Grade 4 · 3.NBT.A.2 · 4.NBT.B.4

Subtracting across zeros

A fossil museum has 7{,}006 fossils and lends out 2{,}849 of them. The ones run short, but the tens and hundreds hold nothing to unbundle. Where does the trade come from?

You need

The standard subtraction algorithmGrade 4

Where you are 7 in Multi-digit subtraction

Number9

Arithmetic5

Addition and subtraction12

Multi-digit subtraction7

▴ 2 above Regrouping in subtraction Subtracting three-digit numbers by place value, including across zero The standard subtraction algorithm

Subtracting across zeros

Counting up to find a difference

When you reach for it

Bahar works at a fossil museum. The museum owns 7{,}006 fossils, and it lends 2{,}849 of them to a school for a year. How many fossils stay at the museum?

The answer is 7{,}006 - 2{,}849. You already know the standard subtraction algorithm: line up the places, start at the ones, and when a place on top runs short, unbundle one unit from the place to its left. That one trade is regrouping in subtraction.

Here the ones run short: 6 ones, and you need to take away 9. The usual move is to unbundle a ten. But the tens place holds 0. So does the hundreds place. There is no ten to unbundle, and no hundred either.

This lesson is about that case: subtracting across zeros. It comes up all the time, because round numbers are full of zeros. A goal of 1{,}000 visitors, a fundraising goal of 10{,}000 dollars, a box of 500 labels: every one of them, minus some amount, asks you to unbundle through zeros.

There are two good ways to think about it. Part 1 walks left to the first place that has something, and passes the trade back down one place at a time. Part 2 changes both numbers first, so that no trade is needed at all.

The procedure

Part 1: walking left to the first nonzero place

Take 7{,}006 - 2{,}849, lined up on the right.

\begin{array}{r} 7{,}006 \\ -\;2{,}849 \\ \hline \end{array}

Ones. The top has 6 ones and you need to take away 9. Not enough. Look left for something to unbundle. The tens hold 0. The hundreds hold 0. The thousands hold 7. That is the first place with anything in it, so the trade starts there.

Pass the trade down. Each step is the ordinary trade between two neighboring places.

Unbundle one thousand into ten hundreds. The thousands go from 7 to 6. The hundreds go from 0 to 10.
Now the hundreds have something. Unbundle one hundred into ten tens. The hundreds go from 10 to 9. The tens go from 0 to 10.
Now the tens have something. Unbundle one ten into ten ones. The tens go from 10 to 9. The ones go from 6 to 16.

Now every column can be subtracted, right to left.

Ones. 16 - 9 = 7.
Tens. 9 - 4 = 5.
Hundreds. 9 - 8 = 1.
Thousands. 6 - 2 = 4.

The difference is 4{,}157. The museum keeps 4{,}157 fossils.

Figure 17{,}006 - 2{,}849 = 4{,}157. The trade starts at the thousands and is passed down through both zeros to the ones.

Read the record in the figure. The thousand that was unbundled did not vanish. Part of it went to the hundreds and stayed there as 9 hundreds, part went to the tens and stayed as 9 tens, and the last ten went to the ones. After all the trades, the top number reads 6 thousands, 9 hundreds, 9 tens and 16 ones, and

6{,}000 + 900 + 90 + 16 = 7{,}006.

The value did not change. Only the bundling changed.

One big trade. There is a quicker way to see the same thing. Read 7{,}006 as tens and ones: 7 thousands is 700 tens, so 7{,}006 is 700 tens and 6 ones. You need more ones, so unbundle one ten. That leaves 699 tens and 16 ones.

Place	7{,}006	After the trades
thousands	7	6
hundreds	0	9
tens	0	9
ones	6	16

And 699 tens is exactly 6 thousands, 9 hundreds and 9 tens, the digits in the last column. So all the trades through the zeros are really a single trade: one ten taken from the 700 tens. This is why every zero you pass through turns into a 9. It gave up one of its ten, and kept nine.

