Jump to content
BrainDen.com - Brain Teasers

bonanova

Moderator
  • Content Count

    6967
  • Joined

  • Last visited

  • Days Won

    65

Everything posted by bonanova

  1. Each weighing has three outcomes: Left side is [lighter than] [equal to] [heavier than] Right side. Three weighings can thus discern among [3]x[3]x[3]= 27 cases. We have only 24 cases: one of 12 balls is heavier or lighter than the rest. So we can solve the problem, so long as ... [1] The first weighing reduces the cases to no more than 9. [2] The second weighing reduces the cases to no more than 3. [3] The third weighing then distinguishes among 3 or fewer cases. First weighing: Set aside four balls. Why? Because, if the first weighing balances, we have 8 [fewer than 9] cases: One of the 4 excluded balls is heavier or lighter. [1] Weigh 1 2 3 4 5 6 7 8 Outcome 1[a] 1 2 3 4 balances 5 6 7 8. Since only one ball is odd, 1 2 3 4 5 6 7 8 must all be normal. We have 8 cases: 9 10 11 or 12 is H or L [shorthand: H=heavier; L=lighter] [2] Weigh 1 2 3 9 10 11 [we know 1 2 3 are normal] If this balances, 9 10 11 are also normal, and 12 is H or L. [3] Weigh 12 [any other ball] If 12 rises, then 12L; if 12 falls, then 12H. If [2] 9 10 11 rises, we have 3 cases: 9L, 10L or 11L [3] Weigh 9 10. If [3] balances, then 11L; If 10 rises, then 10L; If 9 rises, then 9L If [2] 9 10 11 falls, we have 3 cases: 9H, 10H or 11H [3] Weigh 9 10. If [3] balances, then 11H; If 10 falls, then 10H; If 9 falls, then 9H end of Outcome 1[a]: balance. Outcome 1: 1 2 3 4 falls, and 5 6 7 8 rises. This gives 8 cases: 1H, 2H, 3H, 4H, 5L, 6L, 7L, or 8L The second weighing in this case becomes tricky. Remember each of its three outcomes can lead to no more than 3 cases for the third weighing to resolve. Again, we exclude a number of balls and involve the others. We exclude any three of the balls. Why? Because if the other [included] balls balance, we have exactly 3 cases. Without loss of generality we exclude balls 1 2 3. Since that leaves an odd number of balls, 4 5 6 7 8, we need to use one of the normal balls. Finally we choose which three to weigh against the others. And here's the only hard part of this problem. We must mix some of the possibly light balls with some of the possibly heavy balls. Otherwise, one of the outcomes of the second weighing will leave us with more than 3 cases, and the third weighing will not resolve this. [2] weigh 4[H] 5[L] 6[L] 1[normal] 7[L] 8[L] in parentheses I've indicated the POSSIBLE cases that we have determined: 4[H] means 4 is heavier if it's not normal. Outcome 2[a]: 4 5 6 balances 1 7 8 These balls are all normal. We have 3 cases: 1H 2H or 3H. [3] weigh 1 2 If [3] balances, then 1 and 2 are normal, and 3H if 2 falls, then 2H if 1 falls, then 1H Outcome 2b: 4 5 6 falls, and 1 7 8 rises. We have 3 cases: 4H 7L or 8L [3] weigh 7 8 if [3] balances, then 7 and 8 are normal, and 4H if 7 rises, then 7L if 8 rises, then 8L Outcome 2[c]: 4 5 6 rises, and 1 7 8 falls. We have 2 cases: 5L or 6L. [3] weigh 1[normal] 5[L] If 5 rises, then 5L If balance, then 5 is normal, and 6L Now we can go back and take the remaining case Outcome 1[c]: 1 2 3 4 rises, and 5 6 7 8 falls. to distinguish among the remaining 8 cases: 1L 2L 3L 4L 5H 6H 7H and 8H exactly as we analyzed Outcome 1b. Simply substitute H for L and v.v. Problem solved.
  2. bonanova

    Barbera

    The puzzle asks: Who am I to Barbara - NOT who is Barbara to me. Hence the correct answer is ... Daughter. Unless of course I am a male. Then the correct answer is not in the choices listed, it's Son-in-Law.
×
×
  • Create New...