← Carry ten units to the next place · Place-Value Regrouping

Carry ten units to the next place · 12 practice problems

4.NBT.A.24.NBT.A.1

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Generated variants — 12

Freshly produced from the archetype’s parameters — problem, figure, and solution derived together.

Variant 1 easy

Find the digit $\blacksquare$ , where $\blacksquare$ is one of the numbers from $0$ to $9$ .

In $1357$ , there are $1$ thousands, $\blacksquare$ hundreds, $5$ tens, and $7$ ones.

Your answer

Show solution

1 · Understandwhat's really being asked

In the number 1357, written as 1 thousands, 3 hundreds, 5 tens, and 7 ones, find the missing digit in the hundreds place.

Givens

The number is 1357.
It is described as 1 thousands, 3 hundreds, 5 tens, and 7 ones.
The missing digit is a single digit from 0 to 9.

Unknowns

The hundreds digit.

Constraints

Each place value digit is 0 through 9.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #5 Look for a Pattern

The counts here are the standard digits, so this is a direct place-value chart task: line up the digits of 1357 and read off the hundreds digit.

3 · Execute2 carry out the plan

1Line up the digits with their places

#1 Draw a Diagram 4.NBT.A.1

Write 1357 in a place-value chart: the 1 is in the thousands place, the 3 is in the hundreds place, the 5 is in the tens place, and the 7 is in the ones place.

1357 = 1000 + 300 + 50 + 7

Each digit's spot in the number tells you exactly which place it counts, like columns on a chart.

2Read the hundreds digit

#5 Look for a Pattern 4.NBT.A.2

The digit sitting in the hundreds place is 3, so 1357 has 3 hundreds. That means the

\blacksquare

is 3.

\blacksquare = 3

The hundreds count is simply the digit in the hundreds column.

Answer: 3

4 · Reviewdoes it hold up?

Rebuild the number: 1 thousands + 3 hundreds + 5 tens + 7 ones = 1000 + 300 + 50 + 7 = 1357, which matches, so the $\blacksquare$ is 3.

Another way: Subtract the known parts of 1357 to isolate the hundreds contribution of 300, which is 3 hundreds.

Standardsmin grade 4

4.NBT.A.1 Recognize that a digit represents ten times what it represents in place to its right — Understanding the thousands, hundreds, tens, and ones place groupings of the number.
4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Reading the hundreds digit directly from the four-digit number.

💡Takeaway. Each digit lives in its own place column, so the hundreds count is just the digit standing in the hundreds spot!

Variant 2 easy

Find the digit $\blacksquare$ , where $\blacksquare$ is one of the numbers from $0$ to $9$ .

In $3812$ , there are $\blacksquare$ thousands, $8$ hundreds, $1$ tens, and $2$ ones.

Your answer

Show solution

1 · Understandwhat's really being asked

In the number 3812, written as 3 thousands, 8 hundreds, 1 tens, and 2 ones, find the missing digit in the thousands place.

Givens

The number is 3812.
It is described as 3 thousands, 8 hundreds, 1 tens, and 2 ones.
The missing digit is a single digit from 0 to 9.

Unknowns

The thousands digit.

Constraints

Each place value digit is 0 through 9.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #5 Look for a Pattern

The counts here are the standard digits, so this is a direct place-value chart task: line up the digits of 3812 and read off the thousands digit.

3 · Execute2 carry out the plan

1Line up the digits with their places

#1 Draw a Diagram 4.NBT.A.1

Write 3812 in a place-value chart: the 3 is in the thousands place, the 8 is in the hundreds place, the 1 is in the tens place, and the 2 is in the ones place.

3812 = 3000 + 800 + 10 + 2

Each digit's spot in the number tells you exactly which place it counts, like columns on a chart.

2Read the thousands digit

#5 Look for a Pattern 4.NBT.A.2

The digit sitting in the thousands place is 3, so 3812 has 3 thousands. That means the

\blacksquare

is 3.

\blacksquare = 3

The thousands count is simply the digit in the thousands column.

Answer: 3

4 · Reviewdoes it hold up?

Rebuild the number: 3 thousands + 8 hundreds + 1 tens + 2 ones = 3000 + 800 + 10 + 2 = 3812, which matches, so the $\blacksquare$ is 3.

Another way: Subtract the known parts of 3812 to isolate the thousands contribution of 3000, which is 3 thousands.

Standardsmin grade 4

4.NBT.A.1 Recognize that a digit represents ten times what it represents in place to its right — Understanding the thousands, hundreds, tens, and ones place groupings of the number.
4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Reading the thousands digit directly from the four-digit number.