Check by adding back. If the difference is right, then 4{,}157 + 2{,}849 is 7{,}006. Use the standard addition algorithm. Ones: 7 + 9 = 16, write 6, carry 1. Tens: 5 + 4 + 1 = 10, write 0, carry 1. Hundreds: 1 + 8 + 1 = 10, write 0, carry 1. Thousands: 4 + 2 + 1 = 7. The sum is 7{,}006, so the difference is right.

Part 2: taking the same amount from both numbers

Ikaika is collecting coins for the museum’s fundraising goal of 1{,}000 dollars. He has 487 dollars so far. How much more does he need?

That is 1{,}000 - 487. Every place on top except the thousands is 0, so the Part 1 method passes a trade through every place: the thousands go from 1 to 0, the hundreds and tens each become 9, and the ones become 10. Then 10 - 7 = 3, 9 - 8 = 1 and 9 - 4 = 5, so the difference is 513.

Here is another way. Take 1 away from both numbers:

1{,}000 - 487 \quad\text{becomes}\quad 999 - 486.

Now look at 999 - 486. Every digit on top is 9, and no digit below can be bigger than 9. So no column ever runs short and no trade is needed: 9 - 6 = 3, 9 - 8 = 1, 9 - 4 = 5. The answer is 513, the same as before. Ikaika needs 513 more dollars.

Figure 21{,}000 - 487 needs a trade in every place. Taking 1 from both numbers gives 999 - 486, which needs none, and the difference is the same 513.

Why may you change both numbers? A difference is the size of a gap: how far it is from the smaller number up to the larger one. Picture both numbers on a number line. If you slide both ends of the gap one step to the left, the gap moves, but its length stays the same.

Figure 3Sliding both ends of a gap the same distance leaves the gap the same length. That is why 100 - 37 and 99 - 36 are equal.

The figure shows it with smaller numbers. From 37 to 100 is 63 steps. From 36 to 99 is also 63 steps, because each end moved back by one. The same is true for any amount you slide both ends: take the same amount from both numbers (or add the same amount to both), and the difference does not change. This is called compensation: one change makes up for the other. It is a cousin of compensation when adding and subtracting, where you add a round number and then take the extra back off.

Choosing how much to take. The aim is a top number made of 9s at the end. For 1{,}000, take 1 and get 999. The trick also works when the top number does not end in 0. For 7{,}006 - 2{,}849, take 7 from both:

7{,}006 - 2{,}849 = 6{,}999 - 2{,}842.

Now 9 - 2 = 7, 9 - 4 = 5, 9 - 8 = 1 and 6 - 2 = 4. The difference is 4{,}157 again, with no trades.

Both ways give the same answer, because both are fair: Part 1 rebundles the top number, and Part 2 slides the whole gap. Use whichever is easier for the numbers in front of you, and check the answer by adding back.

Worked examples

Example 1

Part 1: one zero in the way

The museum has 605 trilobite fossils. Bahar puts 238 of them on display. How many are left in storage?

Solution

Take away: 605 - 238.

Ones. 5 - 8 runs short. The tens hold 0, so look further left: the hundreds hold 6. Unbundle one hundred into ten tens: hundreds 6 \to 5, tens 0 \to 10. Then unbundle one ten into ten ones: tens 10 \to 9, ones 5 \to 15. Now 15 - 8 = 7.

Tens. 9 - 3 = 6.

Hundreds. 5 - 2 = 3.

\begin{array}{r} 605 \\ -\;238 \\ \hline 367 \end{array}

There are 367 fossils in storage. Seen as one trade: 605 is 60 tens and 5 ones, and unbundling one ten leaves 59 tens and 15 ones. Check: 367 + 238 = 605.

Example 2

Part 1: a zero met in the middle

Ikaika counts 8{,}040 fossil shells in the museum’s drawers. He packs 3{,}276 of them for a traveling show. How many shells stay in the drawers?

Solution

Take away: 8{,}040 - 3{,}276.

