325: “three hundred and twenty-five”, 3 × 100 + 2 × 10 + 5 × 1 = 300 + 20 + 5. This question teaches the moves: click to add, tap a counter on the chart to take it away. Keep the displays off until pupils have said the number.

It slides into the tens column (23 becomes 33) and the tool says why: every counter has its own column. Dragging a counter off the chart, or tapping it, takes it away. Everything can be undone with ↩ or U.

With Exchange on, the ten ones slide together and become one 10 counter. With it off, the ones column holds 10 counters and Count shows a red 10, because 10 is not a single digit. Both are 10.

Make builds the number with the fewest counters (4 ones and 7 tenths) and shows any column it needs, here the tenths. A column can’t be hidden while it has counters in it, so nothing is ever lost.

The row being filled has a blue edge and a blue label, and clicked counters go there (the A and B keys switch too). Dragging works into either row. Each mode keeps its own counters, so Build’s chart is still there when you go back.

The tens column is empty and Count shows a grey 0. The zero holds the place, so the 4 stays in the hundreds; without it you read 47. Count only shows the zeros a written number needs.

“Three thousand and sixty”. No word stands for the empty hundreds or ones: words skip empty places, figures need a 0. That is why pupils who write numbers from words confuse 3,060, 3,600 and 300,060.

Random numbers use every column showing. Ten thousands and hundredths cause the most trouble: pupils lose track of the thousands and say “point twenty-five” instead of “point two five”. N checks.

Each digit after the point is in its own column: 3 tenths and 5 hundredths. “Point thirty-five” suggests 35 of something, the start of the “0.35 > 0.4” misconception. 2.35 is also 2 and 35 hundredths, which is fine because it names the unit.

5,555 has 4 terms; 5,005 has 2 (5,000 + 5). The number of terms counts the non-zero digits, not the size: 9 has one term and 1,001 has two.

Seven hundredths counters: “zero point zero seven”, 7 × 0.01 = 7/100. The 0 in the tenths place does the same job as the 0 in 407.

1,000 + 100 + 10 + 1 + 0.1 + 0.01: each value is a tenth of the one before. The same 1 counter is worth something different in every column, which is the whole idea of place value.

999. Adding 1 sets off a cascade: ten ones make a ten, ten tens make a hundred, but there is no thousands column, so the hundreds column is left holding 10 (Count shows a red 10). Showing the thousands column and pressing ⇄ Regroup finishes the job: 1,000.

10 clicks make a ten, 100 make a hundred, 1,000 would make a thousand (don’t). Each column needs ten of the column to its right, so each needs ten times as many clicks. The prediction before 100 is the point of the question.

Two: ten ones become a ten, which makes ten tens, which become a hundred. 99 + 1 = 100. Pupils who say 910 have written 9 + 1 in the ones column without regrouping.

Four exchanges, ending at 10,000. Any number ending in exactly three 9s, such as 3,999 or 8,999, gives three. The number of exchanges equals the number of 9s at the end.

Yes: ten hundredths become a tenth, ten tenths become a one. After the first exchange you need 9 more tenths for the second. “Ten of these make one of those” works the same across the decimal point.

Four presses: 47 ones → 1 ten and 37 ones → … → 4 tens and 7 ones. Number stays at 47 throughout: regrouping changes how the number is made, not its value.

Turning Exchange on changes nothing by itself: it acts when a counter is added. The 14th tenth sets it off, leaving 1 one and 4 tenths = 1.4. ⇄ Regroup (R) exchanges without adding anything.

Still 325, now made as 2 × 100 + 11 × 10 + 15 × 1 = 200 + 110 + 15. Breaking down is the move pupils need for subtraction. The value stays the same; only how it is made changes.

23 tens and 4 ones, or 22 tens and 14 ones. “234 is 23 tens and 4” is the fact behind 234 ÷ 10 = 23.4 and behind the question “how many tens are in 234?” (23, not 3).

Six: 5 tens + 3, 4 + 13, 3 + 23, 2 + 33, 1 + 43, 0 tens + 53 ones. Each step swaps one ten for ten ones, and the tens can’t go below 0, so the list ends.

