If you work in finance you probably have some aptitude for mathematics. Some of you will even have an exceptional mathematical talent. It’s worth remembering, therefore, that $6m is still on the table from the Clay Mathematics Institute’s Millennium Prize. Over 14 years after the prize was announced, just one of the seven problems has been solved. Each problem comes with a prize of $1m. What better way to compensate for falling compensation in finance?

The problems are listed on the Clay Institute’s website, here. Only the last one – the Poincaré Conjecture, has been solved. Grigori Perelman, the reclusive Russian mathematician who found the solution famously turned down the dollars, stating that he wasn’t interested in money or fame and didn’t want people staring at him, “like an animal in a zoo.”

Self-evidently, solving any of the other Millennium Prize questions will take a lot longer than a few weekends between now and bonus time. The Poincaré conjecture was first postulated in 1904 and it took 98 years before Perelman solved it 2002. The Clay Institute then took another four years to validate his proof, making the payment deferral far worse than any bonus – assuming Perelman actually wanted to get paid.

If you’re disinclined to spend the next 2,000 weekends tackling the 150-year old Riemann hypothesis or the Hodge Conjecture in pursuit of a million dollars, you can always spend a few moments this weekend solving the far easier problems below for the hell of it. Add your answers to the comments box at the bottom of this page for peer review. The first two are comparatively easy. The second two are comparatively harder.

1. You are making chocolate chip cookies. You add N chips randomly to the cookie dough and you randomly split the dough into 100 equal cookies. How many chips should go into the dough to give a probability of at least 90% that every cookie has at least one chip? (Source: Heard on the Street, Timothy Falcon Crack)

2. You have two string-like fuses. Each burns in exactly one minute. The fuses are inhomoegenous, and may burn slowly at first. the quickly, then slowly, and so on. You have a match and no watch. How do you measure exactly 45 seconds? (Source: Heard on the Street, Timothy Falcon Crack)

3. Isaac is planning a nine-day holiday. Every day he will go surfing, or water skiing, or he will rest. On any given day he does just one of these three things. He never does different water-sports on consecutive days. How many schedules are possible for the holiday? (Source: British Maths Olympiad 2013)

4. Every diagonal of a regular polygon with 2014 sides is coloured in one of n colours. Whenever two diagonals cross in the interior, they are of different colours. What is the minimum value of n for which this is possible? (Source: British Maths Olympiad 2014).

## Comments (6)

For question 2 you simply have to burn one fuse at both ends (giving you 30 seconds) and also burning one end of the other fuse (all at the same time – pretty tricky I imagine). After the first fuse has burnt out, you immediately light the remaining end of the second fuse to give you a further 15 seconds, totalling 45 seconds.

For Question 3: Counting or combination should follow this trend. Each surfing can be followed by surfing or rest. Similarly for water skiing. For 9 days the combinations are as follows:

Day 1 2 3 4 5 6 7 8 9

S 1 2 5 12 29 70 169 408 985

W 1 2 5 12 29 70 169 408 985

R 1 3 7 17 41 99 239 577 1393

Total possibility: 3363

Question 3: Every step would require combination with the following principle. Water Skiing or Surfing can result only from either “Rest” or “Water Skiing” and “Surfing” respectively in the previous day. So during addition for Surfing or Water Skiing: add the possibilities of Water Skiing or Surfing with “Rest” from the previous day.

Day 1 2 3 4 5

S 1 2 5 12 29

W 1 2 5 12 29

R 1 3 7 17 41

and so on…

Total: 3363 possibilities.

@Noice Actuary, your solution works provided the string-like fuses are homogeneous. The problems statement says they are inhomogeneous so potentially it might take 59 seconds for one-half of a fuse to burn and then just 1 second for the remaining half. So you can’t light the fuse from both ends and measure 30 seconds when it burns out.

Question 1:

For 1 cookie the number of ways the chips can fall are 1

For 2 cookies are (N+1)/1! ways N chips can fall

For 3 cookies are (N+1)(N+2)/2! ways N chips can fall

For 4 cookies are (N+1)(N+2)(N+3)/3!

… For 100 cookies are (N+1)(N+2)*…*(N+99)/99! ways

Then to calculate all possibilities which have at least 1 chip:

(N-m+1)(N-m+2)*…*(N-1) because initially you put at least 1 chip in each cookie and then you calculate the ways the remaining (N-m) chips can be put on the cookies. (m is the number of cookies)

So:

((N-99)(N-98)*…*(N-1)/99!) / ((N+1)(N+2)*…*(N+99)/99!) = .9

N = 93965 (rounded to the nearest upper integer).

Question 2: @Nan, Noice Actuary solution works. The fact that the fuses are non homogeneous doesn’t affect the fact that they will burn in half the time when lighted in both ends at the same time. It just affects where the burning will end. In your example for the first half burning in 1 second then maybe the fuse will end burning at the beginning of the first half but still in 30 seconds. You are seeing time as distance.

Imagine this fuse: dividing the fuse in non homogeneous parts totaling 60 seconds:

1,2,3,10,3,2,9||5,1,1,10,1,1,11 => the fuse ends burning where marked with || in 30 seconds

Now imagine this other fuse:

30||1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1 => the fuse will end burning where marked with || but still in 30 seconds