Starting with the watch on the left, add 42 minutes to the time shown to give
the time on the next watch to the right.
Hand pointing to 5
Starting with the top clock and moving anti-
clockwise around the others, the hour hand
moves back 1 hour, then 2, then 3 etc, while
the minute hand moves forward 10 minutes
To the 4
Start with the top left clockface, and move
around the others in a clockwise direction.
The value pointed to by the hour hand equals
the value pointed to by the minute
hand, subtracting 4 for the first clockface,
5 for the next, then 6 etc.
Answer Hour hand is pointing to 2.
:Multiply the hour hand value by 2 and
add the minute hand value.
This total is always 15.
Answer: To the 5
Start with the top left clockface,
and move around the others
in a clockwise direction.
The sum of the numbers
pointed to by the hour and
minute hand follows the sequence
14, 15, 16, 17 and 18.
Answer : 11:01
Taking the hour and minute values on each
watch separately, as you move to the right the
hour value increases by 1, 2, 3 and 4, and the
minute value decreases by 11, 22, 33 and 44.
Answer: 9 of Clubs
Taking red cards as positive values and black cards as negative values, in each
column of the diagram, the lower card value equals the sum of the two upper
The suits are used alternately in each column.
Answer: King of Clubs
Start at the top left of the diagram and move to the right, then down one row and to the
left etc. in a snakes and ladders pattern. The value of each card increases by 5 each time,
with their suit following the sequence of hearts, clubs, diamonds and spades.
Answer 7 (any suit)
Taking the value of Aces as 1 and all court cards as 10, In each column of the
diagram, the value of the sum of the 3 cards is always 21.
Answer : Jack of SpadeS
There are 2 sequences in the grid - one determining the value of the card, and one determining the
suit of the card.
Starting on the top left and moving right, then down one row and to the left, then down the final row
and to the right, cards are arranged in order, with their value increasing by 4 each time.
To calculate the suit of each card, start on the top left and move down, then right one row and move
up etc. cards are arranged in the order Hearts, Clubs, Diamonds, Spades.
Answer : 3 of SpadeS
Divide the diagram in half, vertically.
In each half, start at the top left card and move to
the right, then down one row and to the left, and
finally to the right, in a snakes and ladders
The value of the cards in the left hand half
increase alternately by 3 and 4, and the value of
the cards in the right hand half increase
alternately by 4 and 5.
To calculate the suit of each card, start at the top
left of the whole diagram and move down, then
to the right one space and upwards etc. in a
snakes and ladders pattern.
Suits are written in order, following this path,
starting with Hearts, then Clubs, Diamonds and
Answer : D
The number of black dots in each grid
increases by 1 each time, starting with the top
left grid and working to the right, top row then
Adding the three numbers in each
square together gives the numerical
value of the letter at the centre of each
Answer : G
Convert each letter to its numerical value, and
read each pair of values as 2 digit numbers. In
each row, the number in the centre equals the
difference between the 2 digit values on the
left and right.
Answer : E
Explanation In each row, the left
hand grid is symmetrical around
a vertical axis, the central grid is
symmetrical about a horizontal
axis, and the right hand grid is
symmetrical about a diagonal
axis, running bottom left to top
Explanation :The Hammer = 1, the File =
3 and the Axe = 5
Explanation: Multiply the numerical
values of the letters in each pair to give
the 3 digit result in the spaces above.
Explanation :The value at each corner of the
diagram equals the difference between the
sums of the numerical values of the letters in
the boxes adjacent to the corner.
Explanation:Working in rows, add the left and
right hand numbers together, and put the
letter with the reverse alphabetical value of
this sum in the central box.
Split the left and right hand circles in half vertically.
The letter with the numerical value of the sum
of the digits in the left half of the left hand circle
is placed in the top left segment of the central
the letter with the numerical value of the sum of
the digits in the right half of the left hand circle is
placed in the top right segment of the central
Repeat this formula for the 2 halves of the right
hand circle, putting the resulting letters in the
lower segments of the central circle.
MISSING LETTER PUZZLES
Working in rows, add together the numerical
values of the left and right hand letters to give
the numerical value of the central letter.
In each segment of the diagram are a
pair of letters, one of which is the same
distance from the start of the alphabet
as the other is from the end.
MISSING LETTER PUZZLES
Answer : K
values of the letters
segments of the
circle always add up
As you move down, the numerical value of the
letters follows the sequence of Prime Numbers.
Answer : 1
Explanation :Working from left to right, letters in corresponding
segments of the circles move through the alphabet in steps of 2, 3
and 4, with their relative positions moving one place clockwise at
MISSING LETTER PUZZLES
Starting at the top left circle, and moving
right, then down one row and moving left,
in a snakes and ladders pattern, letters
move through the alphabet in steps of 2, 3
and 4, repeating this pattern all the way
Splitting the large square into quarters, each
quarter features the same pattern of letters.
