# What method can be used to solve a numeric finding puzzle involving getting the right sum?

I've been trying to solve this puzzle at no avail. Can somebody help me?

The problem is as follows:

In each circle from the figure below write down an integer number between $1$ to $9$. Each one of them must be different from each other. The resulting sum from those numbers joined by three circles in a straight line and those joined by arrows in the direction indicated be equal to $18$. Which number must be written in the circle painted by an orange shade?

$\textrm{The alternatives given were:}$

• 7
• 4
• 5
• 3
• 6

The only thing I could come up with was to make a sequence of dividing $18$ into different choices for summing that number between $1$ to $9$.

Therefore:

$1+8+9=18$

$2+7+9=18$

$3+6+9=18$

$4+5+9=18$

and the rest would be repeating those numbers in different order like

$5+4+9=18$

$6+3+9=18$

$7+2+9=18$

$8+1+9=18$

So the preceding can't be:

But this other combination can also reach to $18$:

$8+7+3=18$

$8+6+4=18$

However at this point I got tangled up with many choices therefore end up with clear clue to take to get $18$.

How can I solve this problem without just guessing?. Is there an algorithm, formula or method that can be followed orderly to avoid wasting much time?.

I would like someone could re drawn the question and explain me with much details possible.

Please do not just fill out the drawing right away. I know maybe you have the ability to do this, but this is not what I'm looking for but rather a more step-by-step solution which has proven that it works in many similar scenarios.

Edit: I currently replaced the previous picture to this version which is the right one.

So you have $1+8+9, 2+7+9, 3+6+9, 3+7+8, 4+5+9, 4+6+8, 5+6+7$ as the seven possible sums, and it wasn't as bad as you feared. There are six rows of three in the diagram, so one of these sums is unused.

$9$ appears in four sums, $8, 7, 6$ appear in three, $5, 4, 3$ appear in two and $2,1$ in just one sum.

There are four circles in the diagram which appear in three rows, and none appear in four. So these four must accommodate $9,8,7,6$ and it is one of the four sets containing $9$ which is not used.

With these clues and some sensible trial it should be possible to fit the numbers in the grid. Actually it is now quite easy to identify what the number in the red circle must be.

• I haven't found the answer yet. As mentioned a graphical approach would had been better for me to understand as I'm lost where to put the numbers. – Chris Steinbeck Bell Apr 6 '18 at 14:48
• @ChrisSteinbeckBell The corners of the left-hand square have to be taken from $9,8,7,6$ and the centre square therefore has to make a common sum with the two diagonally opposite pairs. So you need to split $9,8,7,6$ into two pairs which add to the same. – Mark Bennet Apr 6 '18 at 15:32
• I tried to fill out the circles by using the clues you mentioned and I got to be $3$ the circle in the orange shade but the numbers around do not seem to sum $18$. Namely $6+8+9+2=25$. Could it be that I am not understanding clearly what you wanted to say? – Chris Steinbeck Bell Apr 7 '18 at 19:35
• @ChrisSteinbeckBell Not sure what your problem is, but you can trace paths longer than $3$ in the diagram. I was assuming that straight lines only were involved, otherwise the problem is inconsistent. You are right that $3$ goes in he middle - you get $9+3+6=8+3+7=18$ – Mark Bennet Apr 7 '18 at 21:07
• I did filled out the circles with numbers using your clues. But when I tried to check if the sum by following the arrows gives $18$ it didn't work. Maybe I'm not getting it right. In comments I cannot add a picture of how I filled the drawing. But I can give you the sequence. Let $6,7,5$ be the first sequence of three circles going from the tip in the left to the circle which is on the most northeastern position. $6,3,9,1$ the sequence from circles going from the same tip but now crossing the circle in orange ending with the circle inmediately following the one which is on the – Chris Steinbeck Bell Apr 8 '18 at 15:43

Let the orange circle be $X$. Let the four circles immediately adjacent to $X$ be $A,B,C,D$ so that $A+X + C = B+X+D = 18$. You can label the rest of the circles what you wish.

===

Notice that $A,B,C,D$ are all involved in $3$ sums. And these are the only four circles that are involved in that many sums.

