Topic started by anu (@ 164.164.86.66) on Thu Aug 23 02:21:08 .
All times in EST +10:30 for IST.

Let this thread be devoted to discussing some of the rare maths puzzles that have captured our imagination. We can discuss the more popular ones too, but pls specify the source if they are taken from the net.

Responses:

• Old responses
  
• From: S H I T K I C K E R (@ 202.86.162.6) on: Sat Oct 13 10:16:45
  
  Out of Rs. 30/-, the bill was Rs. 25/-, the waiter got Rs 2/- and one rupee each was given back to each of the guys. So where's the f**king confusion?
  
  
• From: tracerbullet (@ 192.91.75.29) on: Mon Oct 15 10:04:28
  
  Why add 27+2 = Rs. 29/-? Makes no sense.
  Total spending = Rs.27/- = 9*3 = 25 (bill) + 2 tip)
  
  Analyze this.
  x+x+x+x+..+x (x times) = x^2 (any x)
  differentiating w.r.t x,
  1+1+1+1+..+1 (x times) = 2x
  x = 2x (any x)
  1 = 2
  How?
  
  
• From: anu (@ 203.200.144.141) on: Mon Oct 15 12:27:26
  
  substitute x=2y
  2y+2y+2y+...2y(2y times) = (2y)^2
  
  ie (2y + 2y+ 2y +.... y times) + (2y+2y+....y times) = 4y^2
  ie 2(2y + 2y+.... y times) = 4y^2
  ie 2y+2y+.... y times = 2y^2
  ie 2(y+y+....y times) = 2y^2
  ie y+y+... y times = y^2 which is true (since we can replace y by any other variable)
  
  hence there's no fallacy..
  
  have a nice day :)
  
  
• From: RR (@ krdlfirewall.krdl.org.sg) on: Mon Oct 15 22:40:19
  
  tracerbullet: you cannot apply the differentiating operator to the terms of the sum since it's not a constant sum, i.e. the number of terms depends on x.
  
  
• From: tracerbullet (@ news.ti.com) on: Mon Oct 15 23:57:52
  
  anu:
  >ie y+y+... y times = y^2 which is true (since we can replace y by any other variable)
  This is precisely my premise I have assumed at the start. You have proved it true. Tell us how an untrue statement of 1 = 2 results out of this true statement. Whither lieth the errment?
  
  RR:
  The differentiation operation can be done, as you remember, on constants (gives zer0) and variable expressions. Why, the value of x^2 depends on x; so do you mean you can't differentiate x^2?
  
  
• From: tracerbullet (@ news.ti.com) on: Tue Oct 16 00:23:49
  
  I'm afraid I didn't quite understand what exactly you meant until a moment ago.
  > you cannot apply the differentiating operator to the terms of the sum since it's not a constant sum
  Why do you say so? As far as I know, differentiation can be applied to additions irrespective of how many terms are involved, and of whether the no. of terms is constant or not.
  
  
• From: sree (@ 62.3.1.114) on: Tue Oct 16 08:18:01
  
  I DON’T THINK IT WILL BE CRITICAL PUZZLE. HOWEVER LET ME HAVE AN ANSWER PLEASE.
  
  The milk booth guy has 10 litres of milk with him in a 10 litres can.
  
  He also got two empty cans of 7 litres and 3 litres.
  
  How can he measure 5 litres each to serve his two customers using these three cans?
  
  
  
• From: Sridhar (@ dialup-225-6.bol.net.in) on: Tue Oct 16 11:16:49
  
  Sree,
  
  I have tried.Tell me whether it is correct
  
  A - 10 lt can
  
  B - 7 lt can
  
  C - 7 lt can
  
  D - 3 lt can
  
  Put the milk in such a way that empty A by filling B with 7 litres.Then fill C with 3 litres.
  Put B in D.
  
  Now
  A is 0
  B is 4
  C is 3
  D is 3
  
  C + D = 6
  
  Put
  6 in A
  4 in B
  0 in C
  0 in D
  
  Take D
  
  Put B in D
  
  B will be 1
  D will be 3
  
  Put B in C and D in B and C in D
  
  Now
  A is 6
  B is 3
  D is 1
  c is 0
  
  A B C D
  
  6 3 0 1
  
  
  Put A in B.....B becomes 7
  
  A becomes 2....Put B in D ..B becomes 5
  
  D becomes 3..Put D in A ..A becomes 5
  
  That's
  
  A =5 Litres
  
  B = 5 Litres
  
  I may have taken the circuitous route.A simple step should be there.
  
