HP Forum Archive 14

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Finding unknown variables (OT) Message #1 Posted by Frank on 17 Mar 2005, 7:17 a.m. Greetings, My daughter received this problem to solve. She is presently in 5th grade. We have tried to solve it, to no avail. Thanks you for your help, here it is: c=4 d-a=i b-c=g ac+ic=de sg-de=hi We need to know the values of all the variables. TIA. 123 to delete

Re: Finding unknown variables (OT) Message #2 Posted by John Limpert on 17 Mar 2005, 8:42 a.m., in response to message #1 by Frank It looks more like an arithmetic cryptogram than a set of equations. In an arithmetic cryptogram, each digit in the solution is replaced with a letter. In your example, you are given a starting clue, that C stands for 4. Using the rules of mathematics and logic, the goal is to deduce which letters stand for which digits. For example, E = C + C, therefore E stands for 8.

I think your right. Message #3 Posted by Larry Corrado on 17 Mar 2005, 8:01 p.m., in response to message #2 by John Limpert When I first read the problem, I thought the same thing, that this is really an "arithmetic cryptogram," with letters respresenting digits of numbers. I doubt that fifth graders are solving simultaneous equations in this many unknowns. Ciao, larry

Me too ! Message #4 Posted by Tom Sherman on 18 Mar 2005, 10:50 a.m., in response to message #3 by Larry Corrado John and Larry are right. ac, for example, should probably be read in this puzzle as a two-digit number rather than as a product. The solution then would be: 1= a, 2= h, 3= g, 4= c, 5= i, 6= d, 7= b, 8= e, 9= s May Frank's daughter enjoy mathematics all her life, remembering the good times she had with her dad! Cheers, Tom

Re: Me too ! Message #5 Posted by Bill (Smithville, NJ) on 18 Mar 2005, 11:35 a.m., in response to message #4 by Tom Sherman Tom, Thanks for posting the solution. I had played around with it a little bit yesteday morning, but kept coming up with a contridiction in one of the solution grids I had made. Had to quit and start doing some real work, so I'll now go back and see what I was doing wrong. Bill

Re: Me too ! Message #6 Posted by Frank on 19 Mar 2005, 3:29 p.m., in response to message #4 by Tom Sherman Thanks for those kind words Tom and thanks to everyone that made the effort to enlighten me in my quest to help my daughter. Your effort and time is greatly appreciated.

Re: I think your right. Message #7 Posted by Ed Look on 19 Mar 2005, 10:30 p.m., in response to message #3 by Larry Corrado Oh, I don't know about that... ... one does wonder how Ben Salinas got his start!

Re: Finding unknown variables (OT) Message #8 Posted by Tom Sherman on 17 Mar 2005, 2:28 p.m., in response to message #1 by Frank Frank, If this is a problem in numbers (rather than some sort of word code game), here is one amateur's initial reaction: C is given as a constant, so we have 8 variables and 4 equations, and therefore no unique solution. An infinite number of solutions can be found if we assign enough (but not too many) zero values to the variables. Here is one possibility. Let g, e, h, and d all be zero. Then b=c=4, -a=i, and s = anything. So one solution would be: c= 4, b= 4, a= -1, i= 1, s= 1, g= 0, e= 0, h= 0, d= 0 Not very pretty. I'm sure that more interesting solutions can be found! Cheers, Tom

P.S. Message #9 Posted by Tom Sherman on 17 Mar 2005, 6:21 p.m., in response to message #8 by Tom Sherman Here's one with fewer zero values: c= 4, e= 4, b= 4, h= -4, d= 1, i= 1, s= 1, a= 0, g= 0 s could be any number; so could d and i, so long as they equal one another.

P.S.2 Message #10 Posted by Tom Sherman on 17 Mar 2005, 6:56 p.m., in response to message #9 by Tom Sherman Hence if a and g are zero, all the other variables can be 4 (with a minus for h).

Re: Finding unknown variables (OT) Message #11 Posted by Frank on 17 Mar 2005, 8:15 p.m., in response to message #1 by Frank Thanks for all your input. I concur with the majority and feel that this is an arithmetic cryptogram.

Re: Finding unknown variables (OT) Message #12 Posted by Eric Lundgren on 18 Mar 2005, 1:50 a.m., in response to message #11 by Frank I'll say that E = 4.. Leaving 3eq in 7unkn...a task indeed! :) Eric

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