Edmundo Braverman:

LOL. I'll take the easy one:

3 (412)/(21)=24

This reminds me of that "24" game we played in middle school.. You could also do (12 - 4)*(2+1)

zbb:

#6 211

If I haven't been tricked:

if you count upwards, every ten numbers has a no ending in 5 a no ending in 2

multiply these, and it'll and it'll create a number ending in 0

every ten also has a number ending in 0

except, 9 of those will end in two 0s and one in three

therefore, each ten that you cunt will add two more 0s, but for the exceptions

there are 100 tens in 1000

plus nine and two more for the "exceptions"

nope, sorry.

Magua:

Question #2 (4) You have a digital clock which either displays a time A:BC:DE or AB:CD:EF, depending on the hour. Sometimes, the time displayed will be a palindrome. What is the shortest distance between two palindrome times? The longest distance?

Assumptions if the format is A:BC:DE there is still a 0 where A would be in the character time.

Shortest: 12:00:21 from 11:55:11 etc. = 5:10 unless we can use 01:00:10 from 12:55:21 = 4:49

Longest: 12:55:21 to 10:00:01 = 9:04:40 unless we can use 05:55:50 to 10:00:01 = 4:04:11

Wrong, note these are consecutive times so that might change your answer for the longest. Your shortest is definitely wrong. Also, to clarify, 05:55:50 is NOT palidromic since it will display as 5:55:50. Also, 0:30:30 still displays the 0.

zbb:

Isn't the shortest 9:59:59 to 10:00:01?

Correct, now find the longest. :D

edtkh:

#5: This is akin to the coin flip question where you estimate the expected returns of flipping a coin k times until you land a head. The expected return in this case is essentially the mean of the geometric progression which yields 1/0.5=2.

3: Another solution = 12/2/1*4=24

Your #5 is correct. Congrats.

edtkh:

#4: 8/(3-(8/3)) = 24.

PS: I wouldn't put this at difficulty level 2.

That's correct. They are all difficult questions (well except Eddies) by layman standards. But the spread is necessary since some of the 4s and 5s are REALLY hard. #1 is pretty difficult, for example, but it is still a 2 compared to the 3s,4s,5s in my estimation.

It is difficult for me to judge how hard the 1000! question is, because I had seen the 100! factorial question so 1000! was an easy extension. I found the 8 8 3 3 question pretty straightforward.

edtkh:

The trailing 0's for the 1000! question is certainly way beyond a 3 in my book. As a matter of fact, believe it or not, I have yet to get a correct answer from the candidates I have thrown that question to.

To be fair though, I personally don't think that's as important (nevermind if it's hard; the crux really is that most people fail to see the pattern or identify the root of the problem from which one should use to tackle the question) to the job itself as simple probability questions like #5.

Well, nobody has solved a 4 or a 5 yet. We only have a partial solution for one of the 4s. So who is to say that the 4s or 5s aren't harder than the 1000! question?

Erebus:

What is the angle formed by the hands of the clock when it is 1:45?

The difference between 270 degrees and (360/12)*1.75 degrees is 142.5 degrees

ivoteforthatguy:

Question 9:

My solution is: you can make 2X at the most. I believe the key word here is "guaranteed" which I take to be "riskless." And I am assuming that I am picking my strategy before the game begins.

I would wait until the last card in the deck, by which time I know the color with 100% certainty. Then I bet it all.

That's a nice strategy, and yes you double. But that's not the correct answer, there's a better guaranteed result.

For q9: Maybe I am getting tangled in semantics, but I was looking for the guaranteed return. After the first card is dealt, you immediately have information about the odds in the next draw and you make a Kelly bet. But making any Kelly bet up until the last card is a risky strategy.

If I had to solve this on my feet, I would make a probabilistic argument. Assuming that I had to pick my strategy before the game begins, I could say: if I wait until two cards were left in the deck, there is a .5 chance that it's same-color and .5 that it's mixed. If it's mixed, I have a strategy to win 2X risk-free (i.e., wait a card). If it's same, I have a strategy to win 4X. So the EV is 3X. I can go on inductively and go above 3X, etc.

marketimpact:

Q9: this can't be done by hand and requires coding:

• start of with one dollar

• let D(b,r) be the best you can get with b black and r red cards

• boundaries: D(0,r) = 2^r //bet all each time and double up D(b,0) = 2^b

• situation is symmetrical: D(x, y) = D(y,x)

• and always bet on the color with the most cards remained

• so if r >= b

• x is fraction of money to gamble D(r,b) = max_x min{ won bet, lost bet} = max_x min { (1+x) D(r-1,b) , (1-x) D(r, b-1) }

• i.e. first not-obvious value T(2,1) get T(1,1) = 2 and then T(2,1) = max min { (1+x)2, (1-x)4}

and the min is maximized when the lines intersect at x = 1/3 => T(2,1) = 8/3

You have the right approach. It's pretty easy to get the exact answer with excel, which the monkeys on this board should know how to use.

qtrade:

#1. Expected turns: 11.85 = 8/5 + 7/4 + 6/3 + 5/2 + 4/1. Think of throwing dice.

7. You need 10 seconds. Use conditional expectations and the fact the cube only has 4 distinct corners.

10. exp(log(3)/3) - to see it take log() on both sides

All correct. Good job.

LetsGoSailing:

#24: 135 degrees... oops, just saw thisdude already got it...

Also that's wrong.

physconomist:

#19 (3,3,3,3,3,1), maybe

Nope. Note: I never said they had to be integers.

qtrade:

#17. All the x should be equal to 16/n. To see this take log(x1x2 ... * xn) and consider what happens when you perturb any of the x in the sum of log(x).

18. same as 17 - replace 16 with k.

