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Parker Whitfill's avatar

I can solve all the questions easily

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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.

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amphilomath'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)?

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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.

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amphilomath's avatar

Well, that is an interesting education market mechanism design…

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