on the complex-number set), so you "may" not be allowed to use the direct substitution property.

Now are pretty sure that $$\sqrt 2$$ is one real solution, but $$-\sqrt 2$$ may be another solution (only two solutions are possible, according the algebraic equation $$x^2 = 2$$ without considering sequences); if so, it may be true by some sort of extension. But, directly, you can try to substitute it, and see if the sequence $$\{s_n\}$$, with $$s_1 = -\sqrt 2$$, (defined in my first post) can have a limit point $$2$$.

In fact, for two minus root 2's,
$$(-\sqrt 2)^{-\sqrt 2} = \left[\sqrt 2 e^{i\pi}\right]^{-\sqrt 2} = \left[\sqrt 2^{\sqrt 2} \times e^{\sqrt 2 i \pi}\right]^{-1}$$;
and then for three minus root 2's,
$${(-\sqrt 2)^{(-\sqrt 2)}}^{(-\sqrt 2)} = \left[\sqrt 2^{\sqrt 2} \times e^{\sqrt 2 i \pi}\right]^{-1\times (- \sqrt 2)} = {\sqrt 2^{\sqrt 2}}^{\sqrt 2} \times e^{2 i \pi} ={\sqrt 2^{\sqrt 2}}^{\sqrt 2}$$;
by induction, for the subsequence $$\{ s_{2n-1}\}$$, the limit is $$2$$; for the subsequence $$\{ s_{2n}\}$$, the limit is $$\frac{1}{2} e^{-\sqrt 2 i\pi} \neq 2$$ (the sequence $$\{s_n\}$$, with $$s_1 = -\sqrt 2$$, diverges). Hence $$x = - \sqrt 2$$ is not the solution.

I believe this is just an interview problem, so getting the solution $$x= \sqrt 2$$ is very good enough; the rest is just way too much and heavy.

#### Bimark

Hi, Davis

try to solve $$x^{x^{x^{...}}} = 4$$ by your method.

It seems the answer will be the same $${\sqrt2}$$ . Am I correct?

#### Davis Tang

Hi, Davis

try to solve $$x^{x^{x^{...}}} = 4$$ by your method.

It seems the answer will be the same $${\sqrt2}$$ . Am I correct?

Hi Blmark,

I think the point you raise is very mind-provoking, yet illusionary.

Right, the sequence now seems to have two limits, and so this sequence does not converge by your seemingly good example.

In fact, there is a flaw in your argument, Bob's and mine as well. Why? Let's look into the following logic and arguments.

If you just test my defining recurrence relation numerically (as shown in the following figure), you will find that the sequence $$\{a_n\}$$:
1) looks to approach $$2$$, and
2) seems to have an upper bound $$2$$.

Now let's prove these two observational points.

First, $$a_1 = \sqrt 2 > \sqrt 1 = 1$$, and so by induction we see all $$a_n > 1$$.

Now notice that $$a_1 = \sqrt 2 < 2$$, $$a_2 = \sqrt 2^{\sqrt 2} < \sqrt 2^2 =2$$, $$a_3 = \sqrt 2^{a_2} < \sqrt 2^2 = 2$$.

Quickly, by induction, we realise all $$a_n < 2$$ and thus $$1< a_n < 2$$.

Moreover, $$a_2 = \sqrt 2^{\sqrt 2} > \sqrt 2^1 = a_1; a_3 = \sqrt 2 ^{a_2} > \sqrt 2^{a_1} = a_2$$.

By induction, we see $$a_n > a_{n-1}$$. So $$\{a_n\}$$ is an increasing sequence with an upper bound $$2$$, and thus converges to some real number $$r \leq 2$$.

We are not done yet. In order to show the limit of the sequence $$\{a_n\}$$ is $$2$$, we must show that $$2$$ is the least upper bound of the set $$\{a_n: n \in \mathbb N\}$$. In fact, you can claim the contrary that for some $$\delta > 0$$, $$2 - \delta$$ is the least upper bound; however, as you quickly calculate there is always some big index $$N(\delta)$$ such that $$a_{N(\delta)} \geq 2 - \delta$$ and thus there is no such positive delta, and so $$2$$ must be the least upper bound. That way, you can guarantee that $$\{a_n\}$$ converges to $$2$$ and so $$x = \sqrt 2$$.

Furthermore, for your proposed problem that $${x^x}^{\cdots^x} = 4$$, you can see that $$x=\sqrt 2$$ is impossible and thus rejected, because we can see the sequence constructed that way only converges to some real number not greater than $$2$$. Consequently, the solution you found by my method is not really the solution. So what's wrong? We have to check where the limit lies in.

