- .problem #251 - posted friday, august 25, 2006
- > Nice Dice Twice < (back to top)
- a) Using a pair of standard dice (each labeled 1, 2, 3, 4, 5, 6), what
- are the probabilities of rolling each total from 2 through 12 ?
- b) Is it possible to relabel the dice with positive integers and get
- the very same table of probabilities, for the same totals ?
- > > If so, show how . . . if not, prove it can't be done. < <
- ("relabel" means use different lists of numbers, not just permute the 1 thru 6)
- Show your steps and reasoning. (Time's up on this one)
- Solution: Part a) is a pretty standard result; part b) was more controversial. Some of you sent
- in 'proofs' that the standard dice were the only solution, while others found a new pair that work!
- Mark Rickert wrote in first, and I loved his and others' use of factored polynomials! - Dan
a) 2: 1/36 - 3: 1/18 - 4: 1/12 - 5: 1/9 - 6: 5/36 - 7: 1/6 - 8: 5/36 - 9: 1/9 - 10: 1/12 - 11: 1/18 - 12: 1/36
(Denominators 36, numerators 12345654321.)
b) (1, 2, 2, 3, 3, 4) and (1, 3, 4, 5, 6, 8) - I've solved this before using trial and error. I also happened to- find this cool solution: The probability distribution for standard dice can be generated by powers of
- d(x)=(x+x^2+x^3+x^4+x^5+x^6)^n. The coeff of x^k gives the frequency out of 6^n of rolling k.
Two dice: (x+x^2+x^3+x^4+x^5+x^6)^2 = x^2+2x^3+3x^4+4x^5+5x^6+6x^7+5x^8+4x^9+3x^10+2x^11+x^12
An alternate factorization of the polynomial is (x+x^3+x^4+x^5+x^6+x^8)*(x+2x^2+2x^3+x^4)
The first factor represents a die w faces (1, 3, 4, 5, 6, 8). The other factor, faces (1, 2, 2, 3, 3, 4).- Here is Ed W's list of how to get each roll and associated probabilities:
Total How to Roll # of ways Probability 2 (1,1) 1 1/36 3 (1,2)(1,2) 2 2/36 = 1/18 4 (1,3)(1,3)(3,1) 3 3/36 = 1/12 5 (1,4)(3,2)(3,2)(4,1) 4 4/36 = 1/9 6 (3,3)(3,3)(4,2)(4,2)(5,1) 5 5/36 7 (3,4)(4,3)(4,3)(5,2)(5,2)(6,1) 6 6/36 = 1/6 8 (4,4)(5,3)(5,3)(6,2)(6,2) 5 5/36 9 (5,4)(6,3)(6,3)(8,1) 4 4/36 = 1/9 10 (6,4)(8,2)(8,2) 3 3/36 = 1/12 11 (8,3)(8,3) 2 2/36 = 1/18 12 (8,4) 1 1/36
- WINNERS - Problem 251 . . please vote for dansmathcast on podcast alley! (back to top) . leader board
- * Some contestants got an extra point for responding early, albeit incorrectly... - Dan
- Mark Rickert. . . . . . . . .10 pts - Wow I liked this idea of factoring generating polynomials, good job!
- Nick McGrath . . . . . . . . 7 pts - Also a fine factoring fellow, I'm glad I got you hooked on partitions!
- Ed Wern. . . . . . . . . . . . . 5 pts - Ed did this problem 25 yrs ago, cut the new dice from blocks of wood!
- Oliver Schebanek . . . . . 4 pts - Good job part a); also right that b) can't be diff't if kept to 1, 2, ..., 6.
- Denis Borris . . . . . . . . . 4 pts - Right you are about the new dice; I liked your joke and bumper strip!
- K Sengupta . . . . . . . . . . 4 pts - Very nice answer using polynomials wisely, thanks for your response
- Marcello Cammarata . . 3 pts - How loyal; on a distant vacation and only 2 days "late" with answer!
- Zahi Teitelman . . . . . . . 3 pts - Correct on prob dist and second pair of dice; no proof of uniqueness ok.
- Kirk Bresniker . . . . . . . 3 pts - Thanks for the link to an excellent page containing this problem!
- Jeremy Galvagni. . . . . . 3 pts - Yes, a) is standard high school, b) Sicherman dice visit JG's Stat class!
- Ken Duisenberg . . . . . . 3 pts - Another link to a great page - this problem apparently gets around!
- Hermen Jacobs . . . . . . . 3 pts - Good work for part a), early ans ; turns out b) is possible even same totals
- Etienne Desclin. . . . . . . 3 pts - Nice type layout for a) ;-} but part b) used an illegal zero; see above
- Al Nelson. . . . . . . . . . . . 3 pts - I liked your reasoning on part b) you got close to the answer above.
- Art Morris . . . . . . . . . . . 2 pts - Part a) no problem; part b) assumed there needed to be 6's but not so.
- Garry Malashkin . . . . . 2 pts - Ok for first part; but I'm disallowing the use of a simgle 36-sided die.
- Nats Kroy. . . . . . . . . . . . 2 pts - True if there has to be a 2 on each die then there is no other solution.
- Claudio Baiocchi . . . . . 2 pts - Again there doesn't need to be a 6 on each. And no zeroes but nice try.
- Abhijit Parashar . . . . . 2 pts - Glad to see you back; good list for a); see above soln for part b).
- Phil Sayre . . . . . . . . . . . 2 pts - Hello POP user; aren't you also a grandPOP? Good notation for part a).
- Adam Panagos . . . . . . . 2 pts - Nice try Adam; a) ok but b) you fell for the split 2's trap giving std ans.
- Sara Elsnick (new) . . . . 2 pts - Welcome; nice logic try; hope you can enter again when there's time!
- Allen Druze. . . . . . . . . . 2 pts - Another fan of 12xxx and 12xxx leading only to the std solution
- Keith Downey (new) . . . 2 pts - Thanks for finding my site; good start on part b); see above solution
- Ameya Joshi (new) . . . . 1 pt - A ninth-grader from the other side of the world; very cool, nice try!
- Michael Thwaites (new) 1 pt - Good mix of table and list; right-o on part b) (After 252 up so just 1 pt)
- Giridhar Prasannan. . . 1 pt - Was marked #252 in subject line, after 253 up, but good answer!
- problem #252 - posted sunday, september 10, 2006
- Nearly Multiples ! next-to-last problem of the 2005-06 contest (back to top)
- a) Prove that the square of any odd number is always 1 more than a multiple of 8.
- b) Prove that the square of any odd prime > 3 is always 1 more than a multiple of 24.
- c) If p is prime and is 1 more than a multiple of 4, prove there's always a perfect square
- that's 1 less than a multiple of p. Show your steps and reasoning. "Numbers, primes" are positive
- integers. You may quote famous theorems.
- Solution: We serve up a combo plate of a) TimP, b) EtienneD, and c) GarryM: (Also got some
- great answers from Hermen J, Marcello C, K Sengupta, Al N, 9th grader Ameya J, and others!)
- (a) Prove that the square of any odd number is always 1 more than a multiple of 8. For every x^2,
- there is an x which is odd. x-1 is even, as is x+1. For each pair of consecutive even numbers, one
- of them must be divisible by four, so either x-1 or x+1 (it doesn't matter which one) is so divisible.
