Start with a warm-up or jump to a challenge. Try an idea, change your mind, and ask why it works. Some puzzles have more than one answer.
Use the boards below or a pencil and paper. Each puzzle has a hint and a worked solution to open when you are ready. There is no timer.
Warm-ups
01 · Routes
The garden patrol
Start at S and take your first step to the right. Move only between squares sharing an edge. Visit every open square exactly once, finishing at F. The square marked × is blocked.
Choose the squares in your route, starting with S. Use Undo to take back a step.
Hint
The top-right corner has only two neighbours. Your route must connect it to both.
Solution and why
Follow the numbers in this grid, from 1 to 15:
1 2 3 4
10 9 × 5
11 8 7 6
12 13 14 15
The first step and the corners force the route across the top and down the right. You cannot finish early. The square just left of F must be saved for the last step, which forces the detour through the middle and left side.
02 · Numbers
Six seats, three sums
Use each number from 1 to 6 exactly once. The columns must total 6, 7 and 8, from left to right. The top row must total 10, and the top-left number must be even.
Total 6Total 7Total 8
Hint
Which pair can make 6 with an even number on top? Then look at the numbers left for the other columns.
Solution and why
4 1 5
2 6 3
The first column must contain 2 and 4. The last then needs 3 and 5, leaving 1 and 6 in the middle. Putting 2 at the top left would require the other two top numbers to total 8, which is impossible with these pairs. Putting 4 there forces 1 and 5 beside it.
03 · Symmetry
The hidden badge
Shade squares so the design is unchanged by both a left–right reflection and a top–bottom reflection. The numbers beside the board give the shaded-square totals for each row and column.
Choose a square to shade or unshade it.
1232112321
Hint
A row with just one shaded square must put it in the centre. Apply the same idea to the columns.
Solution and why
· · ■ · ·
· ■ · ■ ·
■ · ■ · ■
· ■ · ■ ·
· · ■ · ·
The outside rows and columns force the four tips. The middle row needs its centre shaded too. The centre column is now full, so rows 2 and 4 must use columns 2 and 4. Both reflections and all ten totals then agree.
04 · Numbers
Triangle totals
Place the numbers 1 to 6 around the triangle, using each once. The three numbers along each side must add to 10. Corner numbers belong to two sides.
Hint
Adding the three side totals counts each corner twice and each midpoint once. What must the three corners add to?
Solution and why
Put 1 at the top, 3 at the bottom left and 5 at the bottom right. The left midpoint is 6, the right midpoint is 4, and the bottom midpoint is 2.
The side totals are 1 + 6 + 3, 1 + 4 + 5, and 3 + 2 + 5: all 10. The numbers 1 to 6 total 21, while the three sides total 30, so the corners must total 9. Rotations and reflections work too.
05 · Logic
The woodland queue
Place Hare, Bear, Fox, Frog and Owl once each, from left to right.
Owl is immediately to Frog’s right.
Exactly one animal stands between Fox and Hare.
Bear is somewhere to Hare’s right.
Neither Frog nor Bear is at either end.
Hint
Treat Frog and Owl as a two-animal block. Try each possible place for the block.
Solution and why
Hare → Bear → Fox → Frog → Owl.
If Frog were second or third, the other clues would force Hare and Fox to the two ends, breaking their spacing rule. So Frog is fourth and Owl fifth. Bear must be second, between Hare and Fox, and Bear’s clue puts Hare first.
Think further
06 · Deduction
A small Sudoku
Fill each empty square with a number from 1 to 4. Every row, every column and each outlined 2-by-2 box must contain all four numbers exactly once.
Fill the empty squares
1
4
4
1
2
4
3
1
Hint
Look at the bottom-left box. Which numbers are missing, and which can go in its top-right square?
The bottom-left box is missing 1 and 4. Row 3 already has 4, so its second square is 1; the bottom-left square is 4. The remaining entries follow from the row, column and box rules. This puzzle has one solution.
07 · Working backward
Double the destination
There are 2 tokens in bowl A, 3 in B and 7 in C. Make exactly two moves to leave 4 tokens in every bowl.
In each move, transfer exactly as many tokens as the destination already holds, doubling its count. The source must have enough tokens, and the destination must not be empty.
A 2
B 3
C 7
Moves: 0 of 2
Hint
Just before the last move, the destination must hold 2 tokens, the source 6, and the untouched bowl 4.
Solution and why
First move 3 tokens from C to B, giving A: 2, B: 6, C: 4. Then move 2 from B to A, giving 4 in every bowl.
Working backward from the target tells you to look for counts 2, 6 and 4 after one move. Only the first move above produces those counts from the starting position.
08 · Placement
Four quiet owls
Place four owls on this board. No two may share a row, a column or a diagonal of any length. Diagonal means a straight line through diagonally touching squares.
Choose a square to place or remove an owl.
Hint
Try putting the first owl in row 1, column 2. Which squares are still available in the next row?
Solution and why
Use columns 2, 4, 1, 3 in rows 1, 2, 3, 4 respectively. No pair shares a row, column or diagonal.
Further challenge: find every solution. The only other one is its mirror image: columns 3, 1, 4, 2. Trying each possible first-row column, and ruling out attacked squares, exhausts the possibilities.
