Math-related Interview Questions - Page 3

my answer requires calculus/finite series

Ambiguity implies what?

The only knowledge you initially seem to have to go on is that the product of their ages is 72. The second clue seems meaningless. However, the final clue tells you in that one of the numbers is strictly smaller than the other two (which could just as easily be the same number as not). Since 72 isn't a perfect cube, then there's a limited number of answers it could be.

a * b * c = 72
a + b + c = D

Possible values are
==============
a = 24, b = 3, c = 1, D = 28
a = 18, b = 4, c = 1, D = 23
a = 12, b = 6, c = 1, D = 19
a = 12, b = 3, c = 2, D = 17
a = 9, b = 8, c = 1, D = 18
a = 9, b = 4, c = 2, D = 15
a = 8, b = 3, c = 3, D = 14
a = 6, b = 6, c = 2, D = 14
a = 6, b = 4, c = 3, D = 13

Because of the first clue, a must be less than 15, and because of the second clue, D must be greater than 15, which leaves us with:

a = 12, b = 6, c = 1, D = 19
a = 12, b = 3, c = 2, D = 17
a = 9, b = 8, c = 1, D = 18

So, the youngest daughter is either 1 or 2, but since a 1-year old is mostly still crawling around the floor, rather than playing piano, her age must be 2, which leaves the other daughter's ages as 3 and 12.

Although the poster above, ]343[, brings up a good point. Mathematician B would know the value of D, but the only reason to ask why they didn't know the answer after being given the second clue would be if there was ambiguity in the results, and that only occurs where D is equal to 14, so the only two outcomes are:

a = 6, b = 6, c = 2, D = 14
a = 8, b = 3, c = 3, D = 14

Since there's a youngest, the answer has to be 6, 6, 2.

This is equivalent to the number of increasing subparts (i'm sure there's a more technical term) of a permutation of length at least 2. It's probably easier to include all increasing subparts and then subtract the expected number of increasing subparts of length 1. When do those occur? when a number has a higher number before it and a lower one after it (or just the relevant one if it's the first or last). Thus there are an expected 2/2+(n-2)/4=(n+2)/4 increasing subparts of length 1. Now to count the total number of increasing subparts. Since my combo is awful I will attempt to biject and probably fail, but interesting problem!

If higher number=higher speed, each jam is started by a number for which the numbers on either side are both higher than it. (If the one on the left was lower it would not be the beginning of a jam, if the one on the right was lower it would not cause a jam at all.) So what's the probability for this to happen for any given number? For the n-2 middle numbers, the probability that out of any 3 adjacent ones the middle is the smallest is clearly 1/3. However, for the one on the right side it cannot start a jam, whereas for the one on the left the probability is 1/2. Thus we have (n-2)/3+1/2=(2n-1)/6.

1 + 1/2 + ... + 1/(n-1) + 1/n

If the slowest car is the kth car from the back it creates a jam behind it. In addition if f(n) is the number of expected jams, there are f(n-k) expected more jams in front of the slowest car. Since each of these is equally probable, we have f(n)=(f(n-1)+f(n-2)+...+f(1))/n+1. Thus nf(n)=f(n-1)+...+f(1)+n. Thus nf(n)-(n-1)f(n-1)=f(n-1)+1, so f(n)=f(n-1)+1/n. Induction trivially yields f(n)=H_n.

X_i = 1 if i-th slowest car is blocked by a jam, 0 otherwise
Y - number of jams (including 'single-car jams')

The following holds (P denotes probability):
P(X_i=1) = 1 - 1/i

explanation: look at subset {1,2,...,i-1} of i slowest cars. The chance that i-th
car has no slower car anywhere in front equals to the chance that i is at the first position of i-permutation, which is 1/i.

Observe that (E denotes expected value):
Y = N - (sum of EX_i)

Then the answer to the simplified problem is:
EY = N - (sum of EX_i) = N - (N - 1 - 1/2 - ... - 1/N) = 1+1/2+...+1/N = H_N

The answer is 7.

Divide your horses into 5 groups of 5. Race each group of 5. Let us call each group A, B, C, D, E with and the rank the horses 1, 2, 3, 4, 5. We then have:

A1, A2, A3, A4, A5
B1, B2, B3, B4, B5
C1, C2, C3, C4, C5
D1, D2, D3, D4, D5
E1, E2, E3, E4, E5

Now we race A1, B1, C1, D1, E1. Let's just say that the final result has A being the fastest, B coming second, C third, etc. Since these are just labels, you can relabel them with something else if you want.

We can eliminate everything in group D and E for obvious reasons. We also eliminate C2 to C5 as we can always form a group of 3 faster horses (e.g. A1, B1, C1). The same logic allows us to eliminate B3 to B5 and A4 and A5.

We have A1, A2, A3, B1, B2, C1 left. We know A1 is the fastest horse of them all. So race the remaining 5 and choose the two fastest from the rest.

I don't actually understand this question. Is each string only allowed through 1 hole?

The fact they don't burn uniformly doesn't matter.

Light string 1 at both ends at the same time and light string 2 at one end. String 1 will fully burn in 30 minutes. String 2 will be half burnt. When String 1 fully burns, light the non-lit end of string 2. String 2 will now burn for 15 more minutes on top of the 30 min it has already burnt.

The answer on Page 1 was correct.

The solution on Page 2 works, but you probably won't get out of jail until the universe ends because the amount of time it would take for everyone to be randomly selected and have visited the room and then have one person go back would take far too long.

I believe the most elegant solution would be this one which also covers every possible variation of the problem.

This a pretty well known question with many explanatory solutions so I won't type one up here.