Working from left to right, letters move
forwards 7 places, then back 2. Repeat this
sequence until the end. ANS : W
Taking the top row of circles, numbers in the
central circle equal the sum of the numbers in
corresponding segments of the left and right
In the bottom row, numbers in the central
circle equal the difference between numbers
in corresponding segments of the left and right
hand circles. ANS: 7
Starting with the 10 at the top, one set
of numbers increases by 3 each time,
written in alternate boxes as you move
down the diagram, and the other set of
numbers decreases by 2, written in the
boxes remaining.ans 2
Starting with the top 3 digit number, the first digit
increases by 2 as you descend, from 1 to 11.
The middle digit decreases by 1 each time, and the
right hand digit alternates between 5 and 3.
Reading each row of the diagram as
a series of 3 digit numbers, the
centre 3 digit number equals the
sum of the top 2 numbers, and the
sum of the lower 2 numbers.
Add together values in corresponding positions of the
top two crosses, and put the results in the lower left
cross. Calculate the difference between values in
corresponding positions of the top two crosses, and put
the results in the lower right cross. Finally, add together
the values in corresponding positions of the lower two
crosses to give the values in the central cross.
In each row of the diagram, the central value
equals the sum of the differences between the
left hand pair of numbers and the right hand
pair of numbers. Ans 7
The inner digit in each segment equals
the sum of the two numbers in the
outer part of the opposite segment.
Crosshatching - finding squares for numbers.
The obvious way to solve a sudoku puzzle is to
find the right numbers to go in the squares.
However the best way to start is the other way
round - finding the right squares to hold the
This uses a technique called
'crosshatching', which only takes a couple of
minutes to learn. It can solve many 'easy' rated
puzzles on its own.
Crosshatching works in boxes (the 3 X 3 square subdivisions of the grid). Look at the top-
left box of our sample puzzle (outlined in blue). It has five empty squares. All the numbers
from 1 to 9 must appear in the box, so the missing numbers are 1,2,3,5 and 9.
We'll ignore 1 for a moment (because it doesn't provided a good example!), and see if we
can work out which square the missing 2 will go into.
To do this, we'll use the fact that a number can only appear once in any row or
column. We start by looking across the rows that run through this box, to see if any
of them already contain a 2. Here's the result:
The first two rows already contain 2s,
which means that squares in those rows
can't possibly contain the 2 for this box.
That's all we need to know, because the
third row only has one empty square, so
that must be the home for the 2.
Now let's see if we can place the 3 for this
This time we end up checking the columns
that run down through the box, as well as
the rows that run across it:
Again, we get a result first time –
there's only one empty square that the 3
can possibly go into.
You can see from this example why it's
called 'crosshatching' - the lines from
rows and columns outside the square
criss-cross each other.
Of course you don't always get a
result first time. Here's what
happens when we try to place the
There's only one 5 already in the
rows and columns that run through
this box. That leaves three empty
squares as possible homes for the 5.
For the time being, this box's 5 (and
its 1 and 9) have to remain unsolved.
Now we move on to the next box:
Here we're crosshatching for 3, the first
missing number in this box.
Note how we treat the 3 we placed in the first
box as if it had been pre-printed on the puzzle.
We still can't place this box's 3 though, so we'll
move on to the next missing number (5), and
In sudoku, accuracy is essential. If the 3 in the first box is wrong,
we'll be starting a chain of errors that may prove impossible to
unravel. Only place a number when you can prove, logically, that
it belongs there. Never guess, and never follow hunches!
In sudoku it pays to look at the same thing in different ways.
By using crosshatching slightly differently, you can often get quicker results.
Instead of looking at a single box and its missing numbers, you can look at a group
of three boxes running across or down the puzzle, trying to place each number from
1 to 9 in as many of the boxes as you can.
In this example we're trying to place 7s in the three right-hand boxes:
The stack of boxes starts out with just one 7 in place
(bottom box). This solves the middle box's 7 (entered
in blue), and entering that immediately solves the
top box's 7 as well.
This 'chain reaction' of solving wouldn't occur in
single-box crosshatching. It happens here because
we're focussing on a single number across multiple
boxes - looking at things differently.
Crosshatching and slicing/dicing are basically the
same thing, but slicing/dicing can be more efficient,
and often feels less laborious than doggedly working
through the empty squares in a single box (although
that's what you will have to do in order to solve
tough puzzles, so be prepared!).
Looking at the top-left box of our original
puzzle, crosshatching produced three possible
squares where the missing 5 could go. Here's
the box, with 5 'pencilled-in' to the corners of
its three possible squares:
1 and 9 were also unsolved for this box. Here's
the box with all its missing numbers pencilled
into their possible squares:
On the left is is the top-left box,
and the one below it. We've
just entered a 5 in the bottom-
right square of the lower box.
Now we remove that square's
candidate list. We also remove
5 from the candidate list at the
top of the same column, and
the left of the same row. Here's
how the boxes look afterwards:
Whenever you fill in a square, remove the number you've used from all
candidate lists in the same row, column and box. Here are the areas we
needed to check for candidate 5s after filling in this square:
Rule 1 - Single-candidate squares.
