Solving Number Triangle Puzzle with Trial and Error

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  • #1
musicgold
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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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  • #2
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.
 
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  • #3
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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  • #4
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.
 
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  • #5
musicgold said:
Thanks. I have corrected the equations.
Good. The second (corrected) equation is all you need.
 
  • #6
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?
 
  • #7
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.
 
  • #8
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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  • #9
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
 
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  • #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.
 
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  • #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.
 
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  • #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?
 
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  • #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.
 
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  • #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?
 
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  • #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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