Subtracting three-digit numbers by place value, including across zero
The apple barn has 642 apples, packed in crates of a hundred, bags of ten and a few loose, and a store orders 278. With only 2 loose apples and 4 bags on hand, how can the order be filled, and how many apples are left?
Where you are 7 in Multi-digit subtraction
When you reach for it
Any time a three-digit amount is taken away, compared with another, or split into a known part and an unknown part, the question is a subtraction within 1{,}000. A number such as 642 is 6 hundreds, 4 tens and 2 ones (Reading and writing numbers to 1,000), which Expanded form writes as 600 + 40 + 2. To take 278 away from it, take hundreds from hundreds, tens from tens and ones from ones. When a place does not hold enough, trade: one ten for ten ones, exactly as in Regrouping in subtraction, or, new here, one hundred for ten tens. When the place you need to trade from is empty, trade twice in a row.
One picture carries the whole lesson. At a pumpkin farm, Wanjiku runs the apple barn. Apples are packed in crates of 100, bags of 10, and a few loose apples. Opening a crate gives ten bags. Opening a bag gives ten loose apples. Neither move changes how many apples there are, and every trade in this lesson is one of those two moves. Base ten blocks say the same thing: a flat is a hundred, a rod is a ten, a unit is a one, and a flat can be swapped for ten rods or a rod for ten units.
The lesson runs over four days. Day 1 takes away with no trade. Day 2 trades a ten, a hundred, or both. Day 3 meets a zero in the tens place. Day 4 meets two zeros and checks every answer by adding back. Each day moves from blocks to quick drawings (a square for a hundred, a stick for a ten, a dot for a one) to writing by place. The column method that crosses out digits comes later, in the standard subtraction algorithm; here every trade is shown with pieces or written as hundreds, tens and ones, so the child can see what each trade means.
The procedure
Day 1: hundreds from hundreds, tens from tens, ones from ones
Wanjiku has 758 apples: 7 crates, 5 bags and 8 loose. A store buys 325 apples, which is 3 crates, 2 bags and 5 loose. Build 758 with blocks: 7 flats, 5 rods and 8 units. Then take away 3 flats, 2 rods and 5 units.
Count what is left by kind. Flats: 7 - 3 = 4. Rods: 5 - 2 = 3. Units: 8 - 5 = 3. That is 4 flats, 3 rods and 3 units, so
Written in expanded form, the same work is
and 400 + 30 + 3 = 433. A place value table keeps the columns straight. Each column is one kind of block.
| hundreds | tens | ones | |
|---|---|---|---|
| 758 | 7 | 5 | 8 |
| take away 325 | 3 | 2 | 5 |
| left | 4 | 3 | 3 |
Why taking place by place is allowed. Taking 325 away means taking away its pieces: 3 hundreds, 2 tens and 5 ones. Taking them one kind at a time removes exactly the same apples as taking them all at once, so the amount left is the same. And 7 hundreds take away 3 hundreds is 4 hundreds for the same reason that 7 - 3 = 4: a hundred is one thing, a crate, and crates are counted like anything else. That is also why the places must match. The store asked for 3 crates; taking 3 loose apples instead would take far too few.
On Day 1 every place of the first number holds at least as much as the same place of the second, so no trade is needed. The child may start at the hundreds or at the ones; when nothing is short, the order does not matter.
Day 2: trading a ten, then a hundred
Now the order from the question: 642 - 278. Wanjiku has 6 crates, 4 bags and 2 loose apples. The store wants 2 crates, 7 bags and 8 loose. Look at each place before taking anything. The ones are short: 2 loose and 8 wanted. The tens are short: 4 bags and 7 wanted. The hundreds have plenty.