- The product of a number divisible by two and one divisible by four will be divisible by eight.
- (x-1)*(x+1) is, of course, x^2-1, one less than the product [square]. Q.E.D.
- (b) Every prime number can be written 3n+1 or 3n-1 (because every prime number is equal to
- either 1 or 2 modulo 3) . . . (3n$1)2 = 9n2 $ 6n + 1 which is 1 mod 3 ($ = + or -)
- So for every prime p, p2-1 is divisible by 3. By (a), for every odd prime p, p2-1 is divisible by 8
As 3 and 8 are relatively prime to each other, for every odd prime p, p?-1 is divisible by 3*8=24
So p2 = 1 (mod 24) or p2 = 24k+1. T.I.J. (That's It Jack ;-)- (c) We need the Wilson's theorem: http://primes.utm.edu/notes/proofs/Wilsons.html
Let's say p = 4*n+1 is prime. By theorem: (4*n)! = -1(mod p).- We have (4*n)! =1*2*...*(2n-1)*2n*(2n+1)*(2n+2)...*(4n-1)*4n
- . . . . . . . . . . . . . =1*2*...*(2n-1)*2n*(p-2n)*(p-(2n-1))*...*(p-2)*(p-1)
. . . . . . . . . . . . . =1*2*...*(2n-1)*2n*( A*p + (-1)^(2n)*2n*(2n-1)*...*2*1)- . . . . . . . . . . . . . = (2n)! * (A*p + (2n)! ) = (2n)!*A*p + ((2n)!)^2 ; so ((2n)!)^2=-1 (mod p) QED
- Al N. also partitioned {1, 2, ..., p-1} into sets {a, -a, 1/a, -1/a} mod p, leaving 2 besides 1 and -1. Cool!
- WINNERS - Problem 252 . . please vote for dansmathcast on podcast alley! (back to top) . leader board
- Marcello Cammarata. . 10 pts - MC is back on top this week and seems a lock for the 2005-06 title!!
- Garry Malashkin . . . . . 7 pts - Nice job on first parts, and then pulling out the Wilson's (not sports gear)
- K Sengupta . . . . . . . . . . 6 pts - I like your link to Prime Pages and yr proof that (6n+1)^2 = 24k + 1.
- Hermen Jacobs . . . . . . . 5 pts - Good job answering b) on second try; bonus for suggesting problem!
- Kirk Bresniker . . . . . . . 5 pts - Thanks for link to MathWorld about the quadratic residue theorem!
- Etienne Desclin. . . . . . . 5 pts - Good mention of Fermat's Little Theorem, thanks for recurring entry!
- Ameya Joshi . . . . . . . . . 4 pts - Fine work this week, the third part was pretty obscure, keep entering!
- Denis Borris . . . . . . . . . 4 pts - I like the (1.2....(p-1))^2 = -1 (mod p) argument; I'm in Wilson's court!
- Al Nelson. . . . . . . . . . . . 4 pts - Moving up on the charts with another good and clever answer (above)
- Nick McGrath . . . . . . . . 4 pts - Valiant run at Marcello this year, great answer this week too; go Nick!
- Akifumi . . . . . . . . . . . . . 3 pts - Welcome back from a short break; now you're near my home town!
- Claudio Baiocchi . . . . . 3 pts - From Italy we get complex idea: c) p^f - 1 = (p + i(p-1))^2 is a sol'n.
- Oliver Schebanek . . . . . 3 pts - Answer as early or late as you want to ; nice job here finding \/-1 mod p.
- Phil Sayre . . . . . . . . . . . 3 pts - Hey Phil, Carmichael is a good number theory reference; i like it 2!
- Michael Thwaites . . . . 3 pts - Good to have you back ; I like the "primitive root mod p" approach.
- Ed wErn. . . . . . . . . . . . . 3 pts - Early entry earns Ed extra exemplary epoint (every element ees e)
- Tim Poe . . . . . . . . . . . . . 3 pts - That's who the mysterious alias was! And the horse you wrote in on!
- Mark Rickert . . . . . . . . 3 pts - Great job on a) and b); nice intutive start on c) using sum of 2 squares
- Nats Kroy. . . . . . . . . . . . 3 pts - Right; 3^2 isn't 1 mod 24, but then again 24^2 isn't 1 mod 3 either!
- Jeremy Galvagni. . . . . . 2 pts - I liked your b) that any 6n +/- 1 works, not just prime ones. Go JG!
- Adam Panagos . . . . . . . 2 pts - Good on b) to use (p-1)(p+1) and divis. by 3 as well as 8, making 24.
- Sara Elsnick . . . . . . . . . 2 pts - Glad u took time away from your poor students' papers. Great job a) b)
- Allen Druze. . . . . . . . . . 2 pts - Liked b) (4n +/- 1)^2 = 8n(2n+1)+1 finding mult of 3 in there!
- Damon Haught. . . . . . . 2 pts - Welcome back (been 18 mos); good ans on a) and b)! (not factorial)
- problem #253 - posted wednesday, september 27, 2006
- Powers Five Apart : last problem of the 2005-06 contest (back to top)
- Find pairs of positive integers (a, b) such that
- and prove you have found all of them. Show steps and reasoning.
- Solution: [I had a bit of a struggle deciding whether Al's proof was "holy" or "holey"; it was the first
- entry that claimed a proof of the completeness of the solutions list (a=3, b=1) and (a=5, b=3). Hmm.
- A similar proof from Akifumi confounded me further, sending me out on a six-mile run to think about it.
I will post winners and solutions soon! - Dan Bach the Math Jock earlier]- Ok I'm back; here's Al's submission, suspect step in red; [there could be an extra common multiple]:
- "There are two solutions for (a,b): (3,1) and (5,3). First note the "obvious" solution (3,1), since
- 2^3 = 3^1 + 5, Any other solution must fit 2^a = 3^b + 5. Subtracting, we get 2^a - 2^3 = 3^b - 3^1 ;
- or 8(2^[a-3] - 1) = 3(3^[b-1] - 1). Since 8 and (2^[a-3] - 1) are relatively prime, as are 3 and (3^[b-1] - 1),
- the above equation is only possible if (2^[a-3] - 1) = 3 and (3^[b-1] - 1) = 8, which gives the second
- solution (5,3) [and we see that 2^5 = 3^3 + 5].
- Marcello; this week's and official season 9 winner, sent this; I edited it a bit:
- "Since (2,4,3,1) and (3,4,2,1) are cyclic residues (mod 5) for integer powers of 2 and 3 respectively, the
- combinations that give a residue zero (mod 5) for 2^a - 3^b are 1) a and b both even and congruent
mod 4; 2) both odd with a=4n+3, b=4m+1; 3) both odd with a=4n+1, b=4m+3.- Case 1: Suppose a=2u and b=2v, then (2^u - 3^v) (2^u + 3^v) = 5, and since 5 is prime it must be
- (2^u - 3^v) = 1 and (2^u + 3^v) = 5. The second equation implies u=v=1, which doesn't satisfy the
- first one, so no solution of type 1 exists.
