Random questions from my math class

 #4

By induction...
Base case is true (2x2) as you showed.

Assuming for a 2^k x 2^k board, you can fill it leaving out one corner of the board:
Create a 2x2 array of those boards, where 3 have the corner facing the centre and 1 has it facing out:

Code:

BB|B0
B0|BB
-----
B0|0B
BB|BB

Place another L into the whole in the middle and you have a 2^k+1 x 2^k+1 with 1 corner piece missing.
Induction complete.

 #6

If 7 numbers are right, the 8th can't be wrong. So 0.

 #11

I don't see what the paradox.
There are 2 equally likely options for a given revealed card Y (assuming all equal sized ranges of real numbers are equally likely - that is, uniform)
2/Y, Y
Y, 2*Y
It could be considered payment for risk - if you stay, you have a 100% chance of keeping your value. If you switch you could lose it, or you could profit.

 #12

I'm assuming you can take coins out of the jars to weigh them.
If not it's kinda trivial - weigh each jar, find whether they are 500g or not, and done with it. If you weigh more than one at once, then you exceed the 600g limit.

Anyway, assuming you can take them out...
Weigh coins from 7 jars.

• None fake: Weigh 2 of the other jars
  • None fake: Weigh the last jar
    • Not fake: none are fake
      
    • fake: You know it's this jar
  • One of the 2 is fake: Weigh one.
    • Not fake: It's the other one.
      
    • Fake: You know it's this jar
• Fake in here: Weigh 4 of the 7.
  • None fake: Weigh 2 of the remaining jars
    • None fake: It's the other one.
      
    • One is fake: Weigh one of the jars.
      • If it's fake, you know it.
        
      • If it's not fake, it's the other one.
  • One is fake: Weigh 2 of these jars.
    • If one is fake, weigh one of them.
      • If this one is fake, done.
        
      • If this one is not fake, it's the other one.
    • If none is fake, weigh one of the other 2.
      • If it's fake, done.
        
      • If it's not fake, it's the other one you didn't weigh.

After 3 weighings you only have 8 possible outcomes (2^3) so you can't always determine which one is fake.

 #13

After much derivation

Code:

1 - 1 way

= 1

2: 3: 4: 5: (...) 14: 15 - 1 = 14 ways
            (...)     1 - 1
         1 - 1
      1 - 1
   1 - 1

= sum(1 to 14) 1


3: 4: 5: 6: ... = 13+12+11+...+1 = (13+1)*13/2 = 91 ways
         2: 6: (...) 15 = 11
            1 - 1
      2: 5: 6: (...) 15 = 12
            1 - 1
         1 - 1
   2: 4: 5: 6: (...) 15 = 13
            1 - 1
         1 - 1
      1 - 1

= sum(i = 1 to 13) sum (j = 1 to i) 1

4: 5: 6: 7:
         3: 7:
            2:
      3: (...) = (11+1)*11/2 = 66 ways
   3: 5: 6: 7: = (12+1)*12/2 = 78 ways
            2: 7:

         2: 6: 7:

      2: 5: 6: 7: ... = 12 ways
            1 - 1
         1 - 1

= sum(i = 1 to 12) sum (j = 1 to i) sum (k = 1 to j) 1

Basically
sum (i = 1 to K) sum ( sum( sum ( ... (15-K times) 1)))) = K'th ball
This works out to be
1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
= 16384

This number is familiar to me (2^14) and makes me wonder what the trick is to rearrange them so they're easier to add up.

 #20

4 square feet.