Ones. 0 - 6 runs short. The tens hold 4, so this is an ordinary trade: tens 4 \to 3, ones 0 \to 10. Then 10 - 6 = 4.

Tens. The tens now hold 3, and 3 - 7 runs short. The hundreds hold 0, so walk left to the thousands. Unbundle a thousand: thousands 8 \to 7, hundreds 0 \to 10. Then unbundle a hundred: hundreds 10 \to 9, tens 3 \to 13. Now 13 - 7 = 6.

Hundreds. 9 - 2 = 7.

Thousands. 7 - 3 = 4.

\begin{array}{r} 8{,}040 \\ -\;3{,}276 \\ \hline 4{,}764 \end{array}

There are 4{,}764 shells left. A zero only matters when the place next to it needs a trade. Here the ones did not need the hundreds at all; the tens did. Check: 4{,}764 + 3{,}276 = 8{,}040.

Example 3

Part 1: three zeros in a row

The museum’s archive holds 30{,}005 fossil fragments. Sipho sends 12{,}368 of them to a university to be studied. How many fragments are still in the archive? Then show that the top number keeps its value after the trades.

Solution

Take away: 30{,}005 - 12{,}368.

Ones. 5 - 8 runs short. The tens, hundreds and thousands all hold 0. The first nonzero place is the ten thousands. Pass the trade down:

ten thousands 3 \to 2, thousands 0 \to 10;
thousands 10 \to 9, hundreds 0 \to 10;
hundreds 10 \to 9, tens 0 \to 10;
tens 10 \to 9, ones 5 \to 15.

Now subtract each column: ones 15 - 8 = 7, tens 9 - 6 = 3, hundreds 9 - 3 = 6, thousands 9 - 2 = 7, ten thousands 2 - 1 = 1.

\begin{array}{r} 30{,}005 \\ -\;12{,}368 \\ \hline 17{,}637 \end{array}

There are 17{,}637 fragments still in the archive.

As one trade: 30{,}005 is 3{,}000 tens and 5 ones. Unbundle one ten and you have 2{,}999 tens and 15 ones, and 2{,}999 tens is 2 ten thousands, 9 thousands, 9 hundreds and 9 tens. The top still totals

20{,}000 + 9{,}000 + 900 + 90 + 15 = 30{,}005.

Check by adding: 17{,}637 + 12{,}368 = 30{,}005.

Example 4

Part 2: a round number, both ways

Find 600 - 274 by passing trades through the zero, and again by taking 1 from both numbers.

Solution

Through the zero. Ones: 0 - 4 runs short, the tens hold 0, so unbundle a hundred (hundreds 6 \to 5, tens 0 \to 10) and then a ten (tens 10 \to 9, ones 0 \to 10). Then 10 - 4 = 6, 9 - 7 = 2 and 5 - 2 = 3. The difference is 326.

Taking 1 from both. The difference stays the same:

600 - 274 = 599 - 273.

Now 9 - 3 = 6, 9 - 7 = 2, 5 - 2 = 3, with no trades. The difference is 326.

The two methods agree, as they must. Check: 326 + 274 = 600.

Example 5

Part 2: how far to a fundraising goal

The museum’s fundraising goal is 10{,}000 dollars, to build a new case for a dinosaur skull. Bahar’s report says 3{,}618 dollars have been raised. How much more is needed?

Solution

“How much more is needed” asks for the gap from 3{,}618 up to 10{,}000, so subtract: 10{,}000 - 3{,}618.

Every place on top but one is 0, which would mean a trade in every column. Take 1 from both numbers instead:

10{,}000 - 3{,}618 = 9{,}999 - 3{,}617.

Now subtract with no trades: 9 - 7 = 2, 9 - 1 = 8, 9 - 6 = 3, 9 - 3 = 6. The museum needs 6{,}382 more dollars.