- Case 2: The solution a=3, b=1 (n=0, m=0) can be checked directly. Suppose another solution exists,
- with n>0, m>0. Then 2^(4n+3) - 3^(4m+1) = 2^3 - 3, i.e. 2^3 (2^4n - 1) = 3 (3^4m - 1) = 3
(3^2m - 1) (3^2m + 1) = 3 (3^m - 1) (3^m + 1) (3^2m + 1). There are three even factors at second- member, but one of (3^m - 1) and (3^m + 1) must be divisible by 4, so second member is divisible
- by 2^4, whilst the first one is only divisible by 2^3. Therefore no other solution of this type exists.
- Case 3: The solution a=5, b=3 (n=1, m=0) can be checked directly. Suppose another solution exists,
- with n>1, m>0. Then 2^(4n+1) - 3^(4m+3) = 2^5 - 3^3, i.e. 2^5 (2^4(n-1) - 1) = 3^3 (3^4m - 1)
= 3^3 (3^2m - 1) (3^2m + 1) = 3^3 (3^m - 1) (3^m + 1) (3^2m + 1). The ( continues . . . )- Since 3^2m + 1 is even, a factor of at least 2^6 appears at second member, against 2^5 at the first
one. The case m= 2 can be dispatched by direct inspection: 3^11 + 5 = 177152, not a power of 2.- Therefore no other solution of type 3 exists. The only solutions are (3,1) and (5,3).
- I had originally found this solution: "It is easy to see that a=3, b=1 ; a=5, b=3 are all the solutions
- of (1) : 2^a = 3^b + 5, with a <= 5. We will show that they are the only ones.
Let a > 5. Then 3^b + 5 is divisible by 2^6 = 64. On the other hand, 3^16 = 1 mod 64, and direct- calculation shows that 3^11 = -5 mod 64, and 3 ^k =/= -5 for all other k = 0, . . . , 15.
- So, b must be of the form: b = 16k + 11. Now, if we divide (1) by 17, using 3^16 = 1 mod 17,
- we get : 3 ^(16k+11) = 3^11 = 7 mod 17. Therefore, 2^a = 3^11 + 5 = 12 mod 17.
On the other hand, the possible remainders of 2^a, divided by 17 are 2, 4, 8, 16, 15, 13, 9, 1
and in particular, 2^8 = 1 mod 17. Therefore 2^a = 12 mod 17 is impossible.
- WINNERS - Problem 253 . . did you vote for dansmathcast on podcast alley? (back to top) . leader board
- Marcello Cammarata. . 10 pts - Winner, season #9 - Pozzo fatto Marcello; buon lavoro! (See ans above)
- Doug Babcock . . . . . . . . 7 pts - Good strategy, reducing powers of 3 mod 64 - see second proof above.
- Al Nelson . . . . . . . . . . . . 5 pts - Finally decided your prf was close, but made dangerous assumption, c above
- Akifumi . . . . . . . . . . . . . 4 pts - Proof seemed ok but same basic assumption as Al; close though!
- Nick McGrath . . . . . . . . 4 pts - Good for trying to fix proof; it was a tricky prob! Thx for yr long support.
- Ed Wern. . . . . . . . . . . . . 4 pts - Excel can tackle a multitude of things, Ed, but not infinity itself ;-}
- Tim Poe . . . . . . . . . . . . . 3 pts - Good try using logs and ruling out certain a's; nice early entry after Ed
- Mark Rickert . . . . . . . . 3 pts - Nice progress proving by digital roots (mod 9) that a = 6n+5, b = 2m+1
- Denis Borris . . . . . . . . . 3 pts - Cool use of diff of sqrs (2^m + 3^[2x+1])(2^m - 3^[2x+1]) = 5
- Kirk Bresniker . . . . . . . 3 pts - Nice use of ratio a/b = log3/log2; i'm unfam with gen exp dioph eqns.
- Phil Sayre . . . . . . . . . . . 3 pts - Right track; a g = log2/log3; a = int, b = floor(g*a) but still not a proof
- Zahi Teitelman. . . . . . . 3 pts - Progress good to 8(2^x = 1) = 3(3^y - 1) ; (3,1) and (5,3) fit this
- Allen Druze. . . . . . . . . . 3 pts - Decided on 2(2^[a-1] - 1) = 3(3^[b-1] + 1) which limited choices...
- Hermen Jacobs . . . . . . . 2 pts - Those are the right solutions; nice try proving the limited extent.
- Claudio Baiocchi . . . . . 2 pts - Partial success knowing properties of more solutions. {also see #179}
- Nats Kroy. . . . . . . . . . . . 2 pts - It's a bit of a mystery when the next prob comes up; good partial ans
- Isha Kotecha . . . . . . . . . 2 pts - Welcome Isha from India; good try usg lots of a values; see sols above
- David Madfes . . . . . . . . 2 pts - Good 2 have u back; nice idea using a table of (a,b) combos; diag!
- K Sengupta . . . . . . . . . . 2 pts - Result of Ellison, 1971: |2^a - 3^b| > 2^a * e^(a/10) > 184 > 5 ; cool!
- problem #254 - posted monday, october 23, 2006
- Vote Four Pumpkins ! first problem of the 2006-07 contest, my tenth season! (back to top)
- Three pumpkins are sitting on my front porch. They are spherical, of radii 4 ft, 3 ft,
- and 2 ft, and they are mutually tangent. I put a fourth pumpkin of radius 1 ft on top,
- in the hollow . . .How high off the porch is the top of the last pumpkin?
- Show your steps and reasoning.
- Solution: Happy 80th birthday to Art Morris, our all-time leader!
- Most of you set up the bases of the pumpkins on the (x, y, 0) plane, choosing (0, 0, r) etc.
- Philippe: (0, 0, 4), (4\/3, 0, 3), (7 / \/3, \/(47/3), 2), a bit tall on z1...
- Claudio: (0, 0, 0)4, (7, 0, 0)3, (30/7, 12\/6 / 7, 0)2 , then rotates the bldg to get horiz bottom plane!
- Kirk : (0, 0, 4), (4 \/2, 0, 2), (7 / \/2, \/(47/2), 3) , and put that 4th pumpkin by dist formula.
- Marcello : Centers form 5-6-7 triangle; top center satisf. 1296 z^2 - 7152 z + 5376 = 0
- Here's Garry M's solution: Let's p1, p2, p3, p4 - pumpkins centers ; Let's this points have coordinates
- p2(0,0,2) ; p3(2*sqrt(6),0,3) [Pythag Thm on vert plane - Dan] ; p4(sqrt(6)/3,sqrt(282)/3,4)
easy to check that distance between p2 and p3 is equal to 5, distance between p2 and p4 is equal to 6- and distance between p3 and p4 is equal to 7 so pumpkins are mutually tangent.
