# Math Help - How to fill in this table?

1. ## How to fill in this table?

Hope it belongs to the right section....Following is a table.Here is the question:
Fill in the remaining areas by real no.s such that the sum of the numbers in any three neighboring squares is constant and sum of all no.s is 210.

|_|_|26|_|_|_|12|_|_|_|

Is it only by trial & error?
Sorry,dont know how to make tables

2. Originally Posted by anshulbshah
Hope it belongs to the right section....Following is a table.Here is the question:
Fill in the remaining areas by real no.s such that the sum of the numbers in any three neighboring squares is constant and sum of all no.s is 210.

|_|_|26|_|_|_|12|_|_|_|

Is it only by trial & error?
Sorry,dont know how to make tables
Filling in the blanks:
.a.b.26.c.d.e.12.f.g.h.

Resulting equations:
a+b+26 = x
b+26+c=x
26+c+d=x
c+d+e=x
d+e+12=x
e+12+f=x
12+f+g=x
f+g+h=x
a+b+26+c+d+e+12+f+g+h = 210

You have nine equations and nine unknowns.

There are several ways to attack this:

You can remove the x by subtracting the adjoining pairs.

a+b+26 = x
b+26+c=x
a - c = 0

26+c+d=x
c+d+e=x
26 - e =0

d+e+12=x
e+12+f=x
d - f =0

12+f+g=x
f+g+h=x
12 - h = 0

That's a start.
and continue with every other equation
a+b+26 = x
26+c+d=x
a+b = c+d

26+c+d=x
d+e+12=x
26+c = e +12

etc..

Or create an augmented matrix and reduce to a triangular matrix.

3. Hello, anshulbshah!

Fill in the remaining cells with real nnumbers
so that the sum of three consecutive numbers is constant
and sum of all numbers is 210.

. . $\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline
\;\;& \;\;& 26 & \;\; & \;\; & \;\; & 12 & \;\; & \;\;& \;\; \\ \hline \end{array}$
We are given: . $\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline
a & b & 26 & c & d & e & 12 & f & g & h \\ \hline \end{array}$

We have: . $a+b+26 \:=\:b+26+c \quad\Rightarrow\quad a \:=\:c$
We have: . $c+d+e \:=\:d+e+12 \quad\Rightarrow\quad c \:=\:12$
We have: . $12+f+g \:=\:f+g+h \quad\Rightarrow\quad h \:=\: 12$
. . Hence: . $a \:=\:c\:=\:h \:=\:12$

Answer (so far): . $\begin{array} {|c|c|c|c|c|c|c|c|c|c|} \hline 12 & b & 26 & 12 & d & e & 12 & f & g & 12 \\ \hline \end{array}$

We have: . $26 + 12 + d \:=\:12+d+e \quad\Rightarrow\quad e \:=\:26$
We have: . $e+12+f \:=\:12+f+g \quad\Rightarrow\quad e \:=\:g$
. . Hence: . $e \:=\:g\:=\:26$

Answer (so far): . $\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline 12 & b & 26 & 12 & d & 26 & 12 & f & 26 & 12 \\ \hline \end{array}$

We have: . $b+26+12 \:=\:26 + 12 + d \quad\Rightarrow\quad b \:=\:d$
We have: . $d+e+12 \:=\:e+12+f \quad\Rightarrow\quad d \:=\:f$
. . Hence: . $b \:=\:d\:=\:f$

The sum of all numbers is 210.

. . $12 + b + 26 + 12 + d + 26 + 12 + f + 26 + 12 \:=\:210$

. . $b+d+f + 126 \:=\:210 \quad\Rightarrow\quad b+d+f \:=\:84$

Since $b=d=f$, we have: . $3b \:=\:84 \quad\Rightarrow\quad b \:=\:28$

. . Hence: . $b\:=\:d\:=\:f\:=\:28$

Answer: . ${\color{blue}\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline 12 & 28 & 26 & 12 & 28 & 26 & 12 & 28 & 26 & 12 \\ \hline \end{array}}$