Check by adding: 6{,}382 + 3{,}618. Ones 2 + 8 = 10, write 0, carry 1. Tens 8 + 1 + 1 = 10, write 0, carry 1. Hundreds 3 + 6 + 1 = 10, write 0, carry 1. Thousands 6 + 3 + 1 = 10, write 0, carry 1. The sum is 10{,}000.

Example 6

Part 2: taking more than 1 from both

Find 4{,}002 - 1{,}765 by taking the same amount from both numbers.

Solution

Taking 1 from both gives 4{,}001 - 1{,}764, and the ones still run short (1 - 4). To make the top end in 9s, take enough to turn 4{,}002 into 3{,}999. That is 3, since 4{,}002 - 3 = 3{,}999. Take 3 from the bottom number too: 1{,}765 - 3 = 1{,}762.

4{,}002 - 1{,}765 = 3{,}999 - 1{,}762.

Now 9 - 2 = 7, 9 - 6 = 3, 9 - 7 = 2 and 3 - 1 = 2. The difference is 2{,}237.

Check: 2{,}237 + 1{,}765 = 4{,}002. Both numbers moved down by 3, so the gap between them did not move at all.

Example 7

Part 2: a goal reached in two steps

Sipho’s class wants to raise 8{,}000 dollars for the museum. They raise 2{,}475 dollars at a bake sale and 1{,}860 dollars at a book sale. How much is still needed to reach the goal?

Solution

First way: add, then subtract. The two sales together raise 2{,}475 + 1{,}860 = 4{,}335 dollars. The amount still needed is 8{,}000 - 4{,}335. Take 1 from both:

8{,}000 - 4{,}335 = 7{,}999 - 4{,}334.

Then 9 - 4 = 5, 9 - 3 = 6, 9 - 3 = 6 and 7 - 4 = 3, so 3{,}665 dollars are still needed.

Second way: subtract one sale at a time. After the bake sale, 8{,}000 - 2{,}475 are needed. That is 7{,}999 - 2{,}474 = 5{,}525. After the book sale, 5{,}525 - 1{,}860 = 3{,}665. Here the tens run short (2 - 6), so a hundred is unbundled: hundreds 5 \to 4, tens 12, and 12 - 6 = 6; then 4 - 8 runs short, so a thousand is unbundled: thousands 5 \to 4, hundreds 14, and 14 - 8 = 6; then 4 - 1 = 3.

Both ways give 3{,}665 dollars. Taking away the two amounts one after the other is the same as taking away their sum.

Boundaries

Larger number on top. As with any whole-number subtraction, the top number must be at least as large as the bottom one. If the walk to the left never finds a nonzero place, the top was too small, and the answer is not a whole number.
The walk stops at the first nonzero place. The trade comes from the nearest place to the left that holds at least 1. Every zero between that place and the column that ran short ends as 9, and the column itself gains 10.
The same amount, from both. Compensation in subtraction needs the same change to both numbers. For 600 - 274, taking 1 from only the top gives 599 - 274 = 325, which is 1 too small: the gap lost a step at one end.
Shifting must keep whole numbers. In this lesson both numbers stay whole numbers, so the amount taken from both can be no more than the bottom number. For 7{,}006 - 3, taking 7 from both would need 3 - 7 on the bottom, which is not a whole number. There, pass the trade through the zeros instead.

Where this goes next

Subtracting across zeros finishes the column method for whole numbers: with it, the standard algorithm works for any whole-number difference with the larger number on top. The same moves carry over to arithmetic with decimals, where 5 - 2.37 is 5.00 - 2.37 and the zeros after the decimal point are unbundled in just the same way. The idea of shifting both numbers to keep a difference comes back again and again, in mental math with money and later in algebra, where adding or subtracting the same amount on both sides keeps an equation true.