In conclusion, the divergence is just illusionary, and we have found a method to certify the convergence of such a sequence and the existence of the $$x$$.

#### Bimark

New member here. Just wanted to throw in my 2 cents.
This is a misleading question, because the presumed true answer is actually false. Play sudoku lately?

[Edited to add a solved Sudoku puzzle]
6 9 1 5 4 7 2 3 8
3 8 4 2 6 1 9 7 5
5 7 2 3 8 9 4 1 6
8 3 6 9 7 5 1 4 2
7 1 9 6 2 4 5 8 3
2 4 5 8 1 3 6 9 7
9 6 7 1 3 2 8 5 4
4 5 8 7 9 6 3 2 1
1 2 3 4 5 8 7 6 9

Agree with you.

Here is some easier example:
1 2 3
3 1 2
2 3 1

#### Bimark

Nice solution Davis.

It is also interesting to find maximum x for which the sequence

$$x^{x^{x^{...}}}$$

converges.

#### Davis Tang

Nice solution Davis.

It is also interesting to find maximum x for which the sequence

$$x^{x^{x^{...}}}$$

converges.

Hey Bimark,

More importantly, the real concern that we have is that when we modify this problem as follows, most non-rigorous people will get tricked:

Find the solution, or solutions, to the equation $$x^{x^{\cdots^x}} = a$$ where $$a\geq 1$$.

Clearly, when $$a= 2$$, we get a solution $$x = \sqrt 2$$; when $$a = 4$$, we get no solution ( we try $$\sqrt 2$$ cannot be one, certainly if, extensively thinking of $$x^4 = 4$$ as just a polynomial to solve, we get 4 roots; however, $$-\sqrt 2$$ is neither possible as shown in the above, and in fact $$+/- i\sqrt 2$$ wouldn't be either because it will make $$\{s_n\}$$ form a divergent sequence).

Now we raise a general concern of people's over-trivialising (or overlooking, or cheating to solve) the problem as follows:

We can write $$x^{x^{\cdots^x}} = a \Longrightarrow x^a = a$$, and so $$x = ^a\sqrt a$$ is one such solution. A problem is that we have known that when $$a = 4$$, one who uses this kind of argument, or the one including taking logarithm, fails without being aware of the fact that algebraically solving an equation can be illusionary and different to approaching it with the sequential consideration like what we have here. It's needless to say for a more general case. Instead of taking the unnecessarily redundant steps to ensure $$x = \sqrt 2$$ for $$a = 2$$ like what I did last time, would we have a better and faster elementary way to accept or reject the solution $$^a \sqrt a$$. It's a real interesting point and study. Of course, from the practical point of view (we are no longer in an ivory tower solving it classically without a computational tool), we will take out any numerical tools to check it instantly. But how do we develop a theoretical sense to hand the solutions or estimation before taking a computational tool? I'd like to hear any smart ideas.

By the way, I have just found the following websites on which it describes what the minimum and maximum values of $$x$$ can be, so the maximum that $$x^{x^{\cdots^x}}$$ can reach is $$\exp$$ and the minimum, surprisingly needed, is $$\exp^{-1}$$.

Hyperpower Function
Tetration - Wikipedia

#### Batman MFE

I know that this is kind've a n00b question, but how important are these questions in interviews? My background is in applied econ and stats so I've never come across these more mathematical challenge questions. Is there a book/resource that I should consult just for these types of questions?

#### Andy Nguyen

I know that this is kind've a n00b question, but how important are these questions in interviews? My background is in applied econ and stats so I've never come across these more mathematical challenge questions. Is there a book/resource that I should consult just for these types of questions?
These are staples of quant interviews. I don't know anyone who has never got asked these type of questions in any of their interviews for "quant type" jobs. If you have a MFE degree, you are expected to ask them regardless of what the role is.
Here is the list of master reading list where we have a section on quant interview books

#### Batman MFE

These are staples of quant interviews. I don't know anyone who has never got asked these type of questions in any of their interviews for "quant type" jobs. If you have a MFE degree, you are expected to ask them regardless of what the role is.
Here is the list of master reading list where we have a section on quant interview books
Thanks for the response, Andy. I appreciate it.

I've been perusing the reading list but I didn't realize that they would include quantitative questions in the interview guides. I'll get to work on these!

Replies
1
Views
6K
Replies
2
Views
97K
Replies
2
Views
6K
Replies
11
Views
9K
Replies
24
Views
18K