💡Takeaway. Each digit lives in its own place column, so the thousands count is just the digit standing in the thousands spot!

Variant 3 easy

Find the digit $\blacksquare$ , where $\blacksquare$ is one of the numbers from $0$ to $9$ .

In $5093$ , there are $5$ thousands, $0$ hundreds, $9$ tens, and $\blacksquare$ ones.

Your answer

Show solution

1 · Understandwhat's really being asked

In the number 5093, written as 5 thousands, 0 hundreds, 9 tens, and 3 ones, find the missing digit in the ones place.

Givens

The number is 5093.
It is described as 5 thousands, 0 hundreds, 9 tens, and 3 ones.
The missing digit is a single digit from 0 to 9.

Unknowns

The ones digit.

Constraints

Each place value digit is 0 through 9.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #5 Look for a Pattern

The counts here are the standard digits, so this is a direct place-value chart task: line up the digits of 5093 and read off the ones digit.

3 · Execute2 carry out the plan

1Line up the digits with their places

#1 Draw a Diagram 4.NBT.A.1

Write 5093 in a place-value chart: the 5 is in the thousands place, the 0 is in the hundreds place, the 9 is in the tens place, and the 3 is in the ones place.

5093 = 5000 + 0 + 90 + 3

Each digit's spot in the number tells you exactly which place it counts, like columns on a chart.

2Read the ones digit

#5 Look for a Pattern 4.NBT.A.2

The digit sitting in the ones place is 3, so 5093 has 3 ones. That means the

\blacksquare

is 3.

\blacksquare = 3

The ones count is simply the digit in the ones column.

Answer: 3

4 · Reviewdoes it hold up?

Rebuild the number: 5 thousands + 0 hundreds + 9 tens + 3 ones = 5000 + 0 + 90 + 3 = 5093, which matches, so the $\blacksquare$ is 3.

Another way: Subtract the known parts of 5093 to isolate the ones contribution of 3, which is 3 ones.

Standardsmin grade 4

4.NBT.A.1 Recognize that a digit represents ten times what it represents in place to its right — Understanding the thousands, hundreds, tens, and ones place groupings of the number.
4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Reading the ones digit directly from the four-digit number.

💡Takeaway. Each digit lives in its own place column, so the ones count is just the digit standing in the ones spot!

Variant 4 easy

Find the digit $\blacksquare$ , where $\blacksquare$ is one of the numbers from $0$ to $9$ .

In $6792$ , there are $6$ thousands, $7$ hundreds, $\blacksquare$ tens, and $2$ ones.

Your answer

Show solution

1 · Understandwhat's really being asked

In the number 6792, written as 6 thousands, 7 hundreds, 9 tens, and 2 ones, find the missing digit in the tens place.

Givens

The number is 6792.
It is described as 6 thousands, 7 hundreds, 9 tens, and 2 ones.
The missing digit is a single digit from 0 to 9.

Unknowns

The tens digit.

Constraints

Each place value digit is 0 through 9.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #5 Look for a Pattern

The counts here are the standard digits, so this is a direct place-value chart task: line up the digits of 6792 and read off the tens digit.

3 · Execute2 carry out the plan

1Line up the digits with their places

#1 Draw a Diagram 4.NBT.A.1

Write 6792 in a place-value chart: the 6 is in the thousands place, the 7 is in the hundreds place, the 9 is in the tens place, and the 2 is in the ones place.

6792 = 6000 + 700 + 90 + 2

Each digit's spot in the number tells you exactly which place it counts, like columns on a chart.

2Read the tens digit

#5 Look for a Pattern 4.NBT.A.2

The digit sitting in the tens place is 9, so 6792 has 9 tens. That means the

\blacksquare

is 9.

\blacksquare = 9

The tens count is simply the digit in the tens column.

Answer: 9

4 · Reviewdoes it hold up?

Rebuild the number: 6 thousands + 7 hundreds + 9 tens + 2 ones = 6000 + 700 + 90 + 2 = 6792, which matches, so the $\blacksquare$ is 9.

Another way: Subtract the known parts of 6792 to isolate the tens contribution of 90, which is 9 tens.

Standardsmin grade 4

4.NBT.A.1 Recognize that a digit represents ten times what it represents in place to its right — Understanding the thousands, hundreds, tens, and ones place groupings of the number.
4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Reading the tens digit directly from the four-digit number.

💡Takeaway. Each digit lives in its own place column, so the tens count is just the digit standing in the tens spot!

Variant 5 medium

Find the number $\blacksquare$ that makes the statement true. $\blacksquare$ may be larger than $9$ .

