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Oct 12, 2012

### Quant at Jane Street was asked...

Nov 22, 2010
 suppose you have a perfectly round disk. You put three legs randomly on this disk to form a table. Supposing the legs are perfectly perpendicular to the disk and are attached to the disk firmly, what is chance that the table will not fall when you flip the disk or in other word when you put the table to stand on its legs? 23 Answers3/4?2/91/4Show More ResponsesI got 1/4th, can you guys post how you got your answers?1/4 is correct1/2, if three legs cover > 180 degrees, the table can stand.heres what I did. I cut the circle into quarters and then said that the table will not stand if all 3 legs are in the same half of the circle. there is (1/2)^3 odds of putting all 3 legs in the same half, and 4 different halfs (top half, bottom half, left half, and right half) for this to happen. --> 1 - 4 * (1/2)^3 = 1/2 anyone see a problem with this?I don't really understand why you said there are 4 different halfs. You could have have cut the circle up into eights and then there are 8 halfs. You could have cut the circle up into 10000ths and then argued that there are 10000 halfs. The way I think you are supposed to do it is this. Place the first leg, it doesn't matter where you place it. Now, what is the chance that the 2nd and third leg are on the same half as the first leg. So, it's 2C2 * (1/2)^2 = 1/4. Chance of this not happening = 3/4. What am I missing?I think its 1/2 also. Same reasoning as the comment above, but I think the 1st AND 2nd leg can be anywhere, because there is no established "half" until there are 2 legs on the table. The only place the 2nd leg cant be is exactly 180 degrees away from the 1st (the probability of which is about 0, because 179.9 degrees in either direction is already fine). So the probabilty of all 3 legs being on the same half is 1 x 1 x 1/2Why not think the opposite from the beginning? What is the probability that it could not stand? It means that the legs are in the same half circle. 3*(1/2)*(1/2) = 3/4 Then the answer is 1/4Brett, why are you multiplying by 3...the odds that the three are on the same side is as follows 1 peg can go anywhere. 50% chance that the next peg is on the same side and another 50% chance that the 3rd peg is on the same side therefore shouldn't it be (1/2)*(1/2) = 1/4 chance that the table falls (aka all pegs on the same side) and there fore probability that the table stands is 1-1/4 = 3/4 so isn't 3/4 your final answern/360, where n is the angle of the 2 legsOk, here is mine, the first one is always not a problem, so let's jump to the 2nd stand In order to drop, the second must lie inside the half circle with the 1st one so, however, if you think of that in the clockwise and anti clockwise location, you will immediately discover that where ever it is, the 2nd is some how within the 180degree ranges. So, the problem leads to the third stand. So, if you can see the big picture, it is not hard to realized that there is only one safe zone for the 3rd stand, which is the opposite region of the one between 1st and 2nd (say 2nd located at n), so, the probability of it being dropped will be n/360 if using only integers => 1- n/360 being the right answer (to me). And if you sum up all the probabilities and combination, you will soon see that 1/2 is the right answer. Hope it can help!Show More ResponsesSorry, it is 3/4It's 1/2. 1/pi * integral from 0 to pi (x/pi)Fix a leg. Join this point to the opposite side of the disk to split the circle into half. If both legs lie in the same half it will topple. Effectively what is the probability of two coins the same in two tosses. We have HH,HT,TH,TT so 1/2. It will stand 1/2 of the time.The answer is 1/4. The answer is not 1/2.It's not 1/2, and it's not 1/4. Salalah had it right. Draw a circle. Draw the epicenter. Draw your first random leg (any point). Draw a line from it, through the epicenter, and to the circumference (your line will be longer than the radius). Draw a second leg (any point). Draw a line from it, through the epicenter, to the circumference. Your third leg needs to fall in the resulting pie piece. The probability of it falling there will be the the angle from point 1 to the epicenter to point 2, divided by 360 degrees.I believe it is the same with cutting the circle 3 times and probability that all of the pieces are smaller than 0.5* circumference because in my opinion, table will stand if the center of mass is inside of the triangle made by three cut. If one of the pieces is bigger than the 0.5, then the triangle could not capture the center so the answer is 1/4.1/4. here is how to get the answer mathematically: angles between every two legs: x,y,z then x+y+z=2pi and restrictions are 0 AO cuts the circle at A1 and A2, with A1 being on the other side of A w.r.t O 3. The 2nd leg is at B --> BO cuts the circle at B1 and B2, with B1 being on the other side of B w.r.t O 4. The 3rd leg is at C. In order that the table doesn't fall, C has to be in the pie OA1B1 5. Let the angle AOB be phi 6. The probability that the table doesn't fall is phi / 2pi 7. Since A and B are uniformly distributed within the circle, the expected value of phi is \int_{0}^{pi} phi * dphi = pi^2 / 2 8. --> the probability that the table doesn't fall is : pi^2 / 2 / 2pi = pi / 4 Am I wrong any where ?

