Odd Question - Hope You Know It

B

ByronGliik

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My problem is I need to identify each letter. I have a couple hints as to what their values are but have not been able to figure it out. I need to know the value of each but more importantly I would like to know the math behind it. If I posted this in he wrong section I apologize! I am new here.

A + B = C
D + E = F

Givens:

1. The difference between "A" and "D" is a value of 2.
2. The difference between "B" and "E" is a value of 1.
3. The difference between "C" and "F" is a value of 3.
4. All numbers are positive real numbers from a value of 0 to 50.

Please Help!
 
ByronGliik said:
My problem is I need to identify each letter. I have a couple hints as to what their values are but have not been able to figure it out. I need to know the value of each but more importantly I would like to know the math behind it. If I posted this in he wrong section I apologize! I am new here.

A + B = C
D + E = F

Givens:

1. The difference between "A" and "D" is a value of 2.
2. The difference between "B" and "E" is a value of 1.
3. The difference between "C" and "F" is a value of 3.
4. All numbers are positive real numbers from a value of 0 to 50.

Please Help!
Click to expand...
My son used to have problems of this sort in his early grades. I think maybe some arithmetic teachers like to "burn" class time occasionally.

In Rule (4), I will assume that "all numbers" refers to "the values of the variables A, B, C, D, E, and F". The Rules (1), (2), and (3) allow for some setup to be done, but the rest will be guess-n-check, ruling out options until you find an option that works. For instance, since A and D differ by 2, then A = D + 2 or D = A + 2. Since B and E differ by 1, then B = E + 1 or E = B + 1. Much the same can be said for C and F.

Pick sets of options, and see where they lead. If you get something that works, you're done. If not, you try something else. Allow yourself plenty of time (and plenty of scratch paper), and work in an orderly fashion through the options until you find a solution.

Have fun! ;)
 
I'd also note that there's not going to be a unique solution for the values of A, B, C, D, E, and F. Thankfully, Rule #4 limits the number of possible solutions... although I'm not sure exactly how many solutions there are, off the top of my head.
 
ByronGliik said:
My problem is I need to identify each letter. I have a couple hints as to what their values are but have not been able to figure it out. I need to know the value of each but more importantly I would like to know the math behind it. If I posted this in he wrong section I apologize! I am new here.

A + B = C
D + E = F

Givens:

1. The difference between "A" and "D" is a value of 2.
2. The difference between "B" and "E" is a value of 1.
3. The difference between "C" and "F" is a value of 3.
4. All numbers are positive real numbers from a value of 0 to 50.

Please Help!
Click to expand...
First let's agree that "The difference between "A" and "D" is a value of 2" means A-D=2 [you could use D-A=2 if you like and it would change the equations but you should be consistent]. So you have, re-arranging the equations a bit [and assuming I haven't made a mistake in the transcription],
(a) 1 A + 1 B - 1 C + 0 D + 0 E + 0 F = 0 from first equation above
(b) 0 A + 1 B + 0 C + 0 D - 1 E + 0 F = 1 from 2.
(c) 0 A + 0 B + 1 C + 0 D + 0 E - 1 F = 3 from 3.
(d) 0 A + 0 B + 0 C + 1 D + 1 E - 1 F = 0 from second equation above
(e) 1 A + 0 B + 0 C - 1 D + 0 E + 0 F = 2 from 1.
This is a system of 5 linear equations with 6 unknowns which has either zero or an infinite number of solutions [disregarding 4. above for the moment]. This is somewhat more advanced than just Arithmetic [the section you have posted this in]. You can get an introduction to the subject at
http://www.mathsisfun.com/algebra/systems-linear-equations.html
among others.

For your particular problem it turns out that you can choose E and F as you wish and compute the other values [subject to 4. above of course]. For example, if you choose E=0 and F=1 then A=3, B=1, C=4, and D=1 which we will write as (3,1,4,1,0,1), that is (A,B,C,D,E,F)=(3,1,4,1,0,1). We also have as a solution (3,6,9,1,5,6) as well as others.


EDIT: Oh, there is nothing magic about choosing to pick E and F. You could pick any two and compute the others but limiting it to E and F would allow for, IMO, a much easier systematic search for solutions if one were inclined to do that.
 
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Denis said:
2498 solutions!
....a, b, c, d, e, f
1: 0, 0, 0, 2, 1, 3
2: 0, 1, 1, 2, 2, 4
...
2497: 48, 2, 50, 50, 3, 53
2498: 49, 1, 50, 47, 0, 47
Click to expand...
You know, that's funny (possibly both ha ha and peculiar). I read the problem as integer solutions also but can't find that now.
 
toddhicks209 said:
I came up with a+b=c to have a value of 1+1=2. The value of d+e=f is 3+2=6.
Click to expand...

Set up the equations [slightly modified from my post above]
(1) B = 1 + E from 2. The difference between "B" and "E" is a value of 1.
(2) C = 3 + F from 3. The difference between "C" and "F" is a value of 3.
(3) D =-E + F from D + E = F
(4) A =-B + C from A + B = C

Now choose E and F [for example start with E=0 and go through F=0,1,2,3,...,50. Then E=1, F=0,1,2,3,..., 50, Then ...]. Use the above formulas to compute A, B, C, and D. Check that 'The difference between "A" and "D" is a value of 2.', i.e. A-D=2 and check whether the values of A, B, C, and D are between 0 and 50 inclusive. If so you have a solution. If not go on to the next trial [in the example, until you get to E=50 and F=50].

Actually you can cut down on some the work since
-from (1) E must be less than 50 otherwise B would greater than fifty
-from (2) F must be less than 48 otherwise C would be greater than 50
-from (3) F must be greater than or equal to E or D would be less than zero

That would change the example above into
Code: 
(a1) E=0; F=0, 1, 2, .............., 47
(a2) E=1; F=    1, 2, 3, .........., 47
(a3) E=2; F=        2, 3, 4, ......, 47
...
(a48)E=47; F=                        47

EDIT: BTW: In the example you gave above, I'm understanding that A=1, B=1, C=2, D=3, E=2, and thus F=5, not F=6. Also, you interpreted 'The difference between "B" and "E" is a value of 1' as E-B=1 rather than B-E=1 as I did. If that is the case the my equations above would change slightly, i.e.
B = E - 1
C = F - 3
and the restrictions on E and F would change. You example would then be choosing E=2 and F=6 in the 'corrected' scheme.
 
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