$2756$ is the same as $0$ thousands, $\blacksquare$ hundreds, $5$ tens, and $6$ ones.

Your answer

Show solution

1 · Understandwhat's really being asked

2756 is re-expressed as 0 thousands, 5 tens, 6 ones, and an unknown number of hundreds. Because the other places do not use their standard digits, find how many hundreds are needed to rebuild 2756.

Givens

The number is 2756.
It is described as 0 thousands, 5 tens, 6 ones, and $\blacksquare$ hundreds.
The stated counts are not all the standard digits, so the $\blacksquare$ may exceed 9.

Unknowns

How many hundreds make the parts add up to 2756.

Constraints

All parts together must equal 2756.
Ten of any place regroup into one of the next higher place.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #11 Work Backwards#1 Draw a Diagram

The counts are non-standard, so I cannot just read a digit. I add the known parts, subtract from 2756 to see what the hundreds must supply, then convert that amount into hundreds using the rule that 10 of one place equal 1 of the next.

3 · Execute3 carry out the plan

1Add up the parts you already know

#11 Work Backwards 4.NBT.A.2

The stated places give 0 thousands, 5 tens, 6 ones. In value that is 0 + 50 + 6 = 56.

0 + 50 + 6 = 56

Pin down everything that is fixed first; whatever is left over has to come from the blank.

2Find what the hundreds must supply

#11 Work Backwards 4.NBT.A.2

Subtract the known part from the whole number: 2756 - 56 = 2700. So the hundreds place must account for 2700.

2756 - 56 = 2700

Working backwards from the total isolates exactly the blank's job.

3Regroup 2700 into hundreds

#8 Analyze the Units 4.NBT.A.1

Each hundred is worth 100, so 2700 divided by 100 is 27 hundreds. That is more than the 7 you would see in the hundreds column of 2756: the thousands are stated below their usual amounts, so those higher units are traded down — ten of each becoming the next place — adding 20 extra hundreds. So

\blacksquare = 27

2700 \div 100 = 27

Because ten of a smaller place make one of the next, a place can legally hold a count bigger than 9.

Answer: 27

4 · Reviewdoes it hold up?

Rebuild the number: 56 + 2700 = 2756, which matches, so $\blacksquare = 27$ .

Another way: Start from the standard 7 hundreds in 2756 and add the 20 hundreds regrouped from the missing higher places, again giving 27.

Standardsmin grade 4

4.NBT.A.1 Recognize that a digit represents ten times what it represents in place to its right — Regrouping between places — trading 1 of a higher place for 10 of the next — so a place can hold more than 9 units.
4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Composing the four-digit number from non-standard place-value parts.

💡Takeaway. Ten of one place make one of the next, so you can trade higher units down — that is why the

\blacksquare

can be bigger than a single digit!

Variant 6 medium

Find the number $\blacksquare$ that makes the statement true. $\blacksquare$ may be larger than $9$ .

$3812$ is the same as $2$ thousands, $\blacksquare$ hundreds, $1$ tens, and $2$ ones.

Your answer

Show solution

1 · Understandwhat's really being asked

3812 is re-expressed as 2 thousands, 1 tens, 2 ones, and an unknown number of hundreds. Because the other places do not use their standard digits, find how many hundreds are needed to rebuild 3812.

Givens

The number is 3812.
It is described as 2 thousands, 1 tens, 2 ones, and $\blacksquare$ hundreds.
The stated counts are not all the standard digits, so the $\blacksquare$ may exceed 9.

Unknowns

How many hundreds make the parts add up to 3812.

Constraints

All parts together must equal 3812.
Ten of any place regroup into one of the next higher place.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #11 Work Backwards#1 Draw a Diagram

The counts are non-standard, so I cannot just read a digit. I add the known parts, subtract from 3812 to see what the hundreds must supply, then convert that amount into hundreds using the rule that 10 of one place equal 1 of the next.

3 · Execute3 carry out the plan

1Add up the parts you already know

#11 Work Backwards 4.NBT.A.2

The stated places give 2 thousands, 1 tens, 2 ones. In value that is 2000 + 10 + 2 = 2012.

2000 + 10 + 2 = 2012

Pin down everything that is fixed first; whatever is left over has to come from the blank.

2Find what the hundreds must supply

#11 Work Backwards 4.NBT.A.2

Subtract the known part from the whole number: 3812 - 2012 = 1800. So the hundreds place must account for 1800.

3812 - 2012 = 1800

Working backwards from the total isolates exactly the blank's job.