Claim: the only numbers that can be involved in $3$ sums are $6,7,8,9$

Proof: If $k$ is involved in three sums, that is; If $k + h+i = k +j+l = k + m+n=18$ then $h+i = j+ l = m +n = 18 - k$.

If $k= 1,2$ then $h+i = j+ l = m+n = 17,16$ but $9,8$ and $9,7$ are the only pairs that add to $17$ or $16$ so you can't have three distinct pairs.

If $k= 3,4$ then $h+i = j+l = m+n = 15,16$ but there are only two pairs that add to each of those sums. $8+7=9+6=15; 8+6=9+5=14$. So, again, you can't have three distinct pairs.

If $k = 5$ then $h+i = j+ l = m+n = 13=9+4=8+5=7+6$. But those are are the only possible three pairs that add to $13$ but none of the $h,i,j,l,m,n$ may equal $k = 5$. So you can't have three distinct pairs that do not also include $k$.

However if $k=6,7,8,9$ is is possible to have three distinct pairs: $k=6$ then $3+9=4+8=5+7=18-k = 12$. If $k=7$ then $2+9,3+8,5+6 = 18-k =11$. If $k=8$ then $1+9=3+7=4+6=18-k =10$ and if $k=9$ then $1+8=2+7=3+6=4+5=18-k =9$.

So $\{A,B,C,D\}=\{6,7,8,9\}$.

...

Now $A + X + C = 18$ and $B + X + D = 18$ so $A+B+C+D+2X =36$. But $A,B,C,D = \{6,7,8,9\}$ so $6 + 7 + 8 + 9 + 2X= 30 + 2x = 36$.

So $X = 3$

====

An altermative. If you add up all the six sums you get:

$3A + 3B + 3C + 3D + 2X + E+F+G +H = 6*18$ and $A +... + H = 1+ ...+9 = 45$ so

$2A + 2B + 2C + 2D + X = 6*18 - 45$. And $A+ X + C = 18$ so

$A + 2B + C + 2D = 5*18 - 45$

$A + 2B + C + 2D + X = 5*18 - 45 + X$

$2B + 2D = 4*18 - 45 + X$. As $B+ X + D = 18$.

$2B + 2D+ 2X = 4*18 - 45+3X$.

$0 = 2*18 - 45 + 3X$

$3X = 45 - 2*18$

$X = 15 - 2*6 = 3$.

• It helps to be a fan of "Sum Sudoku" puzzles. You very quickly learn that only there is only one pair that add up to 17 or 16, and two that add to 15 or 14. so or the three that add to 13 one of them has a 5. – fleablood Apr 6 '18 at 23:05
• May are may not be worth noting, any assignment of $A,B,C,D$ to $9,,8,7,6$ so that $A+C = B +D =15$ will work. Probably not worth noting. – fleablood Apr 6 '18 at 23:09
• @fleblood It seems that I'm not a fan of Sudoku therefore it gets really very troublesome for me to solve these kinds of problems. As maybe I had mentioned before, these kind of problems are better understood if it has some kind of visual aid other than just relying on Latex. About the pairs you mentioned which add up to $17$ and $16$ which are them, the same goes for the rest. At this point I'm lost. – Chris Steinbeck Bell Apr 7 '18 at 19:20

Fellow the image Below:

$A + B + C + D +E +F + G+ H+ X = 1 + 2+3+4+5+6+7+8+9 = 45$

so

$X+ C +F = (A+B+C+D+E+F+G+H+X) - (B+E+H) - (A+D+G) = 45 - 18 -18 = 9$.

so

$2X + A+E +C +F = (X+C+F)+(A+E+X)= 9 + 18 = 27$

so

$A+B+C+D+E+F + 3X = (2X+A+E+C+F) + (B+E+D) = 27 + 18 = 45$

so

$A+B+C + 3X = (A+B+C+D+E+F) - (D+E+F) = 45-18 = 27$.

so

$3X = (A+B+C + 3X) - (A+B+C) = 27-18 = 9$.

So $X = \frac {3X}3 = \frac 93 = 3$.

• Comments are not for extended discussion; this conversation has been moved to chat. – Daniel Fischer Apr 10 '18 at 21:15