Finding unknown variables (OT)



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:






We need to know the values of all the variables. TIA.

123 to delete


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.


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.



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



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.



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.


Oh, I don't know about that...

... one does wonder how Ben Salinas got his start!



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


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.


Hence if a and g are zero, all the other variables can be 4 (with a minus for h).


Thanks for all your input. I concur with the majority and feel that this is an arithmetic cryptogram.


I'll say that E = 4..

Leaving 3eq in 7unkn...a task indeed! :)


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