Start with the ones. Open one bag: now there are 6 crates, 3 bags and 12 loose apples, still 600 + 30 + 12 = 642. Then the tens: 3 bags and 7 wanted. Open one crate: now there are 5 crates, 13 bags and 12 loose apples, still 500 + 130 + 12 = 642. Every place now has enough.
| hundreds | tens | ones | |
|---|---|---|---|
| 642 | 6 | 4 | 2 |
| after trading a ten for ten ones | 6 | 3 | 12 |
| after trading a hundred for ten tens | 5 | 13 | 12 |
| take away 278 | 2 | 7 | 8 |
| left | 3 | 6 | 4 |
Ones: 12 - 8 = 4. Tens: 13 - 7 = 6. Hundreds: 5 - 2 = 3.
Wanjiku hands over 2 crates, 7 bags and 8 loose apples, and keeps 3 crates, 6 bags and 4 loose.
Why a trade is allowed. A hundred is ten tens and a ten is ten ones. Opening a crate changes how the apples are packed, not how many there are. So “6 hundreds, 3 tens, 12 ones” and “5 hundreds, 13 tens, 12 ones” are both names for 642, written in pieces that make the taking away possible. A place may hold more than nine for a while. That is a working form, not a new number.
Why start with the ones. Trading for the ones takes a ten out of the tens place, so the tens change before anyone looks at them. Working up from the ones settles each place once. Starting with the tens can mean going back: in 652 - 258 the tens look fine (5 and 5), but the ones need a ten, which leaves 4 tens, and then the tens need a hundred after all. So the place to compare is the one after any trade below it.
Day 3: a zero in the tens
705 - 268. Wanjiku has 7 crates and 5 loose apples and no bags at all. The order is 2 crates, 6 bags and 8 loose. The ones are short, 5 and 8 wanted, so a bag must be opened. But there is no bag to open.
So open a crate first. That gives 6 crates, 10 bags and 5 loose. Now there are bags, so open one of them: 6 crates, 9 bags and 15 loose. That is still 600 + 90 + 15 = 705.
| hundreds | tens | ones | |
|---|---|---|---|
| 705 | 7 | 0 | 5 |
| after trading a hundred for ten tens | 6 | 10 | 5 |
| after trading a ten for ten ones | 6 | 9 | 15 |
| take away 268 | 2 | 6 | 8 |
| left | 4 | 3 | 7 |
Why it works. These are the same two trades as on Day 2, and each keeps the amount the same. The zero only fixes their order: a ten cannot come out of an empty place, so the hundred has to become tens first. After trading across a zero the tens place always holds 9, the 10 new tens less the one that went to the ones.
A zero in the ones place is nothing new. It is a short ones place like any other, and the tens place has a ten to give, as on Day 2.
Day 4: two zeros, and checking by adding back
800 - 356. Eight crates, no bags, no loose apples. The ones are short and the tens are empty, so the trades run as on Day 3. Open a crate: 7 hundreds, 10 tens, 0 ones. Open a bag: 7 hundreds, 9 tens, 10 ones. Now take away 3 hundreds, 5 tens and 6 ones.
Ones: 10 - 6 = 4. Tens: 9 - 5 = 4. Hundreds: 7 - 3 = 4.
Checking by adding back. The apples handed over and the apples kept make up all the apples Wanjiku started with. So 444 + 356 must come back to 800. Add by place, as in the partial sums method:
since 400 + 300 = 700, 40 + 50 = 90 and 4 + 6 = 10. Look at how the pieces close up. The 10 ones make a new ten, and with the 9 tens that is 10 tens, a new hundred, and with the 7 hundreds that is 8. The trades of the subtraction come back in reverse: the bag that was opened is filled again, and so is the crate. This is the parts and whole idea from number bonds at a bigger size: the part taken and the part left make the whole, which is also why a subtraction can be read as a missing part.
A second check is to count up from 356: 44 more reaches 400 and 400 more reaches 800, so the difference is 44 + 400 = 444.
