Solving Number Triangle Puzzle with Trial and Error

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The discussion revolves around solving a number triangle puzzle using algebra and trial and error. Initial attempts to find solutions using algebra were hindered by an over-determined system of equations, leading to the necessity of trial and error. Participants explored various values for the variables and identified constraints, such as the requirement for certain sums to be multiples of 3. Several potential solutions for different sums (S) were proposed, with specific values yielding valid configurations while others were ruled out due to internal constraints. The conversation highlights the complexity of the problem and the need for a systematic approach to identify all possible solutions.
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Homework Statement
I am trying to solve the puzzle shown below.
Relevant Equations
I could find the 3 equations shown below that appear to describe the situation.
First I tried to solve this with algebra, but there are not enough equations:
a+ b + c + d + e + f + g + h = 36
S = 12 + (d +f + a)/3 ........... ( d +f + a has to be a multiple of 3)
a + b + c = e + f
a + h + g = d + e

So I had to resort to the trial and error to find the solution below:
a, b, c, d : 6, 1, 5, 7
a, h, g, f : 6, 2, 3, 8
e : 4
Is there an elegant way / pattern to find other possible solutions?

Thanks

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musicgold said:
Homework Statement: I am trying to solve the puzzle shown below.
Relevant Equations: I could find the 3 equations shown below that appear to describe the situation.

Is there an elegant way / pattern to find other possible solutions?
Yes, I can think of a way to answer the question without finding all possible combinations. Plus, your first equation is incorrect, IMO.
 
Your system is way over-determined, more variables than equations, so you won't just find allsolution without additional assumptions. Though unless your equations contradict each other, there will be a solution necessarily. Using a matrix will allow you to describe the solution space.
 
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Hill said:
Yes, I can think of a way to answer the question without finding all possible combinations. Plus, your first equation is incorrect, IMO.
Thanks. I have corrected the equations.
 
musicgold said:
Thanks. I have corrected the equations.
Good. The second (corrected) equation is all you need.
 
Hill said:
Good. The second (corrected) equation is all you need.
S = 12 + (d +f + a)/3

So (d +f + a) has to be a multiple of 3. Here are some possibilities.

a d f
1 2 3
1 3 5
1 3 8
...
5 6 7
...
6 7 8

But again there are many possibilities. Do I have to test all of them out?
 
musicgold said:
But again there are many possibilities. Do I have to test all of them out?
No. It does not matter what a, d, and f are. The only important thing, for S's to be different, is what a+d+f is.
 
You'd need all necessary conditions on all your variables, not just S. Notice you can also conclude S is a multiple of 3. Edit: This is wrong; its (a+b+c) which must be multiples of 3, as Hill correctly stated.
In the matrix algebra approach, you will end up with a solution space where you'll have to make some arbitrary choices for your solution basis vectors.
 
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WWGD said:
You'd need all necessary conditions on all your variables, not just S. Notice you can also conclude S is a multiple of 3.
I seem to get a solution for S=16.

When a=1, d=7, f =4
The following values seem to be working out.
a, b, c, d : 1, 2, 6, 7
a, h, g, f : 1, 8, 3, 4
e : 5
 
  • #10
musicgold said:
I seem to get a solution for S=16.

When a=1, d=7, f =4
The following values seem to be working out.
a, b, c, d : 1, 2, 6, 7
a, h, g, f : 1, 8, 3, 4
e : 5
Oops, my bad, will edit. It's 12+ ## \frac{a+b+c}{3}##, and 12 and ##a+b+c##, which must be divisible by 3.
 
  • #11
A quasi brute force method:
Your second equation gives you a small number of S values.
See if there is at least one solution for each of them.
 
  • #12
You can further remove some of these few S values without trying. For example, for each S you have (a+d+f) and (e+d+f). In one case, it makes e=a. Drop it.
 
  • #13
Frabjous said:
A quasi brute force method:
Your second equation gives you a small number of S values.
See if there is at least one solution for each of them.
I am able to get solutions for the following values of S: 19, 17, 16.