Let's p1(x1,y1,z1) ; distance(p1,p2)=3 , distance(p1,p3)=4 and distance(p1,p4)=5 so
x1^2 + y1^2 + (z1-2)^2 = 9, (1)
(x1-2*sqrt(6))^2 + y1^2 + (z1-3)^2 = 16, (2)
(x1-sqrt(6)/3)^2 + (y1-sqrt(282)/3)^2 + (z1-4)^2 = 25 (3)
subtract (1) from (2) and simplify: z1=11-2*x1*sqrt(6) ; substitute this in (1) and (3):
25*x1^2 + y1^2 - 36*x1*sqrt(6)= -72 (4)
25*x1^2 + y1^2 - 86*x1*sqrt(6)/3 -2*y1*sqrt(282)/3 = -56 (5)
subtract (5) from (4) and simplify : y1=(11*x1*sqrt(6) - 24)/sqrt(282)
substitute this in (4) and simplify : 324*x1^2 - 445*x1*sqrt(6) + 870 = 0
two values for x1 : (445*sqrt(6) + sqrt(60630))/648 and (445*sqrt(6) - sqrt(60630))/648
so two values for z1: (149+sqrt(10105.))/54 and (149-sqrt(10105.))/54
get the maximum value (149+sqrt(10105.))/54 [ ~=~ 4.62081]
we must to add 1 (radius) and get answer (203+sqrt(10105))/54 ~=~ 5.62081
- WINNERS - Problem 254 . . you can vote for dansmathcast on podcast alley! (back to top) . leader board
- Mark Rickert . . . . . . . . . 9 pts - Great method, correct answer, first response; I'd like more steps pls!
- Claudio Baiocchi. . . . . . 6 pts - Neat trick rotating the building to level out the pumpkin bases; wow!
- Ed Wern . . . . . . . . . . . . . 5 pts - Nice job for a 3D challenged guy; are you in the flat earth society?
- Marcello Cammarata . . 4 pts - Back again for season ten, it's Marcello the Fellow more Mellow!
- Nick McGrath . . . . . . . . 4 pts - Smart to check the small pumpkin won't fall thru hole. Go ExcelSolver!
- Garry Malashkin . . . . . 4 pts - Good proof; there it is quoted above. I admire your refusal to decimalize.
- Nats Kroy. . . . . . . . . . . . 4 pts - No charge for minor resub; good to use eqns of spheres and projec'ns.
- Jeff Peterson (new) . . . . . 4 pts - Welcome to the dansmathcontest; thanks for nice detailed solution!
- Kirk Bresniker . . . . . . . 4 pts - The only one to work in Heron's area formula as far as I can tell.
- Phil Sayre . . . . . . . . . . . 3 pts - Thanks for attached Python code; I tried to run it but it bit me!
- Denis Borris . . . . . . . . . 3 pts - Yes; (0, 0, 2), (a, 0, 3), (b, c, 4); then use the power of th'quadratic formula!
- Art Morris. . . . . . . . . . . 3 pts - Hi Art; good try but your pumpkins were half size; happy 80th!!
- Philippe Fondanaiche . 3 pts - Nice attached document; great picture; 4th pumpkin was a bit high.
- Hermen Jacobs . . . . . . . 2 pts - Good centers (0, 0, 2), etc; that equation just might give the right z.
- Jacket (new) . . . . . . . . . . 2 pts - Welcome Jacket; good couple of equations; ht of sphere a bit low
- Humpty Dumpty . . . . . 1 pt - I'll add on to your old name totals unless you tell me otherwise ;-}
- Show your steps and reasoning.
- Solution: i) Many of you knew the algorithm for even n:
- Pair up the players, then fix one and rotate the others each day.
- Example: n = 6; five days: fix 1, rotate 2 and others clockwise:
- 1 2 3 . . 1 6 2 . . 1 5 6 . . 1 4 5 . . 1 3 4
- 6 5 4 . . 5 4 3 . . 4 3 2 . . 3 2 6 . . 2 6 5
From Etienne D. (nice table n=10) -- 4 players: A,B,C,D
By combinatorics C(4,2)= 6 matches to be played,
Then 2 matches by day in 3 days: example
Day1: AB and CD ; Day2: AC and BD ; Day3: AD and BC
Each player plays every other one once: RRT ok
All players play once per day: ok6 players: C(6,2)= 15 matches to be played,
3 matches by day, during 5 days:
Day1: AB , CD and EF ; Day2: AC , BE and DF
Day3: AD , BF and CE ; Day4: AE , BD and CF
Day5: AF , BC and DE8 players: A,B,C,D,E,F,G,H - 4 matches/day
C(8,2)= 28 matches to be played,
4 matches by day, during 7 days:
Day1: AB , CD , EF , GH ; Day2: AC , BE , DG , FH
Day3: AD , BF , CG , EH ; Day4: AE , BD , CH , FG
Day5: AF , BC , DH , EG ; Day6: AG , BH , CF , DE
Day7: AH , BG , CE , DF10 players: C(10,2)= 45 matches to be played,
5 matches per day, during 9 days
A B C D E F G H I J
A 1 2 3 4 5 6 7 8 9 i.e. On day3,
B 3 4 5 6 7 8 9 2 the 5 matches:
C 5 6 7 8 9 1 4 AD BC EI FH GJ
D 7 8 9 1 2 6
E 9 1 2 3 8
F 2 3 4 1
G 4 5 3
H 6 5
I 7
J- ii) If n is odd, we introduce an n+1-st player named "bye" , construct a tournament for the
- even number of players n+1, and whoever plays n+1 gets that day off. For example, if n=5
- (podcast question from show #26), then player 6 is 'bye' so using above the 5 days are:
- (2v5, 3v4, 1 off), (1v5, 2v3, 4 off), (1v4, 3v5, 2 off), (1v3, 2v4, 5 off), (1v2, 4v5, 3 off).
- Jeremy's way of looking at it was to draw odd n points around a circle; leave top point alone, join the other n-1
- horizontally in pairs; each segment rep. a game played between those two. Each day you rotate the circle one click;
- there are n days. If n is even, join the top point to a new center point, so top plays center, then rotate daily.
- Philippe F of Paris (PF of PF!) had a similar solution with nice picture; circles, colored segments for n = 4, 6, 8, 10.
- Denis B. mentioned a nice solution at http://mathforum.org/library/drmath/view/54715.html
- WINNERS - Problem 255 . . you can vote for dansmathcast on podcast alley! (back to top) . leader board
- Nick McGrath . . . . . . . . 10 pts - Used to schedule chess tourneys; so he's an old hand at this!
- Denis Borris . . . . . . . . . . 7 pts - Good answer (+joke); thanks to Dr. Math in the math forum link
- Ed Wern . . . . . . . . . . . . . 6 pts - I like the term 'phantom player' in the odd case. Wikipedia rules!
- Zahi Teitelman . . . . . . . 5 pts - Another term is 'idle' player; nice tables ZT, easy to see solution.
- Etienne Desclin . . . . . . . 5 pts - A quotable response from the Belgian table-maker; thanks!
- Nats Kroy . . . . . . . . . . . . 4 pts - Clockwise rotation never hurt anybody (espec. a New Yorker!)
- Mark Rickert . . . . . . . . . 4 pts - Another player on the clockwise bandwagon, and odd fake player.
- Jeremy Galvagni . . . . . . 4 pts - Good circle trick (see above), your geom students are lucky!
- Philippe Fondanaiche . 4 pts - Another nice picture, explained above; merci pour le visuel %;-}
- K Sengupta . . . . . . . . . . 3 pts - Your attachment was hard to read but I got the bye info ok...
- Garry Malashkin . . . . . 3 pts - Liked your recursive build from one sol'n to the next even n.
- Claudio Baiocchi. . . . . . 3 pts - No penalty for being Italian (even with the tourney advantage)
- Marcello Cammarata . . 3 pts - Another smart Italian, with a link to a cool solution; thanks!
