- Jonathan Ediyanto - 7 pts (correct number, 840; flawed solution/explanation)
- Joao Sousa - 3 pts (answered 960, only 28 divisors)
- Sarah DeSanctis - 1 pt - honorable mention
- Joao Sousa - 3 pts (answered 960, only 28 divisors)
- Joao Sousa -
10 pts (first correct answer, good explanation)
- Peter Williams - 7 pts (Dec.1)
- Sarah DeSanctis - 5 pts - (later that day)
- Hoodin Hamidi - 3 pts - (just over the wire)
- Peter Williams - 7 pts (Dec.1)
- does it take to add up to more than pi ?
- Sarah DeSanctis - 10 pts (first correct answer, good explanation)
- Peter Williams - 6 pts (1 pt penalty for too-brief expl and wrote 3/24 not 3/23))
- Joao Sousa - 2 pts - (partly late, no part c)
- Peter Williams - 6 pts (1 pt penalty for too-brief expl and wrote 3/24 not 3/23))
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- Sarah DeSanctis - 10 pts (nice job, have some chocolate milk
on me!)
- Joao Sousa - 2 pts - (too many assumptions made)
- Peter Williams - 1 pt (see, you don't need the house number)
- Joao Sousa - 2 pts - (too many assumptions made)
- Sarah DeSanctis - 5 pts (1: good job, 2:..., 3: ok but generalize)
- Joao Sousa - 4 pts - (good job on #2)
- Joao Sousa - 4 pts - (good job on #2)
- Sarah DeSanctis - 10 pts (credit for "it's 163")
- Joao Sousa - 5 pts - (answered first, nice counting method, but off by 1. Missed it by that much!)
- * The rules are 7 pts for the second
correct answer, so I'm being generous. *
- Joao Sousa - 5 pts - (answered first, nice counting method, but off by 1. Missed it by that much!)
- Answer: 21 (add two numbers to get the next - "Fibonacci Sequence")
- Answer: 13 (the next prime number; there are 25 primes under 100 and 168 under 1000.)
- Answer: 3 (the number of letters in "one, two, . . . , ten")
- Answer: 63 (each one is 1 less than the next power of 2 ; an = 2^n - 1
- Answer: 46,656 (1^1, 2^2, 3^3, . . . , 6^6)
- Answer: 5040 (the factorials; 1!, 2!, . . . , 7!, where e.g.
4! = 4*3*2*1 = 24.
- n! is the product of the first n natural
numbers.
- Or, we could say "times 2, times 3, times 4, etc.")
- n! is the product of the first n natural
numbers.
- Answer: 2310 (the product of the first n-1 primes,
- or, "times 2, times 3, times 5, times 7, times 11")
- Joao Sousa -
6 pts - (Got 4 of 7 sequences perfect - good try on 2 of the
other 3.)
- Leo Bach - 1 pt - (For at least asking a good question!)
- Leo Bach - 1 pt - (For at least asking a good question!)
- Answer: Each person clinks with all others,
so for 4 people it's 3 + 2 + 1 = 6 clinks.
- For 5 it's 4 + 3 + 2 + 1 = 10, and we can use the binomial coefficient nC2 to do each one. People:Clinks: 4:6, 5:10, 6:15, 7:21, 8:28, 9:36, 10:45.
- Answer: We want nC2 = 903 clinks, let's answer (c) first and come back!
- Answer: c(n) = nC2
= n(n-1)/2 = (n^2 - n) / 2 is "n choose 2",
- which can be gotten from Pascal's Triangle or from the arithmetic series
- n-1 + n-2 + . . . + 3 + 2 + 1 = n(n-1)/2 [ there are n/2 pairs that add to n-1 ]. Now to finish #8b...
- nC2 = n(n-1)/2 = 903, (n^2 - n) = 1806, n^2 - n - 1806 = 0, (n - 43)(n + 42) = 0 ,
- n = 43 or -42, so there were 43 people at the clinkfest.
- which can be gotten from Pascal's Triangle or from the arithmetic series
- Joao Sousa -
5 pts - (Be more thorough - and 42 was close but no cigar with
your wine.)
- Answer: 9^a = (3^2)^a = 3^(2a) = (3^a)^2 =
(4)^2 = 16;
- Let's find b: 4^b = (2^2)^b = 2^(2b) = 8 = 2^3 , so 2b = 3 ; b = 3/2.
- So 9 ^ (a - b) = (9^a) / (9^b) = 16 / (9^b) = 16 / (9^(3/2)) = 16 / (3^3) = 16/27.
- Let's find b: 4^b = (2^2)^b = 2^(2b) = 8 = 2^3 , so 2b = 3 ; b = 3/2.
- Answer: Sure, 4^2 = 2^4 =
16, but did you know
there is a whole curve of (a,b) in the ab-plane with a^b = b^a
?! Use an implicit plot to graph x^y - y^x = 0 ; it goes through
any (a,a); also (2,4), (4,2), and a host of other points such
as (2.478,3) approx.
- Check this : 2.478^3 :=: 3^2.478. (The :=: means "approx equal")
- An answer of "no" might be considered correct, if you couldn't find the numbers!
- Check this : 2.478^3 :=: 3^2.478. (The :=: means "approx equal")
- Answer: 4^6 - 6^4 = 4096 - 1296 = 2800. Note a $ b is often positive if a < b. Is it
always?
- Is there a simple condition for when a $ b > 0 ?
- Answer: No. Here's
a proof that $ is "not associative" :
- 2 $ (3 $ 4) = 2 $ (3^4 - 4^3) = 2 $ (81 - 64) = 2 $ 17 = . . .
- . . . = 2^17 - 17^2 = 16384 - 289 = 130783 , while
- (2 $ 3) $ 4 = (2^3 - 3^2) $ 4 = (8 - 9) $ 4 = (-1) $ 4 = . . .
- . . . = (-1)^4 - 4^(-1) = 1 - 1/4 = 3/4 . That's very different!
- 2 $ (3 $ 4) = 2 $ (3^4 - 4^3) = 2 $ (81 - 64) = 2 $ 17 = . . .
- Andy Murdock -
8 pts - First answer was 16/27 , not your approx value of 0.593.
- Joao Sousa - 4 pts - Your 9a) worked out to be 0.4103, and 9c) wasn't finished.
- Joao Sousa - 4 pts - Your 9a) worked out to be 0.4103, and 9c) wasn't finished.
- Andy Murdock - 9 pts - You got it and explained how, but didn't prove it had to be smallest.
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