Worked examples
Examples 1 and 2 belong to Day 1, 3 and 4 to Day 2, 5 and 6 to Day 3, and 7 and 8 to Day 4. Petar and Kainoa belong to a train set club that keeps its track pieces in boxes of 100, packs of 10 and single pieces.
Day 1: Lending track pieces
Petar has 469 track pieces: 4 boxes, 6 packs and 9 singles. He lends 240 pieces to Kainoa. How many pieces does Petar have left?
SolutionThe question is 469 - 240. In expanded form, 469 = 400 + 60 + 9 and 240 = 200 + 40 + 0. Each place has enough, so take place by place.
Hundreds: 4 - 2 = 2. Tens: 6 - 4 = 2. Ones: 9 - 0 = 9, since taking no singles leaves all of them.
Petar keeps 2 boxes, 2 packs and 9 singles. Check by adding back: 229 + 240 is 400 + 60 + 9 = 469.
Day 1: How many more apples
The barn crew picked 587 apples on Monday and 362 on Tuesday. How many more apples were picked on Monday?
SolutionMonday’s apples can be split into two parts: 362 to match Tuesday’s, and the extra. The extra is the part left when 362 is taken from 587, so the question is 587 - 362.
This time start at the hundreds, since no place is short. Hundreds: 5 - 3 = 2. Tens: 8 - 6 = 2. Ones: 7 - 2 = 5.
Monday’s crew picked 225 more apples. Check: 225 + 362 = 500 + 80 + 7 = 587. A comparison like this uses the same subtraction as taking away, because “how many more” asks for the part of the larger amount that the smaller one does not match.
Day 2: Only the tens are short
Kainoa has 527 track pieces and uses 153 of them to build a bridge. How many pieces are left?
SolutionThe question is 527 - 153. Look at each place first. Ones: 7 and 3, enough. Tens: 2 and 5, short. Hundreds: 5 and 1, enough.
Only the tens need a trade, and the trade comes from the hundreds. Open one box: 4 boxes, 12 packs and 7 singles, still 400 + 120 + 7 = 527.
Ones: 7 - 3 = 4. Tens: 12 - 5 = 7. Hundreds: 4 - 1 = 3.
Kainoa has 374 pieces left. Check: 374 + 153 = 400 + 120 + 7 = 527. There was one trade, and it was a hundred: a trade happens only in a place that is short, never as a habit.
Day 2: A trade that empties the tens
The barn has 913 apples and a farm stand orders 457. How many apples are left?
SolutionThe question is 913 - 457. The ones are short: 3 and 7 wanted. Open a bag: 9 hundreds, 0 tens, 13 ones.
Now compare the tens after that trade: 0 bags and 5 wanted, so the tens are short. Open a crate: 8 hundreds, 10 tens, 13 ones. Check that it is still the same amount: 800 + 100 + 13 = 913.
| hundreds | tens | ones | |
|---|---|---|---|
| 913 | 9 | 1 | 3 |
| after both trades | 8 | 10 | 13 |
| take away 457 | 4 | 5 | 7 |
| left | 4 | 5 | 6 |
Check by adding back: 456 + 457 has 6 + 7 = 13 ones, 50 + 50 = 100, and 400 + 400 = 800, and 800 + 100 + 13 = 913.
Day 3: Across a zero at the train club
The club has 403 pieces of straight track. Petar lays 175 of them on the new layout. How many pieces of straight track are not on the layout?
SolutionThe question is 403 - 175: 4 boxes and 3 singles, with no packs, take away 1 box, 7 packs and 5 singles.
The singles are short, 3 and 5 wanted, and there is no pack to open. So open a box first: 3 boxes, 10 packs, 3 singles. Then open a pack: 3 boxes, 9 packs, 13 singles. Still 300 + 90 + 13 = 403.
Ones: 13 - 5 = 8. Tens: 9 - 7 = 2. Hundreds: 3 - 1 = 2.
228 pieces are not on the layout. Check: 228 + 175 is 300 + 90 + 13 = 403.