- Kirk Bresniker . . . . . . . 3 pts - A clever 'matrix method'; day # in (i, j) when i plays j (see Etienne)
- Al Nelson . . . . . . . . . . . 3 pts - Thanks, Coach, for the 'inside-out' approach (show your guts!)
- Adam Morgan (new) . . . 3 pts - Welcome to my long-running contest; thanks for nice answers.
- Tim Poe . . . . . . . . . . . . . 3 pts - Good to mention the 'cyclic' nature of the progression (and odd bye)
- Bob Chen (new) . . . . . . . 2 pts - Welcome 2 dansmath contest; right about odd player sitting out.
- Humpty Dumpty . . . . . 2 pts - Hey, we need you up on that wall! (sorry, American movie reference)
- Michael Thwaites . . . . 2 pts - Nice to have you back; good incidence matrix (not Keanu's movie)
- Allen Druze . . . . . . . . . 2 pts - Right, if n = power of 2 it's easier (hear my podcast #26 tourneys)
- Phil Sayre . . . . . . . . . . . 2 pts - Good ans for even n's; yes it's ok in odd case to sit out a day.
- V Balakrishnan. . . . . . . 2 pts - The requirement that everyone plays was just for the first part.
- David Stigant (new) . . . . 2 pts - Welcome David, and right about the bye; good ans for even n.
- Solution: This time it worked out that nobody got 10 points, but lots of great answers and methods!
- a) Allen D. says: (2x-3)(2x-1)(2x+1)(2x+3) = n^2 ; (4x^2 - 1)(4x^2 - 9) = n^2 . . . let 4x^2 - 1 = y, then
- (y)(y-8) = n^2 ; y^2 - 8y - n^2 = 0 , (discriminant) 64 +(4)(n^2) must equal a square, let's say c^2. . . .
- 64 + 4n^2 = c^2 , factoring out 4 , 16 + n^2 = d^2; 16 = (d+n) (d-n) ; 8 = d+n , 2 = d-n ; 2d = 10 , d = 5 .
n = 3, n^2 = A = 9.- b) Some of you decided that not all the conditions were needed, but they still needed to be checked!
- A popular method was to prove that B = pqr; p^2 < p^2 + q^2 + r^2 = 2331 means p, q, r <= 47.
- You found five ways of doing this and showed how only one fit the sum of divisors = 10560.
- primes : product : sum of div : sum of squares
- 11,19,43 : 8987 : 10560 * (121 + 361 + 1849 = 2331)
11,23,41 : 10373: 12096 (121 + 529 + 1681 = 2331) . . these are pasted
11,29,37 : 11803: 13680 (121 + 841 + 1369 = 2331) . . in from Denis B's
17,19,41 : 13243: 15120 (289 + 361 + 1681 = 2331) . . table solution
23,29,31 : 20677: 23040 (529 + 841 + 961 = 2331)- Here's Etienne's fine solution . . . B = abc where a,b and c are primes; a2 + b2 + c2 = 2331 (1)
The set of divisors of B = {1,a,b,c,ab,ac,bc,abc}; so there are (ab-1) numbers less than B having c as- common factor , (ac-1) having b, (bc-1) having a as c.f.
As we are counting them twice, there are (a-1) numbers less than B having bc as common factor, etc
abc -(ab-1)-(ac-1)-(bc-1) + (a-1)+(b-1)+(c-1) - 1 = 7560 ; abc - ab - ac - bc + a + b + c = 7561 (2)
As the sum of the divisors of B is given, abc + ab + ac + bc + a + b + c = 10559 (3)
(3)-(2) => 2ab + 2ac + 2bc = 2998; add (1) => a2 + b2 + c2 = 2331; get (a + b + c)2 = 5329- => a+b+c=73 ; now add (3)+(2) => 2B + 146 = 18120 => B = 8987
And follow Akifumi's argument, I dare you! - Dan
- WINNERS - Problem 256 . . did you vote for dansmathcast on podcast alley? (back to top) . leader board
- Ed Wern. . . . . . . . . . . . . 9 pts - a) Nice use of quadratic formula square discrim; b) Check #rp = 7560
- Allen Druze. . . . . . . . . . 8 pts - First entry; liked your expl on a); need specifics for b) but right answer
- K Sengupta . . . . . . . . . . 7 pts - a)(4a2-6a-3)(4a2+6a-3)=n2 => n=3 only; b) sd=(p+1)(q+1)(r+1) worked!
- Denis Borris . . . . . . . . . 5 pts - Yes; (n-3)(n-1)(n+1)(n+3); b) only 8987 has right ssq and sd; check rp.
- Nick McGrath . . . . . . . 5 pts - A=9 by the discrim argt; b) (p-1)(q-1)(r-1)= 7560 & use 20thC BASiC
- Mark Rickert . . . . . . . . 4 pts - a) Sqs more than 6 apart soon - good. b) missed 11.23.41 but 8987 yes.
- Marcello Cammarata. . 4 pts - b) Liked the last-digit sq must be 1+1+9; 11.19.43 works & conds check.
- Etienne Desclin. . . . . . . 4 pts - Good table for a), nice alg to find p+q+r = 73 using ssq and mults of c
- Garry Malashkin . . . . . 4 pts - A=nother discriminanter; B'sums 10560, 12096, ..., 23040 for ssq=2331
- Claudio Baiocchi . . . . . 4 pts - a) GCF(n2-1)(n2-9),=8=>A=9; Again ssq=2331 gives 5 triples for B
- Akifumi. . . . . . . . . . . . . 4 pts - Welcome back; liked (2n+1)2 - (2n-1)2 = 8n but dfsq<=6; b) see above
- Art Morris. . . . . . . . . . . 3 pts - a) No sol'n up to 255.257... so A=9. b) last dig sqs 1,1,9; check 7560
- Nats Kroy. . . . . . . . . . . . 3 pts - a) "Les Nombres Remarquables" b) (p-1)(q-1)...=7560 so only 8987
- Wenjun Kang (new) . . . . 3 pts - Great to have you; a) Used GCD to show uniq; b) Good ans need steps
- Tim Poe . . . . . . . . . . . . . 3 pts - a) 6 apart too close for sqs. (Tim- a few more pts and we had Ed, Allen, Poe!)
- Kirk Bresniker . . . . . . . 3 pts - Of the ssq 2331, only 8987 has right sd. (forgot why i had a pic of money)
- Doug Babcock. . . . . . . . 3 pts - Great details in showing A=9; B=11.19.43 by checking it works well
- David Stigant . . . . . . . . 3 pts - Good amt of steps prove A is uniq; b) missed one combo, nice free resub
- Philippe Fondanaiche . 3 pts - Cool; b) pqr-pq-pr-qr+p+q+r-1=7560 and all pos for 10560; B=8987
- Jeremy Galvagni. . . . . . 3 pts - a) (2n-2)(2n-1)(2n+1)(2n+3)= 16n4-40n2+9 only sq if n=0- good!
- Ajit Athle. . . . . . . . . . . . 3 pts - Same a) = (4n2-1)(4n2-9)must be 9; b) trial&error&commonsense!
- Giridhar Prasannan. . . 3 pts - a) ditto, can't be paired up (to be sq); b) Nice check on sum of pos div
- Phil Sayre . . . . . . . . . . . 2 pts - Right answers they were indeed, with factors of both. How'd you get?