Exercises

Part 1: unbundling across zeros

Find 403 - 156.
Find 802 - 365.
Find 6{,}004 - 1{,}537.
Find 9{,}005 - 4{,}678.
Find 3{,}020 - 1{,}486.
Find 7{,}100 - 2{,}654.
Find 40{,}003 - 17{,}526.
Find 60{,}200 - 34{,}815.
Bahar’s museum has 2{,}005 fossil shark teeth. It sends 768 of them to another museum. How many shark teeth are left?
Write 8{,}003 as tens and ones. Unbundle one ten and write it again. Then find 8{,}003 - 4{,}357.
Ikaika writes 6{,}005 - 2{,}847 = 3{,}268. Check by adding back. If it is wrong, give the right difference.

Part 2: taking the same amount from both

Find 1{,}000 - 352 by first taking 1 from both numbers.
Find 700 - 463 by first taking 1 from both numbers.
Find 3{,}000 - 1{,}274.
Find 10{,}000 - 6{,}059.
Find 9{,}000 - 4{,}526.
Find 6{,}003 - 2{,}578 by taking the same amount from both numbers. Say how much you take.
The museum’s fundraising goal for a new sign is 2{,}000 dollars, and 1{,}365 dollars have been raised. How much more is needed?
Sipho hopes the museum will have 20{,}000 visitors this year. So far it has had 13{,}742. How many more visitors would reach the goal?
A goal of 6{,}000 dollars is set. Donors give 1{,}850 dollars and then 2{,}375 dollars. How much is still needed?
Find the missing number: 2{,}486 + \square = 9{,}000.

Reasoning

Explain why 4{,}999 - 2{,}738 can be done with no trades, but 5{,}000 - 2{,}739 needs a trade in every place.
Bahar says 5{,}000 - 2{,}163 and 5{,}001 - 2{,}164 have the same answer. Is she right? Explain, and find the answer.
Ikaika finds 1{,}000 - 487 by working out 999 - 487 instead. What does he get, and why is it off? Use the number line picture in your answer.
In 5{,}004 - 1{,}628, the hundreds and tens both end as 9. Explain why, using 500 tens, and find the difference.

