111, then 112. This question teaches the moves: drag from the tray (or click), drop ones onto ones to stack them, and Σ (or T) to check. Keep Σ off while pupils predict.

Long-press (half a second) duplicates. The bin only appears while you are dragging. Everything can be undone with ↩ or U, and each slide has its own undo.

Box-select picks up everything the box touches. ± flips every selected counter, so 6 yellow become 6 red. Tapping one counter flips just that one. Shift-click adds a piece to the selection or takes it out.

3, because the visible frame is full (10). Covers turn any model into a missing-number problem: build it, cover part of it, ask. Tap again to hide it. The − and + in the bar resize the cover.

24. Slides keep their own pieces, ink and undo, so two representations can sit side by side. 💾 saves the current slide as an image for a worksheet or slide deck.

7, and 3 more make 10. Pupils should see “a full row of 5 and 2 more” rather than count in ones. Flash it under a 🙈 Cover for a second to push them to subitise.

2 move, leaving 3: 10 + 3 = 13. This is bridging through 10. Keeping the moved counters yellow shows where they came from. Squares cycle red → yellow → empty when tapped.

Most find 5 + 1 (a full row and one more) or 3 + 3 (a double) quickest. Scattered counters force counting. The structure of the frame carries the number, which is the point of using one.

Eleven splits of 10 (0 + 10 up to 10 + 0), or nine if both colours must appear. Six splits of 5. Each list is symmetrical: 3 + 7 and 7 + 3 are different pictures with the same total.

Nine splits, from 8 + 0 to 0 + 8 (seven if both colours must appear). Writing each one as an addition and a subtraction (8 − 3 = 5) builds the fact family.

3, 4 and 2 counters. You can make halves, thirds, quarters, sixths and twelfths: the denominators are the factors of 12. Fifths, sevenths, eighths and so on are impossible with 12 counters.

Exactly 5 red happens only about 1 time in 4 (252 out of 1024). Expect most results from 2 to 8; 8 or more red happens about 1 time in 20. “It is fair, so it will be 5 and 5” is the misconception to tackle.

+2. Each yellow and red pair is a zero pair (+1 + −1 = 0), so 3 pairs vanish and 2 yellow are left. 0 Remove zero pairs shows that the total does not change.

Add two zero pairs (2 yellow and 2 red): still −3, but now there are 5 reds. Take them away and 2 yellow are left, so −3 − −5 = 2. This is why subtracting a negative behaves like adding.

Both are −2. For 4 + −6, put down 4 yellow and 6 red and remove 4 zero pairs: 2 red are left. For 4 − 6, start with 4 yellow, add 2 zero pairs so 6 yellow can be taken away: 2 red are left. Different actions, same answer.

At 10 the ⇌ button appears and swaps 10 ones for 1 ten of the same length. Ask “has the amount changed?” The value is still 10, but there are fewer pieces.

2 hundreds, 2 tens and 14 ones, still 234. Σ lists the pieces as they are, not simplified, which is exactly the regrouping needed for a calculation like 234 − 18.

Exchange one ten for 10 ones (13 ones), then take away 7 ones and 2 tens: 116. Pupils who write 124 have taken 3 from 7 (“smaller from larger”), which the model makes impossible.

32 has 4 ways (3 tens 2 ones, 2 tens 12 ones, 1 ten 22 ones, 32 ones) and 45 has 5. In general it is the tens digit + 1. Non-standard partitions like “2 tens and 12 ones” are what make column subtraction work.

9 hundreds, 9 tens and 10 ones: 28 pieces, still 1000. The number of pieces and the value are different things, which is the heart of place value.

The pieces are worth 213, but read column by column they say 231: the ten is in the Ones column and the three ones are in the Tens. Swap them and both totals agree. It is the same mistake as writing a digit in the wrong column.

The tens column is empty, and the 0 in 305 says “no tens”. With digit cards, Σ reads 3_5 and says an empty column needs a 0. Without it the number reads as 35.

1 hundred, 3 tens and 4 ones: 134 before and after. “13 tens” is a correct way to describe 130, and pupils who can say it find the column methods much easier.

1.1. Ten tenths make one. 0.11 comes from adding the 3 and the 8 like whole numbers and then “putting the point in”.

24. The ones become tens and the tenths become ones, each moving one column to the left. The digits move and the point stays still, which corrects the “move the point” rule.

0.5. Five tenths beats four tenths, whatever the hundredths are. Pupils who say 0.45 are comparing 45 with 5 as whole numbers (“longer is larger”).

357. Arrow cards make the expanded form visible: the 5 means 50 because its card is two digits long. Ask why the 7 card is the shortest.

400 and 6. The 0 in the tens place is the 400 card’s own middle zero showing through. For 460 the 60 card covers that zero. Pupils who look for a “0 card” are thinking digit by digit, not value by value.

Largest 7111, smallest 1117. There are 84 of them, because there are no zero cards, so every digit is from 1 to 9. A good place to start systematic listing (fix the thousands digit first).

A 2 × 5 rectangle, the same as the 10 piece: 7 + 3 = 10. 9 + 1 do the same, and 8 + 2 and 6 + 4 sit side by side. Two odd pieces always fit together to make an even number.

10 − 9 = 1. The uncovered holes show the difference, or how many more make 10. Smaller pieces always sit on top, so the gap is always visible.

Odd (3, 5, 7, 9) have a bump; even (2, 4, 6, 8, 10) are complete rectangles. An even number is made of pairs; an odd number is pairs and one more, which is the bump. 1 is the interesting case: it looks like a square, but it is the “one more” on its own, so it is odd.