- Zahi Teitelman. . . . . . . 2 pts - Good argt to show A=9 unique; no second part but thanks for entry
- Humpty Dumpty . . . . . 2 pts - Decent argument about rep facs for sqs. Resub 23.29.31 ssq ok not sd
- Aclaira (new) . . . . . . . . . 2 pts - Welcome! (n+1)(n+3)(n+5)(n+7) = n4+16n3+...+105; shows A=9. Good!
- Deborah Fradelis. . . . . 2 pts - a) 39.41.43.45 has non-rep primes so not sq; nice job on b) check #rp
- Ameya Mannikar (new) 1 pt - Welcome another Ameya; integers must be all +, all -, or 2 of each, yes.
- Jon Stearn . . . . . . . . . . 1 pt - Nice2haveUbach; yes A=9 but B=20677 won't work (is it from ssq=2331?)
- problem #257 - posted thursday, december 14, 2006
- ? nEver Say Never ? ...we assume n is an integer... (back to top)
- a) Prove that n^4 + 2n^3 + 2n^2 + 2n + 1 is never a square for n > 0
- b) Prove that 1 + 1/2 + 1/3 + . . . + 1/n is never an integer for n > 1
- Show your steps and reasoning.
- Solution: a) The first part turned out to be easier than I thought it was. I had expected you'd have
- to show, as Jin Won did, that (n^2 + n)^2 < n^4 + 2n^3 + 2n^2 + 2n + 1 < (n^2 + n + 1)^2 and so our
- polynomial was stuck between two consecutive squares. Clever argument.
- But no, you guys had to take the direct approach: just factor: n^4 + 2n^3 + 2n^2 + 2n + 1 itself
- = (n^2 + 1)^2 + 2n(n^2 + 1) = (n^2 + 2n + 1)(n^2 + 1) = (n + 1)^2 (n^2 + 1) ;
- this is a perfect square iff (if and only if) (n^2 + 1) is, which it isn't since n^2 is. T.I.J. (That's It, Jack.)
- b) This is apparently well-known; the idea is (as K Sengupta here, and others sent in:)
- "Let f(n) = 1 + 1/2 + 1/3 + . . . + 1/n......(i). Let us choose an integer k such that 2^k<= n<= 2^(k+1)
Consider the lowest common multiple of 1,2,...., 2^k -1, 2^k+1,....,n. This number will be of the form- G*2^(k-1) where G is an odd integer. Multiply both sides of the equation by G*2^(k-1), we obtain:
- G*2^(k-1)*(1 + 1/2 + 1/3 + . . . + 1/n) = G*2^(k-1)*f(n).......(ii)
When multiplied out, all the terms on the LHS of (ii) will be integers, except one, that is:- G*2^(k-1)*(1/(2^k))= G/2, which is not an integer, since G must be odd.
So the LHS of (i) is not an integer, and accordingly, hence neither is the RHS of (i).- Consequently, 1 + 1/2 + 1/3 + . . . + 1/n is never an integer for n > 1."
- WINNERS - Problem 257 . . listen and learn! dansmathcast is free and clear! (back to top) . leader board
- K Sengupta. . . . . . . . . . 10 pts - a) Nice touch proving it's not a square even for neg.n. b) See above!
- Nick McGrath . . . . . . . 7 pts - Same time as Nats to the min; you spelled factoriZe like an American!
- Nats Kroy . . . . . . . . . . . 6 pts - Thanks for link to solution from 1915, or at least title; needed more steps
- Mark Rickert . . . . . . . . 5 pts - a) Popular factor way seems to work; b) great strategy; H(n) = odd/even
- Kirk Bresniker . . . . . . . 5 pts - Right that the harmonic series diverges, so skips over every integer!
- Akifumi. . . . . . . . . . . . . 5 pts - a) f(-1)=0 so (n+1) is a factor. b) L=lcm(1..n); H(n) = (L/1 +...+L/n)/L
- Ed Wern. . . . . . . . . . . . . 4 pts - a) Nice job see quote above; b) I had to stretch to believe this part
- Etienne Desclin. . . . . . . 4 pts - b) After 1/7 next 7 in denom is 14; but 11 and 13 appear; n < p < 2n thm.
- Garry Malashkin . . . . . 4 pts - a) Prf n^2+1 never sq. b) 2^k <= n < 2^(k+1) mult thru by 2^k etc.
- Denis Borris . . . . . . . . . 4 pts - b) Wrote H(n) with denom of n! rather than lcm; good idea worked
- Doug Babcock. . . . . . . . 3 pts - a) Thorough! b) True for n=1..6; for n>6 Bertrand's prime betw n and 2n.
- Aclaira. . . . . . . . . . . . . . 3 pts - a) (n^2+an+1)^2 expand and equate coeffs; gets a=1 nice! b) B < B
- Marcello Cammarata. . 3 pts - b) lcm = M; mult H(n) thru by M/2; then all integer except one term.
- Mo Money (new) . . . . . . . 3 pts - Good to have you; Let y = lgst prime <n; then n! dnd n!(1+...+1/n)
- Stefan Gatachiu (new) . . 3 pts - Welcome to the contest; nice use of Mean Value Thm on ln(x) in b!
- Zahi Teitelman. . . . . . . 3 pts - a) Usual n^2+1 not a sq. b) Yes; s(n)*L is not an integer if L = lcm
- Philippe Fondanaiche . 3 pts - Right; the n^2+1 on a) and the lgst power of 2 on b). Bien fait, P.F.
- Jin Won Park . . . . . . . . 3 pts - That's what I had in mind in a), squeeze betw 2 consec sqs. Wlcm bk!
- Ameya Mannikar . . . . . 2 pts - Good steps factoring part a); now you see how to do part b), eh?
- Sara Elsnick . . . . . . . . . 2 pts - Welcome back; try these any time; good answer on part a) here.
- Claudio Baiocchi . . . . . 2 pts - Two minutes to factor one polynomial; it musta been a hard one!
- Phil Sayre . . . . . . . . . . . 2 pts - Tried (n^2 + an + b)^2 expand; then b^2 = 1, a = 1; one works, good!
- Allen Druze. . . . . . . . . . 1 pt - Good answer; nice use of prime fac in b); (after 258 was up so just 1 pt.)
- problem #258 - posted wednesday, december 27, 2006
- The Math Contest (back to top)
- Twenty-one girls and twenty-one boys entered a math contest.
- (1) Each contestant solved at most six problems, and: (2) For each pair of a girl and
- a boy, there was at least one problem that was solved by both the girl and the boy.
- Prove that there was a problem that was solved by at least three girls and at least
- three boys. Show your steps and reasoning.
- Solution: Not as big a crowd this time, right at New Year's! A couple of you found and
- linked to solutions online; there was an alternate wording of the problem in terms of a
- 21 x 21 grid with numbers in it; here are a couple of examples, one is like the one I found.
- Garry M sent: "For each problem say it's type (A) if two boys (or less) solved it and type (B) otherwise.