24 tenths; 240 hundredths, because each tenth is 10 hundredths. Other ways include 1 one and 14 tenths, or 2 ones, 3 tenths and 10 hundredths.

100 ones, 10 tens, 1,000 tenths. Each column to the right needs ten times as many counters for the same value. A column holds 100 at most, so the tenths answer is reasoning, not building.

No: the ones column has more than 9 counters, and Count shows a red 12. In the standard partition every column has 0 to 9 counters, and the counts are then the digits. 2 × 10 + 12 × 1 = 20 + 12 = 32.

12: one 1 counter, or b tenths and 100 − 10b hundredths for b = 0 to 10. Listing by the number of tenths is the systematic route; 100 hundredths fills a column exactly.

19: only 9 hundreds and 10 tens. 28: 8 hundreds and 20 tens, or 9 hundreds, 9 tens and 10 ones. Each break-down swaps 1 counter for 10, adding 9, so starting from one thousand counter the totals go 1, 10, 19, 28 …; 20 is never reached.

A < B. Both have 3 hundreds, so compare the tens: 4 < 5. Once a higher column differs, the lower columns don’t matter, even though A has more ones.

A = 3,201 (6 counters) beats B = 462 (12 counters). The number of counters says nothing about size until you know which columns they are in. Pupils who compare by “how many” are ignoring place value.

0.7 > 0.65: 7 tenths beat 6 tenths, whatever the hundredths. Break A’s tenths down with 1 → 10 to see 70 hundredths against 65. “Longer is larger” is one of the most common decimal misconceptions.

Identical counters: 2 ones and 5 tenths, and A ? B shows =. 2.50 has 0 hundredths, so there is nothing to put in that column. A zero at the end of a decimal changes nothing; a zero in the middle (2.05) does.

Start at the highest column and move right until the rows differ. The random pairs differ in one of the top three columns, so the decision comes early and the remaining digits are distractions.

For example 1,000 and 999, or 3,000 and 2,999. A difference of 1 that changes every digit has to cross a …000 / …999 boundary. Pupils who expect a difference of 1 to change only the ones digit meet regrouping here.

For example 600 + 400, 750 + 250, 999 + 1. When neither ends in 0, the ones digits add to 10 and the tens and hundreds digits add to 9, because of the carries.

After combining: 3 hundreds, 11 tens, 12 ones. Regroup: 12 ones → 1 ten and 2 ones (12 tens), then 12 tens → 1 hundred and 2 tens. 422. The two exchanges are the two carried 1s of the written method.

Each carried 1 is one exchange: ten ones became a ten, then ten tens became a hundred. The written method does in digits what the counters do. 👁 Answer checks 422.

Count the columns whose digits, plus any carry, reach 10 or more. Pupils who look only at the digits miss cascades: 95 + 5 needs two exchanges even though 9 + 0 in the tens is less than 10.

6.23. Hundredths: 5 + 8 = 13, so ten hundredths become a tenth. Tenths: 7 + 4 + 1 = 12, so ten tenths become a one: that exchange crosses the point. Ones: 3 + 2 + 1 = 6. The point changes nothing about exchanging.

Only 1 + 0 needs none. 0.6 + 0.4 needs one exchange, 0.65 + 0.35 needs two (hundredths, then tenths). Any two decimals that make a whole number must fill a column to 10 somewhere.

999 + 1: three exchanges in a cascade, giving 1,000. 999 + 999: 54 counters (18 in each column), also three exchanges, giving 1,998.

7 tens and 13 ones is 83, correct but not in standard form. Writing the counts as digits gives 713, a common error. Regrouping makes every column a single digit so the number can be written.

Take away removes 2 tens and 2 ones from each row: A has 3 tens, B has 5 ones left over. A has no ones, so break a ten down (1 → 10 under A’s tens) and take away again: 25.

They took the smaller ones digit from the larger. The counters show A has only 2 ones, so 7 can’t come out without breaking a ten. 52 − 27 = 25.