Day 3: How many more pieces are needed
Kainoa needs 520 pieces to build a big loop. He has 284. How many more pieces does he need?
SolutionThe pieces he has and the pieces he still needs are the two parts of 520, so the question is 284 + \square = 520, and the missing part is 520 - 284.
Here the zero is in the ones place. The ones are short, 0 and 4 wanted, and the tens have a ten to give: 5 hundreds, 1 ten, 10 ones. Now the tens are short, 1 and 8 wanted, so trade a hundred: 4 hundreds, 11 tens, 10 ones, still 400 + 110 + 10 = 520.
Ones: 10 - 4 = 6. Tens: 11 - 8 = 3. Hundreds: 4 - 2 = 2.
Kainoa needs 236 more pieces. Check: 284 + 236 has 4 + 6 = 10 ones, 80 + 30 = 110 and 200 + 200 = 400, and 400 + 110 + 10 = 520.
Day 4: Six full crates
The barn has exactly 600 apples, 6 full crates. A school orders 247 apples for a field trip. How many apples are left in the barn?
SolutionThe question is 600 - 247. The ones are short and the tens are empty, so open a crate, then a bag: 6 hundreds become 5 hundreds, 10 tens, 0 ones, and then 5 hundreds, 9 tens, 10 ones.
Ones: 10 - 7 = 3. Tens: 9 - 4 = 5. Hundreds: 5 - 2 = 3.
Check by adding back: 353 + 247 has 3 + 7 = 10 ones, 50 + 40 = 90, and 300 + 200 = 500, and 500 + 90 + 10 = 600. Check again by counting up from 247: 3 more is 250, 50 more is 300, 300 more is 600, and 3 + 50 + 300 = 353.
Day 4: A thousand pieces
The club buys a set of 1{,}000 track pieces and lays 365 of them. How many pieces are still in the set’s boxes?
SolutionThe question is 1{,}000 - 365. A thousand is ten hundreds, so think of it as 10 boxes, 0 packs and 0 singles. The singles are short and there are no packs, so open a box: 9 boxes, 10 packs, 0 singles. Then open a pack: 9 boxes, 9 packs, 10 singles. Still 900 + 90 + 10 = 1{,}000.
Ones: 10 - 5 = 5. Tens: 9 - 6 = 3. Hundreds: 9 - 3 = 6.
There are 635 pieces still in the boxes. Check: 635 + 365 has 5 + 5 = 10 ones, 30 + 60 = 90 and 600 + 300 = 900, and 900 + 90 + 10 = 1{,}000. Every place filled up and passed one piece along, which is the two trades undone in turn.
Boundaries
- The first number must be at least as large as the second. 325 - 758 has no answer among the whole numbers, since there is nowhere above the hundreds to trade from. When the first number is the larger one, the hundreds are never short once the ones and tens are settled (Exercise 22).
- A trade is one piece for ten of the next smaller kind. A hundred becomes ten tens and a ten becomes ten ones. A hundred is also 100 ones, but nobody draws 100 dots; across a zero the two trades in a row do the same job in two small steps.
- A trade changes the pieces, never the amount. Every row of a quick drawing before the crossing out is the first number. A place may hold 10 to 19 while working, but the amount left always has a single digit in each place, because a place is only traded into when it is short: it then holds less than ten more than what is taken from it.
- Within 1{,}000. The largest first number here is 1{,}000, ten hundreds. Past that the numbers have a thousands place, and the same trades continue one place further up.
- Like from like. Hundreds are taken only from hundreds, tens from tens and ones from ones. A digit means nothing on its own; it counts crates, bags or loose apples according to its place.
Where this goes next
The same trades, recorded by crossing out a digit and writing a smaller one above it, are the standard subtraction algorithm. Zeros in longer numbers, and a shortcut that reads 800 as 80 tens, are the subject of subtracting across zeros. The adding companion is the partial sums method, which is also the tool for checking by adding back. Counting up gives a second way to find a difference, often quicker when the first number has zeros. Three-digit differences then turn up in story problems of every kind: taking away, comparing, and finding a missing part.
