Math Cross Puzzle

[nelynge]

The instructions are simple, fill in the blanks with integer numbers between 1 and 9 according to the arithmetic operations indicated. The precedence of the operations must be sequencial. For example: 4 - 2 / 8 + 2, then the solving order must be ((4 - 2) / 8) + 2.

Trying to solve:
1. Put some variables into the cross puzzle

2. Write down the equations:

• a + b = 9

• c + d = 5

• p + q = 5

• e - j = 2

• f + k = 7

• h - n = 5

• i - o = 2

• ( (ef - g + 5) / h ) + i = 9

• ( (jk / l) + m + n - o = 5

• (ac + l) - (g + p) = 7

• ( (5b / d) + m + q = 11

How do I take into account that the numbers are positive integers between 1 and 9
{a, b, c, d, e, f, g, h, i, j, k, l, m, n, p, q} must be a number between 1 and 9

Another thing that I'd noted is that some of the equations filter the possible values that a variable can be, for example:
p + q = 5
then {p, q} must be a number between 1 and 4

What areas of mathematics would lead me to a good solution of this problem?
What would be the steps to solve it.

 

[LivingMath]

Perhaps someone has a better answer, but I suspect the correct branch of mathematics for solving this problem would correctly be called trial and error.

 

[nelynge]

Perhaps someone has a better answer, but I suspect the correct branch of mathematics for solving this problem would correctly be called trial and error.

Hmmm, Is there a section on the site for trial and error?

Another thing that I've done (Don't know if it's the best way to tackle the problem)?
(5b / d) + m + q = 11

{m, q} must be integers between 1 and 9, then the minimal value that they can be added together would be: 2, so
(5b / d) + 2 <= 11

then
(5b / d) <= 9

(b / d) must be a positive integer no matter what, then the possible values for (b/d) would be 1, 2, or 3, etc.
so the only value possible for (b / d) would be 1 otherwise (5b / d) + m + q would be far from 11, then we conclude that:
b = d
(5 x 1) + m + q = 11

and
m + q = 6

then we can replace b = d on the 2nd equation c+d=5 and sum it with the 1st equation and we would have
a - c = 4
Don't know if this would help me to solve it tough, any comments about this aproach?

 

[LivingMath]

When I said trial and error was the correct branch of mathematics for solving this problem, what I meant was that you need to start plugging in numbers and see if you can get it to make sense.

Your general approach of finding the possible ranges for each value is the correct way to go. Then, using the restrictions you make, try putting some numbers into the equations and see if it makes sense. If not, try again with the other possible values. As you do this, you will be able to further restrict the possible range of each variable.

 

[nelynge]

When I said trial and error was the correct branch of mathematics for solving this problem, what I meant was that you need to start plugging in numbers and see if you can get it to make sense.

Your general approach of finding the possible ranges for each value is the correct way to go. Then, using the restrictions you make, try putting some numbers into the equations and see if it makes sense. If not, try again with the other possible values. As you do this, you will be able to further restrict the possible range of each variable.

Ok, start plugging the numbers, with the addition of possible range of the variables. Now, aren't there any other theoretically aproach for solving the whole problem or for narrowing the possible range of the variables? Anyone else?

 

[Denis]

Ok, start plugging the numbers, with the addition of possible range of the variables. Now, aren't there any other theoretically aproach for solving the whole problem or for narrowing the possible range of the variables? Anyone else?

Well, PERHAPS, but your problem has 29 solutions! In a b c d e f g h i j k l m n o p q order:
01: 5 4 1 4 5 1 1 9 8 3 6 6 4 4 6 3 2
02: 5 4 1 4 6 1 2 9 8 4 6 6 3 4 6 2 3
.....
28: 6 3 2 3 8 4 9 7 5 6 3 6 3 2 3 2 3
29: 6 3 2 3 8 5 9 6 3 6 2 6 3 1 1 2 3

Wrote a looper program, starting with:
loop a from 1 to 8 : b = 9 - a
loop c from 1 to 4 : d = 5 - c
loop e from 3 to 9 : j = e - 2
.....

Doesn't matter what "theoretical approaches" are devised, if there are
more than 1 solution, then you'll keep going in circles!

 

[nelynge]

Well, PERHAPS, but your problem has 29 solutions! In a b c d e f g h i j k l m n o p q order:
01: 5 4 1 4 5 1 1 9 8 3 6 6 4 4 6 3 2
02: 5 4 1 4 6 1 2 9 8 4 6 6 3 4 6 2 3
.....
28: 6 3 2 3 8 4 9 7 5 6 3 6 3 2 3 2 3
29: 6 3 2 3 8 5 9 6 3 6 2 6 3 1 1 2 3

Doesn't matter what "theoretical approaches" are devised, if there are
more than 1 solution, then you'll keep going in circles!

Hmmm, OK, thanks, I was thinking about making a program in an Excel sheet with formulas, but I wanted the other "manual" way... :evil:

But thanks for your opinion, 29 results, that would mean that there wasn't a possibility to get something like a = 8.

Another interesting finding looking into your solutions sets is that b = d .

Theorically speaking, this would be an "Equation System", if we had 17 equations with 17 variables, then the answer would be easy to find, if we had 16 equations with 17 variables, then the answer would be expressed as an equation with 2 variables, right? In our case we had 11 equations with 17 variables, but we had more information or constraint: the number must be integers between 1 and 9, this extra information, is there any way to express this extra information and use it for the solution of the problem? This extra information lead me to find that b = d, there aren't any other way to narrow the values of some other variables, for example, as I can see in your solutions, 5 < a < 6

I know I'm just going around, but maybe someone has the same curiosity. For example take into account the first 2 equations:

1) a + b = 9
2) c + d = 5

we know that a must be an integer between 1 and 8, and the same applies to b, we can't express that as an inequation (or we can)
1 <= a <= 8
1 <= b <= 8

the same for c and d
1 <= c <= 4
1 <= d <= 4

and with the finding of b = d
we could get that
a - c = 4
with this we can say that a must be an integer between 5 and 9, and for c, c must be an integer between 1 and 4
5 <= a <= 9
1 <= c <= 4
From the first equation we knew that a must be between 8 and 1, and with the third one we know that a must be between 5 and 9, so we can conclude that a must be between 5 and 8, something like an intersection of 2 inequations.
( 1 <= a <= 8 ) intersects ( 5 <= a <= 9 ) => ( 5 <= a <= 8 )
with this we had narrow down a to 4 possible values 5,6,7,8 and looking into the results of the program a has 2 values 5 and 6, and that is between 5 and 8.

So everything I said until know is to know if someone has more ideas about this, aside from brute force, my approach is somewhat intelligence and brute force.... thanks for everyone who reads this