One hundred comes off each row first. Then break a hundred into 10 tens and a ten into 10 ones, so A is 1 hundred, 9 tens and 10 ones; take away again: 142. Zeros in A are where pupils lose track in the written method.

3.45. A has no hundredths, so a tenth has to become 10 hundredths: 5.2 made as 5.20 in the column method is the paper version of that break.

A column needs a break when B’s digit is bigger than what A has there after earlier breaks. Watch for chains through zeros: 4,005 − 1,237 needs three breaks (thousands, hundreds, tens).

No breaks when every digit of A is at least the digit of B in the same column, such as 586 − 243. A break in the ones and tens: 321 − 189 = 132.

Three breaks (thousand → hundreds, hundred → tens, ten → ones), giving 999. It is the 999 + 1 cascade run backwards: subtraction undoes addition exchange for exchange.

340: every counter moves one column left, so 3 tens become 3 hundreds. “Add a zero” gives the right digits for whole numbers but fails for decimals: 3.4 × 10 is 34, not 3.40. “Every digit moves one place left” always works.

3.4 → 34 → 340 → 34 → 3.4. Multiplying and dividing by 10 undo each other. The chain under the chart keeps the record.

0.05: the 5 tenths move one column right and become 5 hundredths. 0.50 comes from “add a zero”, and 5 from multiplying instead of dividing. Ask where the counters will go before anyone presses.

Every counter moves exactly two columns right: 2,050 ÷ 100 = 20.5. The digits keep their order and their gaps; only the columns change.

Back to 7: the moves are +1, +1, −1, +2, −1, −2 columns, which total 0. Counting columns moved is quicker than working out each step (70, 700, 70, 7,000, 700, 7).

0.0123: the 3 thousandths would need a ten-thousandths column, which this chart doesn’t have. The tool refuses rather than lose counters; on paper the number just gets another decimal place.

0.06 needs two presses (to 6) and 0.006 needs three. Each press moves a counter one column left, so the number of presses is the number of columns between the counter and the ones.

4.12 < 4.9: 1 tenth against 9 tenths. "Point twelve" treats the decimal part as a whole number. Words reads it as "four point one two", which helps.

They meant 406 (4 hundreds and 6 ones); they wrote 4,006 (4 thousands and 6 ones). Both use 10 counters, in different columns. Writing “400” and then “6” side by side causes it; 400 + 6 in expanded form fixes it.

15 tenths = 1.5; ⇄ Regroup shows it as 1 one and 5 tenths. 0.15 is 15 hundredths. The error treats “tenths” as meaning “after the point”.

They did 4 − 0 = 4 instead of 0 − 4: you can’t take 4 from 0, so they took 0 from 4. Take away leaves B’s 4 ones with nothing to come from; break a ten down. 60 − 24 = 36.

43 tens and 2 ones: break each hundred into tens to see it. 3 is the tens digit, not the number of tens. The difference matters for division: 432 ÷ 10 = 43.2.

5.06 < 5.6. The 0 in 5.06 keeps the 6 in the hundredths column; 5.6 has 6 tenths. A zero at the end of a decimal can go (5.60 = 5.6); a zero in the middle can’t.

Largest 20,000 (20 thousands), smallest 0.2 (20 hundredths). 1,000 is impossible: each break-down adds 9 counters (question 27). Closest is 1,000.01: 9 hundreds, 10 tens and 1 hundredth.

Every difference is a multiple of 99, such as 532 and 235 with a difference of 297 = 3 × 99. (100a + 10b + c) − (100c + 10b + a) = 99(a − c), so the tens digit never matters.

Kaprekar’s routine: within seven steps you reach 6174, which gives itself (7641 − 1467 = 6174). Keep zeros as digits: 1000 gives 1000 − 0001.

Six: 5, 14, 23, 32, 41 and 50 (any hundreds counter takes the total past 100). Swapping a 1 counter for a 10 counter adds 9, so every total is 5 more than a multiple of 9, and its digits add to 5. This is the idea behind the test for divisibility by 9.

The link stores the mode, columns, counters and displays, but not the undo history. Good questions ask for a prediction first, have an answer the tool can check, and say which displays to keep off.