Answers

247. Hundreds 4 \to 3, tens 0 \to 10 \to 9, ones 3 \to 13. Then 13 - 6 = 7, 9 - 5 = 4, 3 - 1 = 2.
437. Hundreds 8 \to 7, tens 0 \to 10 \to 9, ones 2 \to 12. Then 12 - 5 = 7, 9 - 6 = 3, 7 - 3 = 4.
4{,}467. Thousands 6 \to 5, hundreds and tens each 0 \to 10 \to 9, ones 4 \to 14. Then 14 - 7 = 7, 9 - 3 = 6, 9 - 5 = 4, 5 - 1 = 4.
4{,}327. Thousands 9 \to 8, hundreds and tens become 9, ones 5 \to 15. Then 15 - 8 = 7, 9 - 7 = 2, 9 - 6 = 3, 8 - 4 = 4.
1{,}534. Ones: tens 2 \to 1, 10 - 6 = 4. Tens: 1 - 8 runs short and the hundreds hold 0, so thousands 3 \to 2, hundreds 0 \to 10 \to 9, tens 11, and 11 - 8 = 3. Hundreds 9 - 4 = 5. Thousands 2 - 1 = 1.
4{,}446. Ones: the tens hold 0, so hundreds 1 \to 0, tens 0 \to 10 \to 9, ones 10, and 10 - 4 = 6. Tens 9 - 5 = 4. Hundreds: 0 - 6 runs short, so thousands 7 \to 6, hundreds 10, and 10 - 6 = 4. Thousands 6 - 2 = 4.
22{,}477. Ten thousands 4 \to 3, the three zeros become 9s, ones 3 \to 13. Then 13 - 6 = 7, 9 - 2 = 7, 9 - 5 = 4, 9 - 7 = 2, 3 - 1 = 2.
25{,}385. Ones: hundreds 2 \to 1, tens 0 \to 10 \to 9, ones 10, and 10 - 5 = 5. Tens 9 - 1 = 8. Hundreds: 1 - 8 runs short and the thousands hold 0, so ten thousands 6 \to 5, thousands 0 \to 10 \to 9, hundreds 11, and 11 - 8 = 3. Thousands 9 - 4 = 5. Ten thousands 5 - 3 = 2.
1{,}237 shark teeth. 2{,}005 - 768: thousands 2 \to 1, hundreds and tens become 9, ones 15. Then 15 - 8 = 7, 9 - 6 = 3, 9 - 7 = 2, 1 - 0 = 1.
8{,}003 is 800 tens and 3 ones. After one trade it is 799 tens and 13 ones. Then 13 - 7 = 6, and 799 tens gives the digits 7, 9, 9: 9 - 5 = 4, 9 - 3 = 6, 7 - 4 = 3. The difference is 3{,}646.
Wrong. 3{,}268 + 2{,}847 = 6{,}115, which is 110 too big. Redoing it: thousands 6 \to 5, hundreds and tens become 9, ones 15; then 15 - 7 = 8, 9 - 4 = 5, 9 - 8 = 1, 5 - 2 = 3. The right difference is 3{,}158, and 3{,}158 + 2{,}847 = 6{,}005.
648. 999 - 351: 9 - 1 = 8, 9 - 5 = 4, 9 - 3 = 6.
237. 699 - 462: 9 - 2 = 7, 9 - 6 = 3, 6 - 4 = 2.
1{,}726. 2{,}999 - 1{,}273: 9 - 3 = 6, 9 - 7 = 2, 9 - 2 = 7, 2 - 1 = 1.
3{,}941. 9{,}999 - 6{,}058: 9 - 8 = 1, 9 - 5 = 4, 9 - 0 = 9, 9 - 6 = 3.
4{,}474. 8{,}999 - 4{,}525: 9 - 5 = 4, 9 - 2 = 7, 9 - 5 = 4, 8 - 4 = 4.
3{,}425. Take 4 from both, since 6{,}003 - 4 = 5{,}999: then 5{,}999 - 2{,}574, and 9 - 4 = 5, 9 - 7 = 2, 9 - 5 = 4, 5 - 2 = 3.
635 dollars. 2{,}000 - 1{,}365 = 1{,}999 - 1{,}364 = 635. Check: 635 + 1{,}365 = 2{,}000.
6{,}258 more visitors. 20{,}000 - 13{,}742 = 19{,}999 - 13{,}741 = 6{,}258.
1{,}775 dollars. The gifts total 1{,}850 + 2{,}375 = 4{,}225, and 6{,}000 - 4{,}225 = 5{,}999 - 4{,}224 = 1{,}775.
6{,}514. The missing number is 9{,}000 - 2{,}486, by the fact family, and that is 8{,}999 - 2{,}485 = 6{,}514. Check: 2{,}486 + 6{,}514 = 9{,}000.
In 4{,}999 - 2{,}738 the top digits in the ones, tens and hundreds are 9, the largest digit, so none of those columns can run short, and 4 - 2 is fine too. In 5{,}000 - 2{,}739 the top ones digit is 0, smaller than 9, and the tens and hundreds are 0 too, so the trade has to come from the thousands and pass through every place on the way down.
Yes. Adding 1 to both numbers slides both ends of the gap one step to the right, which does not change its length. Both are 2{,}837: 5{,}000 - 2{,}163 = 4{,}999 - 2{,}162 = 2{,}837, and 5{,}001 - 2{,}164 = 2{,}837.
He gets 512, which is 1 too small. He moved only the top end of the gap back one step, so the gap from 487 to 999 is one step shorter than the gap from 487 to 1{,}000. He must take 1 from 487 as well, so both ends slide together.
3{,}376. 5{,}004 is 500 tens and 4 ones. The ones need a trade (4 - 8 runs short), so one ten is unbundled, leaving 499 tens and 14 ones. Now 499 tens is 4 thousands, 9 hundreds and 9 tens, so the hundreds and tens digits are both 9. Each zero gave up one of the ten it received and kept nine. Then 14 - 8 = 6, 9 - 2 = 7, 9 - 6 = 3, 4 - 1 = 3.