The wording on that was confusing. You CHOOSE n. I'll edit to clarify.

qtrade:

#17. All the x should be equal to 16/n. To see this take log(x1x2 ... * xn) and consider what happens when you perturb any of the x in the sum of log(x).

18. same as 17 - replace 16 with k.

if this is the case (and I believe it is), and (16/n)*n =16 must be true, then the maximum is (16/n)^n for n a whole number. Which I get at (16/6)^6 = 359.5939643347

monkeysama:

#17 A: 2^8. The proof is left as an exercise for the reader.

This is very incorrect, monkey.

Pfalzer:

#8. 32 seconds?

You must be misreading the question. The answer for #8 should be less than or equal to the answer for #7, which was 10. #8 is not asking how long until all corners have been traversed.

jhconnel:

#18.

You have to maximize (k/n)^n. Taking the derivative you get (ln(k/n)-1)(k/n)^n setting this equal to 0, ln(k/n) = 1 so n = k/e. So the max for a given k equals e^(k/e), minding rounding errors if you want n to be an integer.

I'll accept this. The correct solution is:

k/e+1 >= n >= k/e - 1. Find all integers in this range, and test which one produces a higher value. This is because n must be an integer, and k/e is not an integer.

zbb:

#28 I'm sure there are innumerable ways to solve. Mine is somewhat crude:

EVERY string of nine consecutive integers is divisible by 9

x + (x+1) + (x+2) ... (x+8) = 9x + 36

Therefore, the sum of two or more of these 9-integer strings back-to-back must also be divisible by 9

There are 999,999,999 integers with 9 digits or less That means a whole number of these 9-integer strings (111,111,111 of them) consecutively, starting with 1+2...+9

each of them are divisible by 9, as is their sum

Correct, although technically there are 1,000,000,000 positive integers with 9 digits or less. Good job.

jhconnel:

#23: 2520

31: 12.25

32: 6,12 both pr = 1/9

33. 30, Ecash = 1.667

23 is wrong, the others are correct. Good job on 31-33.

jhconnel:

Sorry meant 2521

Are you sure about that?

acljnle0:

wait does question 40 have to do with the eigenvalue of the transition matrix?

I don't give hints. I only clarify definitions (e.g. integer or not).

vectorkid:

#8 - 81/14 seconds.

My explanation may be hard to follow, so draw a picture of cube if that's the case.

There are 4 distinct pairs of opposite corners. When you've been to one of the corners of every pair, you have either gone in a square or in a x-y-z plane (with origin as the fourth corner). Once you've gone in a square, solve x = 1+2/3x to get x = 3 expected seconds until you hit a "furthest corner from any corner you've been to." Similarly, once you've gone in a x-y-z plane, solve x = 1 + 1/3(1 + x) to get x = 2 expected seconds.

When you've been to one of the corners of three distinct pairs (i.e. the 3 corners form vertices of an isosceles right triangle), you are either on the right angle vertex or one of the two 45 degree vertex. Let u be the expected number of seconds when you're on a 45 degree vertex and v be the expected number of seconds when you're on the 90 degree vertex. Then solve u = 1 + 1/3v + 1/3(3) and v = 1 + 2/3u + 1/3(2) to get u = 23/7 and v = 81/21. Note the 3 and 2 comes from above paragraph.

Suppose you've gone to one of the corners of two distinct pairs. Then solve x = 1 + 2/3(23/7) + 1/3x to get x=67/14. Hence, answer is 1 + 67/14 = 81/14.

I don't think that's right because your final answer isn't a factor of three (or have the denominator a factor of 3).

vectorkid:

#8 - 81/14 seconds.

My explanation may be hard to follow, so draw a picture of cube if that's the case.

There are 4 distinct pairs of opposite corners. When you've been to one of the corners of every pair, you have either gone in a square or in a x-y-z plane (with origin as the fourth corner). Once you've gone in a square, solve x = 1+2/3x to get x = 3 expected seconds until you hit a "furthest corner from any corner you've been to." Similarly, once you've gone in a x-y-z plane, solve x = 1 + 1/3(1 + x) to get x = 2 expected seconds.

When you've been to one of the corners of three distinct pairs (i.e. the 3 corners form vertices of an isosceles right triangle), you are either on the right angle vertex or one of the two 45 degree vertex. Let u be the expected number of seconds when you're on a 45 degree vertex and v be the expected number of seconds when you're on the 90 degree vertex. Then solve u = 1 + 1/3v + 1/3(3) and v = 1 + 2/3u + 1/3(2) to get u = 23/7 and v = 81/21. Note the 3 and 2 comes from above paragraph.

Suppose you've gone to one of the corners of two distinct pairs. Then solve x = 1 + 2/3(23/7) + 1/3x to get x=67/14. Hence, answer is 1 + 67/14 = 81/14.

This is the right approach, but you have errors so it is incorrect. You're close.

jhconnel:

Sorry read the problem wrong #23 : 2519

34: ((n+1)/2)^j

Good job.

vectorkid:

Why is #33 6 or 12. You're expected to win 1/9 (12) if you pick 12, but you're expected to win 2/36 (30) if you pick 30, which is a greater EV.

33 IS 30, I think you misread the solutions.

breakinginnew:

27: Every number 1-N will be assigned so probability that none match up is (n-1)/n * (n-2)/n ...* 1/n = (n-1)(n-2)...(1)/(n^n)= (n-1)!/(n^n)

The answer is 1-that...which is basically just 1.

naw, that's not right.

vectorkid:

#27: 1-1/e

1 - P(nobody is assigned the correct x_i) = 1 - (1-1/n)^n = 1- 1/e as n goes to infinity. Recall from compounded interest (1+r/n)^n = e^r as n goes to infinity.

good job.