Exercises
Part 1: no trade
- A pile has 6 flats, 8 rods and 4 units. Take away 3 flats, 1 rod and 2 units. Write the subtraction and find what is left.
- Find 896 - 534.
- Find 475 - 205.
- Wanjiku has 968 apples and sells 520. How many apples are left?
- Write 759 and 413 in expanded form. Use them to find 759 - 413.
Part 2: trading a ten, a hundred, or both
- Look at each place first. Which place needs a trade in (a) 563 - 238, (b) 563 - 381, (c) 563 - 297? Then find all three differences.
- Find 734 - 186. Write each trade in a place value table.
- Find 451 - 175.
- A quick drawing shows 3 squares, 15 sticks and 6 dots. What number is it?
- Petar has 825 track pieces and Kainoa has 368. How many more pieces does Petar have?
- Find 382 - 145 and check by adding back.
Part 3: a zero in the first number
- Find 506 - 234.
- Find 302 - 127.
- Find 608 - 459. How many ones are there after both trades?
- Wanjiku has 4 crates and 7 loose apples. She sells 158 apples. How many are left?
- Find 750 - 316.
Part 4: two zeros, and checking
- Find 500 - 263 and check by adding back.
- Find 900 - 418 and check by adding back.
- Find 700 - 9.
- The club owns 1{,}000 track pieces, and 572 are on the layout. How many pieces are in the boxes?
- Find the missing number: 600 - \square = 275.
Reasoning
- The first number is larger than the second. Can the hundreds place still be short after the ones and tens have been traded? Explain.
- Show that 9 hundreds 0 tens 2 ones, 8 hundreds 10 tens 2 ones, and 8 hundreds 9 tens 12 ones are all the same number.
- Find two subtractions of three-digit numbers with a difference of 248: one that needs no trade and one that needs two trades.
- A three-digit number ends in two zeros, and 1 is taken away from it. Why does the answer always end in 99?
Answers
- 684 - 312 = 372. Flats 6 - 3 = 3, rods 8 - 1 = 7, units 4 - 2 = 2, so 3 flats, 7 rods and 2 units are left.
- 896 - 534 = 362. Hundreds 8 - 5 = 3, tens 9 - 3 = 6, ones 6 - 4 = 2.
- 475 - 205 = 270. Hundreds 4 - 2 = 2, tens 7 - 0 = 7, ones 5 - 5 = 0.
- 448 apples. 968 - 520 = 448: hundreds 9 - 5 = 4, tens 6 - 2 = 4, ones 8 - 0 = 8.
- 759 = 700 + 50 + 9 and 413 = 400 + 10 + 3. Then 700 - 400 = 300, 50 - 10 = 40, 9 - 3 = 6, so 759 - 413 = 346.
- (a) Only the ones: 563 - 238 = 325, after trading to 5 hundreds, 5 tens, 13 ones. (b) Only the tens: 563 - 381 = 182, after trading to 4 hundreds, 16 tens, 3 ones. (c) Both: 563 - 297 = 266. The ones trade leaves 5 tens, fewer than 9, so the tens trade too, giving 4 hundreds, 15 tens, 13 ones.
- 734 - 186 = 548. The ones are short (4 and 6): 7 hundreds, 2 tens, 14 ones. The tens are then short (2 and 8): 6 hundreds, 12 tens, 14 ones. Then 14 - 6 = 8, 12 - 8 = 4, 6 - 1 = 5.
- 451 - 175 = 276. Trade a ten: 4 hundreds, 4 tens, 11 ones. Trade a hundred: 3 hundreds, 14 tens, 11 ones. Then 11 - 5 = 6, 14 - 7 = 7, 3 - 1 = 2.