The arrangement doesn't seem to work for S = 18, S=14, S =15. There appears to be some internal constraint that I am not able to see.

I noticed the following constraints:
1. e must be < 9. We know that S = d+e+f . S - d - f =< 9.
2. b+c, g+h need to provide the values required to create a S.

S = 14 is not possible because, at the only possible a, d, f (1, 2, 3), e has to greater than 9.
S = 15 is not possible because the remaining numbers are not meeting the second constraint.

Any other way to understand the constraints in this situation?

Also, I am not close to any of the answer choices given in the problem.
 
  • #14
musicgold said:
I am able to get solutions for the following values of S: 19, 17, 16.

The arrangement doesn't seem to work for S = 18, S=14, S =15. There appears to be some internal constraint that I am not able to see.

I noticed the following constraints:
1. e must be < 9. We know that S = d+e+f . S - d - f =< 9.
2. b+c, g+h need to provide the values required to create a S.

S = 14 is not possible because, at the only possible a, d, f (1, 2, 3), e has to greater than 9.
S = 15 is not possible because the remaining numbers are not meeting the second constraint.

Any other way to understand the constraints in this situation?

Also, I am not close to any of the answer choices given in the problem.
15 works. Hint: try a=1.
18 does not work because d+e+f=18=a+d+f requiring a=e
 
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  • #15
Frabjous said:
15 works. Hint: try a=1.
Not sure what I am missing. :oldconfused:

For S=15, a, d, f : 1, 3, 5
e = 15 - 3- 5 = 7

We are now left with 2, 4, 6, and 8.
left leg: 1 + b + c + 3 = 15 -- no odd numbers left to make this work
right leg: 1 + h + g + 5 = 15 -- no odd numbers left to make this work
 
  • #16
musicgold said:
Not sure what I am missing. :oldconfused:

For S=15, a, d, f : 1, 3, 5
Did you skip 1, 2, 6?
 
  • #17
Hill said:
Did you skip 1, 2, 6?
Ahhh....sorry!:sorry:

Thanks
 
  • #18
WWGD said:
Your system is way over-determined
Under-determined?
WWGD said:
Oops, my bad, will edit. It's 12+ ## \frac{a+b+c}{3}##, and 12 and ##a+b+c##, which must be divisible by 3.
Eh? Seems to me it's ##S=12+ \frac{a+d+f}{3}##, as in post #1. This gives ##14\leq S\leq 19##.

My approach was to combine that with ##S=d+e+f## to obtain that ##e-a=36-2S##.
Since the difference of any two of the numbers is 1 to 7, we are left with S limited to being one of 15, 16, 17, 19.
This does not prove they are all possible. The easy way to finish it is to note that only option E satisfies this.
 
  • #19
haruspex said:
Under-determined?

Eh? Seems to me it's ##S=12+ \frac{a+d+f}{3}##, as in post #1. This gives ##14\leq S\leq 19##.
Could you please explain how you reaching to ##14\leq S\leq 19##

haruspex said:
My approach was to combine that with ##S=d+e+f## to obtain that ##e-a=36-2S##.
Since the difference of any two of the numbers is 1 to 7, we are left with S limited to being one of 15, 16, 17, 19.
e - a has to be a negative number for S=19
 
  • #20
musicgold said:
Could you please explain how you reaching to ##14\leq S\leq 19##e - a has to be a negative number for S=19
My bad, over-determined.
 
  • #21
musicgold said:
Could you please explain how you reaching to ##14\leq S\leq 19##
##1+2+3\leq a+d+f\leq 6+7+8##
musicgold said:
e - a has to be a negative number for S=19
Yes. So?
 
  • #22
musicgold said:
Could you please explain how you reaching to ##14\leq S\leq 19##
What are the three smallest numbers you can possibly choose for a, d, and f ? ##\displaystyle 1,\,2,\,3 \ ## .

What are the three largest numbers you can possibly choose for a, d, and f ? ##\displaystyle 6,\,7,\,8 \ ## .

What does each of the above give for ##\displaystyle S=12+ \dfrac{a+d+f}{3} \ ## ?

I see that @haruspex beat me to it.
 

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