Two rods: 1 + 5, 2 + 4 and 3 + 3 (5 trains if order matters). Three rods: 10 trains if order matters (1+1+4 in 3 orders, 1+2+3 in 6, 2+2+2 in 1). This is partitioning before the numbers.

With orange as 1, yellow is 1/2 and red is 1/5. With dark green as 1, red is 1/3, yellow is 5/6 and orange is 1 2/3. A fraction is always a fraction of something.

White (10 of them), red (5), yellow (2) and orange (1): the factors of 10. Rods of 3, 4, 6, 7, 8 and 9 cannot, because those numbers do not divide 10.

55. Two staircases make a 10 by 11 rectangle (110 squares), and half of that is 55. This is the triangular number formula: 1 + 2 + … + n = n(n + 1) ÷ 2.

1/3. The bigger the denominator, the more pieces the whole is cut into, so each piece is smaller. Pupils who choose 1/4 are comparing 4 with 3.

Same size: 2/4, 3/6, 4/8, 5/10 and 6/12. Two different strips: only 1/3 + 1/6. With 🧲 on, strips click end to end and under each other.

1/2, 1/4, 1/5, 1/8 and 1/10 stop. 1/3, 1/6, 1/7, 1/9, 1/11 and 1/12 recur. Denominators that stop have no prime factors except 2 and 5. 1/7 repeats six digits: 0.142857142857…

Sixths: 3/6 + 2/6 = 5/6. The strip that fits both is the common denominator. Pupils who answer 2/5 have added the tops and the bottoms: the strips show 2/5 is shorter than 1/2 alone.

3/4 is closer to 1 (it is 1/4 away; 2/3 is 1/3 away). The strips show fractions as lengths and the line shows them as points, and pupils need both.

5x + 4. Tiles of different shapes are unlike terms: an x tile is not a whole number of 1 tiles long. Pupils who say 9x are adding everything together.

x. Each +1 and −1 cancel. Adding 3 to both sides of x − 3 = 5 gives x = 8, and the tiles show why “do the same to both sides” works.

Take 3 ones from each side (2x = 8), then split each side into two equal groups: x = 4. Keep Σ hidden until pupils have predicted each step.

1 x², 5 x tiles and 6 ones: x² + 5x + 6. Each tile is part of the area, so the rectangle is the grid method made physical. Use ↻ to stand x tiles upright.

(x + 2)(x + 4). The ones must make a 2 by 4 block: a 1 by 8 block would need 9 x tiles. If no rectangle is possible, the expression does not factorise.

2x + 3. One x and one −x cancel, and two +1 and −1 pairs cancel. Watch for pupils who drop the sign of −x: the red tile makes the sign part of the term.

Largest to smallest: £2, 50p, 2p, 10p, £1, 20p, 1p, 5p. Size does not follow value: 2p is bigger than £1. Shapes and sizes help blind and partially sighted people tell coins apart.

With four coins or fewer there are 4 ways: 50p; 20 + 20 + 10; 20 + 20 + 5 + 5; 20 + 10 + 10 + 10. With any number of coins there are 451 ways, which is why the limit matters.

83p is 50 + 20 + 10 + 2 + 1: five coins. 88p, 89p, 98p and 99p each need six coins, the most for any amount under £1.

£3.88. £27.65 needs 6 pieces: £20, £5, £2, 50p, 10p and 5p. Taking the largest note or coin that fits, every time, always gives the fewest pieces with UK money.

Each face should come up about 5 times, but anything from 1 to 9 is normal in 30 rolls. After 600 rolls each face will be close to 100, and the proportions close to 1/6. The same number can come up twice in a row.

7 can be made 6 ways out of 36 (1+6, 2+5, … 6+1); 2 only one way (1+1). In 36 rolls a 7 usually comes up between 2 and 10 times, so pool results across the class.

A difference of 1 is most likely (10 ways out of 36), then 2 (8 ways). 0 happens 6 ways (doubles). 6 is impossible: the biggest difference is 6 − 1 = 5.

Even, 7 times out of 8: the product is only odd if all three dice are odd (1/2 × 1/2 × 1/2 = 1/8). There are 40 different products, from 1 to 216.

13. The dark green rod crosses the 10 after 3 more, so 7 + 6 = 7 + 3 + 3 = 13. The rods show the jumps and the line shows where you land.

Common routes: +20 then +7 (66, 73); +4 to 50, then +20, then +3; or +30 then −3. An empty line has no scale, so it shows the strategy rather than exact distances.

2 °C, then −4 °C. Rises move right and falls move left. Pupils who count the mark they start on get one too many: count the jumps, not the marks.

They have written 400 and then 6 next to it, giving 4006. 406 needs three columns: 4 hundreds, 0 tens and 6 ones. Σ reads the digit cards column by column.

3 − 7 = −4. To take 7 yellow from 3, add 4 zero pairs, then take the 7 yellow: 4 red are left. Subtraction is not commutative; “the difference” only gives the size, not the sign.

Equal only when x = 2. 2x is bigger when x is more than 2 and smaller when x is less than 2. The x tile is deliberately not a whole number of 1 tiles long, so “x” cannot be read as a fixed number.

Dienes and ten-frames show tens as collections you can count; place value counters and arrow cards show a ten as one object worth 10; Cuisenaire shows 23 as a length. Pupils need to move between all of them.

Hiding ones in a Dienes model is easy to reason about; hiding part of an arrow card number is harder because the cards overlap. Covers work on every manipulative.

There is no single answer. Strong models show the structure of the calculation (the bridge through 10, the zero pairs, the exchange), not just the two numbers and the answer.

🔗 copies a link that opens the tool with the current slide’s pieces. The best pupil questions ask for a prediction before anything is moved.