3 = 6
Look For 3 Squares Totaling 6
a quick glance, the top right block seems to be a good starting
point because 15 - 9 = 6 and we are looking for three different
numbers that add up to six - and there are only three that
will do that, namely 1 plus 2 plus 3
we now know that each box b, e and f must be a 1 or
a 2 or a 3
therefore follows that boxes d, h and i cannot be 1, 2
A Given Pair
you are given two numbers side by side then you are able to find
the sum of the pairs in the row or column under them.
this example, 1 and 3 are given, therefore...
+ e = 15 -1 -3 = 11
+ h = 24 -11 = 13
bonus in this example is that we are given that h = 8 and therefore...
= 13 -8 = 5
narrows down the possible d and e values
Tip 4. Common
two overlapping blocks have three possible numbers for each block,
then the square where they overlap will have the value that is
common to each block.
a given number layout like this, square e will overlap with the
blue and pink blocks.
b, c and e must be 2 or 5 or 8
Squares e, h and i must be 2 or 3 or 6
e cannot be a 3 or 6 to match the blue
Square e cannot be a 5 or 8 to match the pink
e must be 2 which is common to all blue and pink
blocks (squares b, c, e, h and i)
the 2 was found, the solution to this 'extreme' puzzle was easy.
of 'lows to highs' using Tip 4 puzzle.
left to be allocated = 2, 3, 5, 6, 8, 9
b, d and e must total 16 so to start-
= 19 - no
2+6+9 = 17 - no
2+6+8 = 16 - ok
2+5+9 = 16 - ok
2+5+8 = 15 - no
2+5+6 = 14 - no
3+6+8 and 3+5+9 are too much
3+5+8 = 16 - ok
3+5+6 = 14 - no
or 9 - too much
Tip 5. Lows
finding the possible numbers for each square, start with the lowest
number added to the highest two numbers and then work down.
pick the second lowest number added to the highest two numbers
and work down again.
quickly you will reach the 'middle of the available numbers and
there is no need to go further as you will only be repeating yourself
with the same numbers reversed.
working logically to a system, though tedious it may be, possible
permutations will not be missed.
find it helps to write in the numbers left
to place -
2 3 4 6 8 9
was Sujiko #357 in the Daily
Telegraph, reproduced with thanks.
Tip 6. One
for when there is only one more the difference on a side as marked
here in red. Squares b+c+e = 18-5=13
Squares a+b+e = 21-7=14 , one more.
Squares b and e must be common so a and c must have a difference
sujiko puzzle is part solved at the bottom; e+h+i must
be 6 or 8 or 9 plus 5 = 28 and running through the possibilities
for square g it must be 1 making i = 9 and 6 or 8 for e
leaves only 2, 3 and 4 for the top row and b is common to top
left and top right blocks and a is therefore one more than c.
glance at e will show that if it was 6 the max it could make a
top three squares was only 13 so e must = 8
b was 3 then a and c would be 2 or 4 which is two more so b is
not 3. If b is 4 then a and c is 2 and 3 - one more so ok. But,
with e = 8 plus b = 4 it makes c=1 and we know g=1 already.
b must be 2; c must be 3 and a must be 4
- one more