### Intern at Jane Street was asked...

Mar 14, 2011
 If you had only 5 and 11 cent stamps. Whats the smallest number that would be impossible to make with those stamps.12 Answers49 i think i dont really remember now.I think you phrased that wrong budAnd I think it's 39Show More ResponsesCan anyone restate the question in a non-ambiguous manner?5*11-5-11=39 (Frobenius number)how bout 1?The problem seems to be the Postage Stamp Problem (look it up!). In this case, it is "what is the largest number which you cannot obtain by a combination of 5 or or 11 value postage stamps?" (implicit in the question is the fact that after a certain value, you can obtain every value by such a combination) As someone said above, it is 39 for these particular numbers. 40, 45, 55, etc. are all multiples of 5. 41 is 11 + 8*5, 42 is 2*11 + 4*5, etc. Basically once you can obtain 40-44, you can obtain 45-49 by adding 5, and then you can obtain 50 or higher by adding 10, and so forth.Why doesn't 17 work?starting with 11c, you can make 16c, 21c, 26c, etc by adding 5c each time, or all numbers congruent to 1 mod 5 is makeable. Do the same starting from 22c, and 33c. At this point, all numbers congruent to 0, 1, 2, 3, mod 5 are makeable. Once you start at 44c, all numbers congruent to 4 mod 5 are makeable, so the answer is the 4 mod 5 number below 44, which is 39.Well, isn't 39 the *largest* number the postage stamp can't make?I used a 15 by 15 Sieve and got 39 as the answer.The question is supposed to be: what is the largest number that cannot be generated by adding fives and elevens. To solve the problem, consider what is the minimum number with the last digit as 0, 1, 2,..., 9 that can be generated by adding fives and elevens: Last digit Minimum Achievable Maximum Unachievable 0 10 0 1 11 1 2 22 12 3 33 23 4 44 34 5 5 NA 6 16 6 7 27 17 8 38 28 9 49 39 Hence, maximum unachievable = 39

### Trader at Jane Street was asked...

May 2, 2012
 What is the sum of the digits of all the numbers from 1 to 1000000? This is different from the sum of the numbers. For instance the sum of the numbers from 1 to 10 is 55 whereas the sum of the digits is 46.16 AnswersThe main idea is that if you write all the numbers from 0 to 999999 down as six digit numbers (possibly prepending zeros) then all digits appear the same number of times. So, its digit appears exactly 6 x 1000000/10 = 600000 times. so the result is 600000x 45 +1 (+1 for the number 1000000)Maybe I'm reading it wrong, but isn't the sum of the digits of 1-10 just 11? 1 digit for 1-9, 2 digits for 10-99? So it's 9+90*2+900*3+9000*4+90000*5+900000+6no, I think the easy way to solve it in your head is to remember that when adding all digits 1 to 100, you have 50 pairs: 1+100, 2+99, etc. Each pair is 101, times 50 is 5050. 1 to 1000 would be 500500 so 1 to 1,000,000 would be 500,000,500,000 Pretty cool, huh?Show More Responsesscott, dude you should add digits not the numbers, so 99+2 = 18+2 =20. not 101Scott's answer from May 24 is the correct way to think about it if summing the numbers.The answer is 27,000,001 - if you do it programatically the operation is a simple map reduce - simply map a digit sum function across the list of values [1, 1000000] and then reduce an addition operator across the result. Python Proof: >>> sum(map(lambda n: sum(map(int, str(n))), xrange(1, 1000001))) 2700000121085156Don't mind the above answer. I read the question wrong.Each digit will appear 1+10+100+1000+10000+100000 times. so the answer is 111,111*45+1=4444440+555555+1=4999996nb is right. Another way you can think about it easily is if you want the digits from 1 to 1,000,000 then each digit should appear 1/10 * 1,000,000 times. So 100,000 x 6 (for each place value) x 45 (the sum from 1 to 9) + 1 (for the 1 million) 27,000,001I think John Doe is right.27,000,001 is what I got. Think of each number as a 6 digit number. The average number each digit could be from 000,000 to 999,999 is (9+0)/2=4.5. Since the average of each number is 4.5 and there are 6 digits the average sum of the digits for a 6 digit number should be 4.5*6=27. There are 1 million numbers from 000,000 to 999,999 so the sum of the digits from 000,000 to 999,999 is 27,000,000. Subtract the digits of 000,000 which is just 0 and add the digits of 1,000,000 which is just 1 to get 27,000,001.For arbitrary a=1,...,9 accounts the how many a's appear in the six digit number. 1 * C(6,1)*9^5 + 2* C(6,2)*9^4 + 3 * C(6,3) * 9^3 + 4*C(6,4)*9^2 + 5*C(6,5)*9 +6*C(6,6) = 600000. 600000*(1+2+3+...+9)= 27000000. Then add the 1 from 1000000. The ans is 27000000.Show More Responses2.7 x 10^7 + 1general rule: to 10^n answer is 10^n * n * 4.5+1 here n=6, result is 27 000 0011-1000000 same as sum of digits in 0 - 999999 then plus 1, treat 0 - 999999 as a 6 digit random number, then the digit of sum is sum of digit for each number: 1000000*(4.5*6)=27000000, so answer is 27000001.
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