Consider a table with 21 rows (boys) and 21 columns (girls). We can fill this table as follows:- for each pair (boy,girl) at least one problem was solved by both of them (condition (1)),
- choose an arbitrary problem from them and assign problem's type to the cell. By the Pigeonhole Principle,
- there is a column with at least 11 cells (A) (case 1) or there is a row with at least 11 cells (B) (case 2)
( otherwise total cell's quantity is equal to 21*10 + 21*10 = 420 < 21^2=441 )
Suppose for every problem there are two boys (or less) who solved it and two girls (or less) who solved it.
Case 1: Consider the column with at least 11 cells of type (A).
This means that each of this at least 11 problems was solved by at most two boys.
Therefore we can found at least 6 distinct problems solved by the girl.
But from condition (2) we can say that this girl solves only this six problems and
so there are at most 12 boys solve a problem also solved by the girl.
So we come to the contradiction with condition (1). Case 2: by analogy with case 1 Q.E.D.- And DBor found a similar table solution using coloring:
- 1. Make a table, listing girls down the left side and boys along the top. In each cell, place a letter representing one problem that
- the girl in that row and the boy in that column both solved. (Every cell will be filled, because of the hyp about boy-girl pairs.)
2. Now look in each row for letters that occur more than two times. (no row will contain more than six different letters.)
3. In that row, color every cell containing those letters red. You will observe that at least 11 cells in each girl's row are colored.
4. Thus more than half the cells in each row--and thus in the entire table--are red, meaning they were solved by at least three boys.
5. Now repeat the procedure for each of the boys, coloring cells blue to represent problems solved by at least three girls.
6. Because more than half the cells are red and more than half blue, at least one must be both red AND blue.
7. The letter in this cell must then be a problem that was solved by at least three girls and at least three boys.
- WINNERS - Problem 258 . . listen to dansmathcast - I dare you! - (back to top) . leader board
- Nick McGrath . . . . . . . 10 pts - I read your steps carefully and finally agree with your proof. Go Nick!
- K Sengupta . . . . . . . . . . 7 pts - Liked both your methods, and proof it's false for 20 boys and 20 girls.
- Garry Malashkin . . . . . 5 pts - Good proof; you have been quoted. I edited slightly to squeeze space.
- Denis Borris . . . . . . . . . 4 pts - D-tective can find solutions to the darndest things on those internets!
- Philippe Fondanaiche . 3 pts - Nice solution, "holds water"! Merci pour tes bons mots sur ma site!
- Kirk Bresniker . . . . . . . 3 pts - That argument seems to work; thanks for the new year's early entry!
- Tony Wang (new) . . . . . . 2 pts - Welcome Tony; good try, proving if avg # solved is 3.333 then true.
- Claudio Baiocchi . . . . . 1 pt - One more than requested; percentagewise that's infinitely better!
- problem #259 - posted monday, january 15, 2007
- Ah, Now We Know ! (back to top)
- Two smart mathematicians (Ben and Jen) are told that a rectangle with integer
- sides L and W has been drawn, having perimeter less than 200, where L > W > 1 .
- Ben is told the area, Jen is told the perimeter. They now say:
- Ben: "I can't determine the dimensions." Jen: "I knew that."
- Ben responds, "Now I can determine them." Jen: "So can I."
- Given these are true statements, what are the length and the width?
- Show your steps and reasoning. (Time's up on this one.)
- Solution: Hi all you dansmathheads! This time we had a tough question that is known by Philippe as
- an 'impossible problem.' The beauty is the non-specificity of the statements leading to a unique solution.
- * Mark R. sent in a play-by-play (without explicit lists!) of how many qualified (L, W) pairs
- remained after each of the four statements: "Since the perimeter is less than 200, W+L is less than 100.
- Therefore 2 <= W <= 49, and 3 <= L <= 97. I wrote a program to solve this. . .
Step 1 - Initialize a list of all possible dimensions subject to the constraints above (2304 items in list).
Step 2 - Per Ben's first statement, mark all unique products as non-solutions.
Step 3 - Per Jen's first statement, for each of the unique product items identified in step 2,- . . . mark all other items that have a matching sum.
Step 4 - Eliminate all items marked in steps 2 and 3 (145 items remain).
Step 5 - Per Ben's second statement, eliminate all non-unique products (86 items remain).
Step 6 - Per Jen's second statement, eliminate all non-unique sums- . . . Finally 1 item remains . . . The width is 4, and the length is 13.
- * Nats Kroy's (re)sub was amazing - good reasoning, thorough table (partly included here):
- "The dimensions are 4 x 13. Ben is told the area is 52. Jen is told that the perimeter is 34.
If the dimensions were both prime numbers then Ben would have had only one possible case and- would have known the dimensions. Thus, when Ben says he does not know the dimensions on round
- 1 it implies that the dimensions are not both prime.
Then when Jen says that he knew that Ben couldn't know the dimensions, Jen is also implying that he- knew that the sum of the dimensions could not be written as the sum of two prime numbers. Goldbach's
- conjecture states that any even number > 2 can be written as a sum of two primes. Thus to avoid the
- possibility of prime dimensions the sum of the dimensions = 1/2 of the perimeter could not be even.
- Also the sum of the dimensions could not be the sum of 2 + a prime.
Thus at the end of round 1 both Ken and Jen know that the sum of the dimensions could be neither- even nor any of: 5, 7, 9, 13, 15, 19, 21, 25, 31 33, 39, 43, 45, 49, 55, 61, 63, 69, 73, 75, 81, 85, 91, 99.
Since Ben knows that the area is 52, the possible dimensions are 2 x 26 or 4 x 13. However, the dimen-- sions could not be 2 x 26 which has an even sum. Thus, Ben knows that the dimensions have to be 4 x 13.
Since Jen knows the perimeter is 34. The sum of the dimension is 17 and the possible boxes satisfying- the conditions of the problem with a perimeter = 34 are 2 x 15, 3 x 14, 4 x 13, 5 x 12 , 6 x 11, 7 x 10, 8 x 9.
The possible boxes that Jen was considering are listed in the table below. The second column lists the- areas of each box. Jen knows that there are at least 2 possibilities with each area in column 2. Otherwise,
- he could not have been certain that Ben would not know the dimensions. The "possible" dimensions that
- would result in the areas in column 2 are listed in column 3. However, for some of these "possible' case
- the exclusions mentioned above imply that they would have been rejected by Ben. These rejected cases
- are listed in column 4 with the sum of their dimensions which caused their rejection.
The remaining "possible" cases that Jen feels that Ben would have had to consider are in the fifth column.
Dimensions with perimeter 34 Area ; Cases that Jen knows that Ben would have to reject after round 1 ;- Possibilities for Ben after round 1 : 2 x 15 ; 30 [2 x 15, 3 x 10, 5 x 6 ; Reject Sum = 13
2 x 15, 5 x 6, 3 x 14 ; 42 [2 x 21, 3 x 14, 6 x 7 . . . ]
The only case where Ben has only one possibility in column 5 after round 1 is the 4 x 13 box.- When Ben says that he knows the dimensions it tells Jen that that the box is 4 x 13."
- WINNERS - Problem 259 . . listen to dansmathcast - I dare you! - (back to top) . leader board
- K Sengupta. . . . . . . . . . 10 pts - Nice well-written solution attachment, like a good term paper! ;-}
- Mark Rickert . . . . . . . . 7 pts - Thanks for the active "whittling-down" process; program written in ??