09 · Planning
Measure four litres
You have empty 3-litre and 5-litre jugs, a tap and a drain. Get exactly 4 litres in the 5-litre jug.
A move fills a jug completely, empties it completely, or pours from one jug to the other until the source is empty or the destination is full. There are no measuring marks.
0 L
3-litre jug
0 L
5-litre jug
Moves: 0
Hint
Fill the big jug and pour into the small one. Can you make use of the 2 litres left over?
Solution and why
Fill the 5-litre jug.
Pour into the 3-litre jug; 2 litres remain in the big jug.
Empty the small jug.
Pour the 2 litres into the small jug.
Fill the big jug again.
Top up the small jug with 1 litre, leaving 4 litres in the big jug.
That takes six moves. Further challenge: can you prove that five moves cannot be enough? List the reachable pairs of amounts, one move at a time, discarding pairs you have seen before; 4 litres first appears at move six.
10 · Counting
How many routes?
Travel from S to F, following the streets. You may move only right or down, one street segment at a time. You cannot pass through the dot marked ×.
How many different routes are possible? Two routes are different if their sequences of moves differ.
Hint
Write each route as a word made of R and D. Every complete route has three R’s and two D’s.
Solution and why
There are 4 routes: DDRRR, RRDDR, RRDRD and RRRDD.
Without the closed dot, choosing the two down-steps among five moves gives 10 routes. There are 2 ways to reach the closed dot and 3 ways to continue from it, so 6 routes use it. Removing those leaves 4.
Challenges
11 · Networks
The one-stroke house
Start at B. Follow every road exactly once and finish at D. You may revisit a point, but you cannot reuse a road.
B is already selected. Choose the next point on your route.
Roads: A–B, A–C, B–D, C–D, C–E, D–E and B–C. Crossings are not extra stopping points.
Route: B
Hint
Count the roads meeting at each point. Why do the two points with odd counts matter?
Solution and why
One route is B → A → C → E → D → C → B → D.
Each intermediate visit to a point uses an arrival and a departure. Only the start and finish can have an unpaired road. Here B and D each have three roads, while A and E have two and C has four. This explains the required endpoints.
12 · Seeing structure
Remove two sticks
The picture contains four small squares and one large square. Remove exactly two sticks so that exactly one small square and one large square remain. Count squares of both sizes; do not move the remaining sticks.
Choose a stick to remove it; choose it again to put it back.
Hint
Keeping the large square means keeping its entire outside border. Which inside sticks can you remove?
Solution and why
Remove the left half of the middle horizontal line and the upper half of the middle vertical line. Only the bottom-right small square and the large outside square remain.
Choose one of the two inner horizontal sticks and one of the two inner vertical sticks. Each such pair breaks three small squares and preserves the outside border, giving four solutions.
13 · Information
The heavier pebble
Nine pebbles look identical. Exactly one is heavier; the other eight weigh the same. Identify the heavier pebble using a balance scale twice.
Put equal numbers of pebbles on the two pans. The scale tells you which pan is heavier, or whether they balance. Reset starts a new round. Check your answer only after making two weighings.
Weighings: 0 of 2
Hint
Divide the pebbles into three groups of three. A balance has three possible outcomes.
Solution and why
Weigh pebbles 1–3 against 4–6. If one pan is heavier, its group contains the mystery pebble; if they balance, the mystery pebble is among 7–9.
From the suspect group, weigh one pebble against another. The heavier one is the answer; if they balance, it is the third pebble.
Why two weighings suffice: each result chooses among three possibilities. Two weighings can distinguish nine possibilities, exactly the number of pebbles. One weighing cannot guarantee an answer.
14 · Planning and proof
The arrow swap
Swap the two groups: finish with three ◀ arrows on the left, the empty cell in the middle, and three ▶ arrows on the right.
Each arrow moves only in the direction it points. It may slide into the next cell if empty, or jump over one arrow pointing the other way into an empty cell. No other jumps are allowed.
Choose an arrow to move it. If you get stuck, reset and try another plan.
Moves: 0
Hint
Begin by moving the right arrow nearest the gap. Alternate short runs of jumps with single slides.
Solution and why
Number the cells 1–7 from left to right. Move the arrow currently in each of these cells:
3, 5, 6, 4, 2, 1, 3, 5, 7, 6, 4, 2, 3, 5, 4.
That is 15 moves. Each of the three right arrows must cross each of the three left arrows, requiring nine jumps. The arrows must travel 24 cell-lengths in total. The nine jumps account for 18, so six single-cell slides are also needed. Thus every successful solution uses 15 moves.
15 · An impossibility proof
Two missing corners
Remove the top-left and bottom-right squares from a 4-by-4 board. Can you cover the remaining 14 squares with seven dominoes?
Each domino covers exactly two squares sharing an edge. You may rotate dominoes, but they must not overlap or extend beyond the board. Try to explain your answer.
××
Hint
Colour the board like a chessboard. How many light and dark squares remain? What does each domino cover?
Solution and why
No. The removed corners have the same colour. There are 6 squares of that colour left and 8 of the other.
Every domino covers one square of each colour. Seven dominoes would therefore need 7 of each, which this board does not have. This rules out every possible arrangement at once.
Further challenge: does this argument still work for an 8-by-8 board with opposite corners removed? Yes: it leaves 30 squares of one colour and 32 of the other.