- 456. The squares are 300, the 15 sticks are 150 and the dots are 6, and 300 + 150 + 6 = 456. Trading ten of the sticks for a square gives 4 squares, 5 sticks and 6 dots.
- 457 more. 825 - 368 = 457: trade a ten (8 hundreds, 1 ten, 15 ones), then a hundred (7 hundreds, 11 tens, 15 ones); 15 - 8 = 7, 11 - 6 = 5, 7 - 3 = 4.
- 382 - 145 = 237. Trade a ten: 3 hundreds, 7 tens, 12 ones; then 12 - 5 = 7, 7 - 4 = 3, 3 - 1 = 2. Check: 237 + 145 = 300 + 70 + 12 = 382.
- 506 - 234 = 272. The ones are fine; the tens are 0 and 3 wanted, so trade a hundred: 4 hundreds, 10 tens, 6 ones. Then 6 - 4 = 2, 10 - 3 = 7, 4 - 2 = 2.
- 302 - 127 = 175. Trade a hundred (2 hundreds, 10 tens, 2 ones), then a ten (2 hundreds, 9 tens, 12 ones). Then 12 - 7 = 5, 9 - 2 = 7, 2 - 1 = 1.
- 608 - 459 = 149, with 18 ones after both trades. 6 hundreds 0 tens 8 ones becomes 5 hundreds 10 tens 8 ones, then 5 hundreds 9 tens 18 ones. Then 18 - 9 = 9, 9 - 5 = 4, 5 - 4 = 1.
- 249 apples. She has 407 apples and 407 - 158 = 249: open a crate (3 crates, 10 bags, 7 loose), then a bag (3 crates, 9 bags, 17 loose); 17 - 8 = 9, 9 - 5 = 4, 3 - 1 = 2.
- 750 - 316 = 434. The zero is in the ones, and the tens have a ten to give: 7 hundreds, 4 tens, 10 ones. Then 10 - 6 = 4, 4 - 1 = 3, 7 - 3 = 4.
- 500 - 263 = 237. Two trades give 4 hundreds, 9 tens, 10 ones; then 10 - 3 = 7, 9 - 6 = 3, 4 - 2 = 2. Check: 237 + 263 = 400 + 90 + 10 = 500.
- 900 - 418 = 482. Two trades give 8 hundreds, 9 tens, 10 ones; then 10 - 8 = 2, 9 - 1 = 8, 8 - 4 = 4. Check: 482 + 418 = 800 + 90 + 10 = 900.
- 700 - 9 = 691. Two trades give 6 hundreds, 9 tens, 10 ones, and only the ones lose anything: 10 - 9 = 1.
- 428 pieces. 1{,}000 - 572 = 428: ten hundreds become 9 hundreds, 9 tens, 10 ones; then 10 - 2 = 8, 9 - 7 = 2, 9 - 5 = 4.
- \square = 325. The part taken and the part left make 600, so the missing number is 600 - 275 = 325. Check: 600 - 325 = 275.
- No. Once the ones and tens are settled, the tens and ones left over make at most 9 tens and 9 ones, which is 99, less than one hundred. If the hundreds were short by even one, the first number would be at least a hundred less in the hundreds and at most 99 more in the tens and ones, so it would be smaller than the second number.
- 900 + 0 + 2 = 902, 800 + 100 + 2 = 902 and 800 + 90 + 12 = 902. Each form comes from the one before by a single trade: a hundred for ten tens, then a ten for ten ones.
- Many answers. No trade: 379 - 131 = 248. Two trades: 512 - 264 = 248, where the ones (2 and 4) and then the tens (0 and 6) are short.
- With 0 tens and 0 ones, taking 1 needs a trade across the zero: a hundred becomes ten tens, then a ten becomes ten ones. That leaves 9 tens and 10 ones, and taking 1 leaves 9 tens and 9 ones. For example 400 - 1 = 399.