When a square has just one candidate, that number goes into the square.
Here's the mid-left box again, as it was before we entered the 5:
The mid-right and bottom-right squares each have only one candidate, so we can put those
numbers into the squares.
Some squares will be single-candidate from the start of the puzzle. Most, however, will
start with multiple candidates and gradually reduce down to single-candidate status.
This will happen as you remove numbers that you've placed in other squares in the same
row, column and box, and as you apply the last three rules described below.
Rule 2 - single-square candidates.
When a candidate number appears just once in an area (row, column or box), that
number goes into the square.
Look at the mid-left box again:
The number 6 only appears in one square's candidate list within this box (top-
middle). This must, therefore, be the right place for the 6.
The remaining three rules let you remove numbers from candidate lists, reducing
them down towards meeting one of the first two rules.
Rule 3 - number claiming.
When a candidate number only appears in one row or column of a box,
the box 'claims' that number within the entire row or column.
Here's the top-left box again:
The number 1 only appears as a candidate in the top row of the box.
This means there will have to be a 1 somewhere in the first three
squares of the puzzle's first row (i.e. the ones that overlap with the
box). That in turn means that 1 can't go anywhere else in that row,
outside of this box.
You can read across the row, and remove 1 from any candidate lists outside of
this box, even though you haven't actually placed 1 yet.
In this example, we can remove the 1 from the right-hand square's candidate list.
This makes the square single-candidate (7) - square solved!
Claims also work during crosshatching. Here we're crosshatching the top-right box for 1:
We can rule out the top row, because the top-
left box has already claimed that row's 1. This
lets us place the 1 in the bottom-right square
of the box.
Rule 4 - pairs.
When two squares in the same area (row, column or box) have identical two-
number candidate lists, you can remove both numbers from other candidate lists in
Here's the second row of the puzzle:
Two of the squares have the same candidate list - 67. This means that between them,
they will use up the 6 and 7 for this row.
That means that the other square can't possibly contain a 6. We can remove the 6 from
its candidate list, leaving just 9 - square solved!
The squares in a pair must have exactly two candidates. If one of the above squares had
been 679, it couldn't have been part of a pair.
Rule 5 - triples.
Three squares in an area (row, column or box) form a triple when:
None of them has more than three candidates.
Their candidate lists are all full or sub sets of the same three-candidate list (explained
You can remove numbers that appear in the triple from other candidate lists in the
Here's the fourth row of the puzzle:
Note the three squares in the middle, with candidates of 23, 23 and 234. These form a
234 is the full, three-candidate list, and 23 is a subset of it (i.e. all its numbers appear
in the full list). Because there are three squares, and none of them have any candidate
numbers outside of those in the three-candidate list, they must use up the three
candidate numbers (2, 3 and 4) between them.
This lets us remove the 4 from the other two candidate lists in this row, solving their
It's worth looking hard for subset triples. In this example, the 23 lists make an obvious
pair (see above), but it's the triple that instantly solves the two outside squares (once
you've dispensed with them, you can treat the 23s as a pair again, and use them to
solve the 234!). A subset (or 'hidden') triple is often the pattern that will unlock a
seemingly impossible puzzle.
Note - the squares in a pair or triple don't have to appear next to each other, or in any
particular order. In the example above, the triple could have occurred in, say, the first,
third and fifth empty squares of the row, with the 234 in the middle.
• the triple rule can be true even if none of the squares have three candidates. Take
these three candidate lists:
• 13 16 36
• All three lists are subsets of the list 136. Between them, these three squares will
use up the 1, 3 and 6 for the area they're in. These triples can be hard to spot
though, so it's probably best to start by looking out for three-candidate squares.
• (Special thanks to Edward for pointing out that the members of a triple can all be
subsets of the full list!)
• ◊ If all-subset triples still don't seem right, think of it this way:
The crucial thing is that the number of squares equals the number of candidates in
the full list (so three squares all with subsets of '136' (a three-candidate list) makes
It doesn't matter if some (or all) of the squares don't have the full candidate list.
What matters is that between them they cover the list, the whole list and nothing
but the list. That means they must use up all three of the list's numbers between
In case all that's put you off, here's an example
of a more obvious triple - they do exist!:
A set of N squares in an area forms a group when:
None of the squares has more than N candidates
Their candidate lists are all full or sub sets of the same N-candidate list.
Pairs (N=2) with subset members tend not to survive long, because a subset of a
two-number list is a single candidate and thus solvable.
However a single-candidate square can function perfectly well as a member of a
pair, or even a triple. This often has the same effect as solving the single-candidate
square then updating its surrounding candidate lists, but can be quicker.
On the left is an example (and from an 'easy' rated puzzle, too!)
The top three unsolved squares, with candidates of 26, 23 and 2, form a triple
(N=3) with a full list of 236.
That lets us remove the 3 and 6 from the bottom square's candidate list, reducing
it to just 8 - square solved!
You also sometimes see 'quadruplets' (N=4) - four squares, none with more than
four candidates, and all full or subsets of a four-candidate list.