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Parker Whitfill's avatar

I can solve all the questions easily

Kris Shaw's avatar

The math question is fun. Seems like there are two kinds of (i,j)-pairs to look for.

(1) Those that leave only segments that consist of consecutive elements in multiples of 4.

m=1 : (1,2) (1,6) (5,6)

m=2 : all those from before, and also (1,10) (5,10) (9,10)

m=3: all those from before, and also (1,14) (5,14) (9,14) (13, 14)

So the number of ``good'' combos of this sort is

1+2+ ... + m + (m+1) = (m+1)(m+2)/2 = (m^2 + 3m + 2)/m

(2) Those that allow for `skipping sequences'.

m=2 : removing (2,9) allows for arithmetic sequences (1 3 5 7) and (4 6 8 10)

m=3: removing (2,13) allows for arithmetic sequences (1 4 7 10) , (3 6 9 12) and (5 8 11 14)

BUT also, (2,9) still works and so does (6 13), which is just (2 9) shifted over by 4.

The removing pairs of the skipping type are therefore:

m=2 : (2,9)

m=3: (2,9) (6,13) and (2 13)

m=4: (2,9) (6,13) (10,17) and (2 13) (6 17) and (2 17)

So the number of ``good'' combos of this sort is

1+2+ ... + (m-2) + (m-1) = (m-1)m/2 = (m^2 - m)/m

Overall, the number of ``good'' combos is

(m^2 + 3m + 2)/m + (m^2 - m)/m = m^2 + m +1

Now, the total number of combinations is

nC2 = n! / 2!(n-2)! = (n-1)n/2 = (4m+2)(4m+1)/2 = 8m^2 + 6m+1

Therefore, the probability of drawing a ``good'' pair (i,j) is

Pm = (m^2 + m +1) / (8m^2 + 6m+1) > (m^2 + m +1) / (8m^2 + 8m+8) = 1/8

QED.

Jack Huang's avatar

This map is so bizarre to me. Why is the exam harder for the lower Yangtze valley, arguably the most educated part since antiquity (at least Song dynasty)?

Cheryl Wu's avatar

I think precisely because it’s rich and educated-they have the best resources so that the exam is harder to adjust for that. Otherwise they will provide too many students to the top schools.

Jack Huang's avatar

Well, that is an interesting education market mechanism design…