u-can-do

EQUATE-sums walkthrough

by G R Burgin

The Puzzle

We have been given one value, 1, in the top left corner square, and the totals of each of the horizontal, vertical and diagonal lines.  We need to fill in the blank squares with numbers so that the totals are as shown. 

Start by naming each blank square as shown below. (We know A=1, so we can just leave the 1 in there) We now need to find the values of B to I.

We can state the values of each letter in 8 equations, like this:

We can also narrow down the possible vaues for each letter using a table, as shown below (We know that A = 1, so it can’t be 2, or 3, or 4, etc, and similarly, B can't be 1, and neither can C, etc):

A	1
B		2	3	4	5	6	7	8	9
C		2	3	4	5	6	7	8	9
D		2	3	4	5	6	7	8	9
E		2	3	4	5	6	7	8	9
F		2	3	4	5	6	7	8	9
G		2	3	4	5	6	7	8	9
H		2	3	4	5	6	7	8	9
I		2	3	4	5	6	7	8	9

Let’s have a look at the first line of the puzzle: 1 + B + C = 6

The value of B + C together must be 5:  B + C = 6 – 1 = 5

This means that B + C could be either 1 + 4 or 2 + 3

Looking at our grid, we can see that neither B nor C can be 1, as it’s already been taken by A.  This means the second option applies, and B or C is 2, and B or C is 3, and no other letters can be 2 or 3.  We can update the table:

A	1
B		2	3
C		2	3
D				4	5	6	7	8	9
E				4	5	6	7	8	9
F				4	5	6	7	8	9
G				4	5	6	7	8	9
H				4	5	6	7	8	9
I				4	5	6	7	8	9

Let’s look at the next line: D + E + F = 15. We have quite a few possible combinations here:

A quick glance at the table above shows us that we must discount any option that includes a 1, 2 or a 3.  This leaves only one option!  D, E & F can only be 4, 5 or 6.

A	1
B		2	3
C		2	3
D				4	5	6
E				4	5	6
F				4	5	6
G							7	8	9
H							7	8	9
I							7	8	9

We can update the table to show this, and can now see that G + H + I = 24 can only be 9 + 8 + 7.

Now let’s look at the first vertical line:
1 + D + G = 12 
or
D + G = 12 – 1 = 11

Looking at the possible values for D + G:
9 + 2 (won’t work as neither D nor G can be 2)
8 + 3 (won’t work as neither D nor G can be 3)
7 + 4 (this would work if G was 7 and D was 4)
6 + 5 (this won’t work as G can’t be 6 or 5)

So we know that G is 7 and D is 4

A	1
B		2	3
C		2	3
D				4
E					5	6
F					5	6
G							7
H								8	9
I								8	9

Next, B + E + H = 15 can be solved using the methods shown above, or like this:

Letters	B	E	H	Total
All the possible combinations of values are listed - only one gives us the total of 15	2	6	8	16
	2	6	9	17
	3	6	8	17
	3	6	9	18
	2	5	8	15
	2	5	9	16
	3	5	8	16
	3	5	9	17

So B = 2, E = 5 and H = 8. Therefore C = 3, F = 6 and I = 9.

A	1
B		2
C			3
D				4
E					5
F						6
G							7
H								8
I									9

The puzzle is solved!

 

 

 

 

Contact: mathspuzzles@gordonburgin.com

© G R Burgin 2009