- Denis Borris . . . . . . . . . 5 pts - The same as the classic puzzle "Mr. Sum, Mr. Product"; that's it!
- Nick McGrath . . . . . . . 4 pts - Nick found a link to someone else on these internets who solved it!
- Claudio Baiocchi . . . . . 4 pts - I like the example of the perimeter 26 and area 22 figure not being it.
- Tim Poe . . . . . . . . . . . . . 4 pts -Right ; there are 336 non-pq nos who satisfy it; with suitable def of "it"
- Nats Kroy . . . . . . . . . . . 3 pts - Correct reasoning explaining how 13 x 4 does work; liked the resub.
- Marcello Cammarata. . 3 pts - Enjoyed seeing Goldbach conjecture 'not' relied on in your proof.
- Garry Malashkin . . . . . 3 pts - Right; if exactly onee of a type that elim many at each phase.
- Philippe Fondanaiche . 3 pts - Credited Hans Freudenthat 1969 and Martin Gardner ~1980!
- Haley Buxton (new) . . . . 3 pts - Welcome Haley; that was a strong effort; not sure if 1st stmt is unnec.
- Alan O'Donnell. . . . . . . 2 pts - Good to see you again; nice early entry with some cases eliminated.
- Etienne Desclin. . . . . . . 2 pts - Resub cut the number of cases you allowed in half, but need unique
- Phil Sayre . . . . . . . . . . . 2 pts - 'Jen is a whiz'. Cheez, I think you're right; she divided up a huge list fast!
- Tony Wang. . . . . . . . . . 1 pt - Your example makes some of the statemts true but not all unique.
- problem #260 - posted wednesday, january 31, 2007
- White, Black, What ? (back to top)
- Three bags contain 6 marbles each; all white or black: one bag has 5 white and
- 1 black, another has 4 white and 2 black, the third has 3 white and 3 black. One
- white marble is drawn from one bag and one black marble is drawn from another
- (it is not known which bags). What is the probability of drawing a white marble from
- the remaining bag? Show your steps and reasoning.
- Solution: Hello people, thanks for your patience; problem 265 is now up and podcast 31 is in the works.
- This one took me a while to sort out, wow! Here is Tim Poe's solution, but the answer 25/36 is incorrect,
- or at least disputed. See if you can find an error in his reasoning:
- "I assume the drawn marbles were drawn at random. (i.e., a marble was drawn at random from a random
- bag, and was white; a second marble was drawn at random from one of the two rem.bags, and was black.)
- The odds of drawing a white marble from bag 1 is 5/6; from bag 2 is 4/6; and from bag 3 is 3/6.
- Since the bag was chosen at random, the odds of drawing a white marble on the first draw are:
5/6*1/3 + 4/6*1/3 + 3/6*1/3 = 5/18 + 4/18 + 3/18 = 12/18.- Given that this did occur, the odds are 5/12 that it came from bag 1, 4/12 from bag 2, and 3/12 from bag 3.
- If the original marble came from bag1, then the odds of drawing a black on draw 2 (from bags 2 or 3)
- are 2/6*1/2 + 3/6*1/2 = 5/12. The odds would be 2/5 that it came from bag 2 and 3/5 it came from bag 3.
- If the orig marble from bag 2, then 1/6*1/2 + 3/6*1/2 = 4/12; 1/4 that it was from bag 1 and 3/4 from bag 3.
- If the original came from bag 3, then 1/6*1/2 + 2/6*1/2 = 3/12; 1/3 that it was from bag 1, 2/3 from bag 2.
- The odds that you're drawing from bag 1 is 4/12*3/4 + 3/12*2/3 = 3/12 + 2/12 = 5/12.
- The odds that you're drawing from bag 2 is 5/12*3/5 + 3/12*1/3 = 3/12 + 1/12 = 4/12
- The odds that you're drawing from bag 3 is 5/12*2/5 + 4/12*1/4 = 2/12 + 1/12 = 3/12. So the odds of a
- white marble are 5/12*5/6 + 4/12*4/6 + 3/12*3/6 = (25+16+9)/72 = 50/72 = 25/36 = 25 chances out of 36."
- WINNERS - Problem 260 . . listen to my show, dansmathcast - (back to top) . leader board
- Nick McGrath . . . . . . . 10 pts - Nick's prowess shows with his "fave subject" probability!
- Mark Rickert . . . . . . . . 7 pts - Very nice table-centric approach; I wish I were that organized.
- Al Nelson . . . . . . . . . . . 6 pts - I bumped you up a point for your nice quotable solution, above.
- Garry Malashkin. . . . . 5 pts - I follow the Hwww, Hbbw, etc. Nice use of Bayes' Theorem!
- Claudio Baiocchi. . . . . 5 pts - Yes, of 25 cases, 7 leave out 3,3 9 for 4,2 and 5,1, giving 17/25.
- K Sengupta. . . . . . . . . . 5 pts - Nice use of concept of permutations and conditional probability.
- Nats Kroy . . . . . . . . . . . 5 pts - Tortured path of resubmitting incorrect ans after early correct one
- Marcello Cammarata. . 4 pts - "For each pair of bags build a 6x6 square table..." it works!
- Philippe Fondanaiche . 4 pts - Calculated the bayesian probability: Pr{W(3) / W(1) and B(2)}.
- David Stigant . . . . . . . . 4 pts - Welcome back, David; and 204/300 is good enough for me!
- Jin Won Park . . . . . . . . 3 pts - A couple days after some, but a nice detailed solution, thanks!
- Tim Poe . . . . . . . . . . . . . 3 pts - At first your answer seemed plausible but I'm now proven wrong.
- Ed Wern . . . . . . . . . . . . 3 pts - Also considered "looking in and drawing" 1st two, giving P = 2/3.
- Etienne Desclin. . . . . . . 2 pts - This argt used conditional probs but shouldn't be equally weighted.
- Denis Borris . . . . . . . . . 2 pts - Again this argt seems to hold water but there has to be a leak...
- Yakov Macak . . . . . . . . 2 pts - Welcome back mate. Thanks for wishing Art Happy 80th bday!
- Ken Duisenberg . . . . . . 2 pts - "Regardless of which bags, prob is:1/3 * (5/6 + 4/6 + 3/6) = 2/3"
- Ravi Raja . . . . . . . . . . . 2 pts - The same way I did: Case I A,B,C; Case II A,C,B; Case III B,A,C...
- Hermen Jacobs . . . . . . . 2 pts - Also decided to give the three bags equal 'weight' which gives 4/6.
- Phil Sayre . . . . . . . . . . . 2 pts - Wrote (1/6)*[(3+4+3+5+4+5)/6]=24/35=2/3 but meant 24/36.
- Jon Stearn. . . . . . . . . . . 2 pts - Had 5/18 + 4/18 + 3/18 = 2/3; Thanks for the pictures, hello Jon!
- Brady Wern . . . . . . . . . 2 pts - Your answer of 17/18 is actually what I got my first time through!
- Kyle Misiaszek (new) . . 2 pts - Welcome to my contest; sorry I'm so slow these days (ok, years)
- Allen Druze . . . . . . . . . 1 pt - Good entry( 25/36); post-proverbial deadline of "next problem up"
- Zahi Teilelman . . . . . . 1 pt - Thanks for your submission; the 17/108 was a major part of it.
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