  
  
• From: Just a thought (@ 64.124.150.137.safeweb.com) on: Tue Oct 16 14:45:55
  
  Here is probably the same answer as Sridhar with a slight twist. Sridhar tries this with 4 cans, I assumed the question to have 3 containers: 1 original 10l, one 7l and one 3l. However, there is still customer's containers. The restriction I assumed was that you cannot take back milk from customers container - you can only pour into it and you don't know the volume of the customer's container. (i.e, in Sridhar's case taking milk out of C, where C can be assumed customers can, is prohibited).
  
  Wondering if there is a better answer when I get 5l and 5l in container A and B (i.e, 10l and 7l) without ever using customer's/any other container.
  
  10l 7l 3l C1 C2
  10 0 0 0
  3 7 0 0
  3 4 3 0
  6 4 0 0
  6 1 3 0
  6 0 3 1
  0 6 3 1
  0 7 2 1
  2 7 0 1
  2 4 3 1
  2 0 3 5
  0 0 0 5 5
  
  
• From: Just a thought (@ 64.124.150.133.safeweb.com) on: Tue Oct 16 15:38:14
  
  tracerbullet
  
  When you add these x times, the "x" times is still the variable with respect to which you are differentiating. However, you seem to treat that as a constant which is what is wrong.
  
  If instead of x times you said 'c' times (which is a constant independent of variable with which you differentiate), then you obviously won't arrive at the erroneous result. x+x+x = 3x, differentiate it you will arrive at 3 = 3.
  
  
• From: RR (@ krdlfirewall.krdl.org.sg) on: Tue Oct 16 22:48:33
  
  > differentiation can be applied to additions irrespective of how many terms are involved, and of whether the no. of terms is constant or not.
  
  No. The distributive rule doesn't hold for variable sums.
  
  
• From: sree (@ 62.3.1.114) on: Wed Oct 17 00:28:30
  
  hey thanks, but it can be done in 5 steps.
  
  
• From: sree (@ 62.3.1.114) on: Wed Oct 17 00:31:29
  
  oh ! there will be only three cans of course...
  
  
• From: tracerbullet (@ 192.91.75.29) on: Wed Oct 17 04:50:48
  
  Jat,
  The x in "x times" is NOT a constant, and I have no idea where you think I have treated it a constant. It's x, not c.
  Yeah, if it's a constant, there's no problem differentiating. But whassa problem with it being a var?
  
  RR,
  It does. Atleast for apparently all other cases.
  Consider
  x+x+x+x....+x (y times) = xy (any y)
  dx+dx+dx...+dx(y times) = xdy+ydx
  Divide by dx
  1+1+1......+1 (y times) = x(dy/dx)+y
  Since it's any y, ie y independent of x,
  dy/dx = 0
  So,
  1+1+1......+1 (y times) = y
  Fits, na? :)
  (That's almost a total giveaway)
  
  
• From: satheesha (@ dialpool-210-214-210-238.maa.sify.net) on: Thu Dec 13 10:48:08
  
  Dear friends
  x+x+... xtimes=x^2, this statement is true for all the real nos.
  but the give function x^2 in the form x+x+...+xtimes is a divergent series, and term-by-term differentiation of a divergent series is not allowed, so the contradiction you are getting 2=1, For term-by-term differentiation series must be convergent and may be needed to be uniform convergence as i think. So do check yourself again friends
  byeeeeeeee
  contact me at "satheesha123@rediffmail.com"
  thanks and bye
  --satheesha.
  
  
• From: tracerbullet (@ proxyle02.ext.ti.com) on: Fri Jan 18 13:50:00
  
  WARNING: Answer for qn. posted by self on Mon Oct 15 10:04:28 below. Don't read if you wanna solve.
  ---------------------------------------------
  Giving away since none's really got the ans.
  It's actually pretty simple. Differentiation is valid only for continuous functions, and
  x+x+x+x+..+x (x times) is not a continuous fn. (like it is undefined for x = 4.2). So differentiation of LHS is senseless.
  ---------------------------------------------
  
  
• From: hipoz (@ tv124154.wright.edu) on: Fri Jan 18 15:08:53
  
  >>Tip paid to the bearer is included in the 9*3=27.
  
  The Re 1/- is in the pocket of one of the three friends.<<
  
  How is it possible?
  
  EACH person has given out $9 each (as they got back $1 each. $10-$1=$9). They have given $2 as a tip to bearer.but where is the remaining $1.
  
  
  
• From: hipoz (@ tv124154.wright.edu) on: Fri Jan 18 15:22:05
  
  ok kewl,
  i got it.....tip + spending = $27 ....and they got remaining money...cool man..
  
  

Want to post a response?