3Regroup 1800 into hundreds

#8 Analyze the Units 4.NBT.A.1

Each hundred is worth 100, so 1800 divided by 100 is 18 hundreds. That is more than the 8 you would see in the hundreds column of 3812: the thousands are stated below their usual amounts, so those higher units are traded down — ten of each becoming the next place — adding 10 extra hundreds. So

\blacksquare = 18

1800 \div 100 = 18

Because ten of a smaller place make one of the next, a place can legally hold a count bigger than 9.

Answer: 18

4 · Reviewdoes it hold up?

Rebuild the number: 2012 + 1800 = 3812, which matches, so $\blacksquare = 18$ .

Another way: Start from the standard 8 hundreds in 3812 and add the 10 hundreds regrouped from the missing higher places, again giving 18.

Standardsmin grade 4

4.NBT.A.1 Recognize that a digit represents ten times what it represents in place to its right — Regrouping between places — trading 1 of a higher place for 10 of the next — so a place can hold more than 9 units.
4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Composing the four-digit number from non-standard place-value parts.

💡Takeaway. Ten of one place make one of the next, so you can trade higher units down — that is why the

\blacksquare

can be bigger than a single digit!

Variant 7 medium

Find the number $\blacksquare$ that makes the statement true. $\blacksquare$ may be larger than $9$ .

$4205$ is the same as $1$ thousands, $\blacksquare$ hundreds, $0$ tens, and $5$ ones.

Your answer

Show solution

1 · Understandwhat's really being asked

4205 is re-expressed as 1 thousands, 0 tens, 5 ones, and an unknown number of hundreds. Because the other places do not use their standard digits, find how many hundreds are needed to rebuild 4205.

Givens

The number is 4205.
It is described as 1 thousands, 0 tens, 5 ones, and $\blacksquare$ hundreds.
The stated counts are not all the standard digits, so the $\blacksquare$ may exceed 9.

Unknowns

How many hundreds make the parts add up to 4205.

Constraints

All parts together must equal 4205.
Ten of any place regroup into one of the next higher place.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #11 Work Backwards#1 Draw a Diagram

The counts are non-standard, so I cannot just read a digit. I add the known parts, subtract from 4205 to see what the hundreds must supply, then convert that amount into hundreds using the rule that 10 of one place equal 1 of the next.

3 · Execute3 carry out the plan

1Add up the parts you already know

#11 Work Backwards 4.NBT.A.2

The stated places give 1 thousands, 0 tens, 5 ones. In value that is 1000 + 0 + 5 = 1005.

1000 + 0 + 5 = 1005

Pin down everything that is fixed first; whatever is left over has to come from the blank.

2Find what the hundreds must supply

#11 Work Backwards 4.NBT.A.2

Subtract the known part from the whole number: 4205 - 1005 = 3200. So the hundreds place must account for 3200.

4205 - 1005 = 3200

Working backwards from the total isolates exactly the blank's job.

3Regroup 3200 into hundreds

#8 Analyze the Units 4.NBT.A.1

Each hundred is worth 100, so 3200 divided by 100 is 32 hundreds. That is more than the 2 you would see in the hundreds column of 4205: the thousands are stated below their usual amounts, so those higher units are traded down — ten of each becoming the next place — adding 30 extra hundreds. So

\blacksquare = 32

3200 \div 100 = 32

Because ten of a smaller place make one of the next, a place can legally hold a count bigger than 9.

Answer: 32

4 · Reviewdoes it hold up?

Rebuild the number: 1005 + 3200 = 4205, which matches, so $\blacksquare = 32$ .

Another way: Start from the standard 2 hundreds in 4205 and add the 30 hundreds regrouped from the missing higher places, again giving 32.

Standardsmin grade 4

4.NBT.A.1 Recognize that a digit represents ten times what it represents in place to its right — Regrouping between places — trading 1 of a higher place for 10 of the next — so a place can hold more than 9 units.
4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Composing the four-digit number from non-standard place-value parts.

💡Takeaway. Ten of one place make one of the next, so you can trade higher units down — that is why the

\blacksquare

can be bigger than a single digit!

Variant 8 medium

Find the number $\blacksquare$ that makes the statement true. $\blacksquare$ may be larger than $9$ .

$6479$ is the same as $6$ thousands, $0$ hundreds, $\blacksquare$ tens, and $9$ ones.

Your answer

Show solution

1 · Understandwhat's really being asked

6479 is re-expressed as 6 thousands, 0 hundreds, 9 ones, and an unknown number of tens. Because the other places do not use their standard digits, find how many tens are needed to rebuild 6479.

Givens

The number is 6479.
It is described as 6 thousands, 0 hundreds, 9 ones, and $\blacksquare$ tens.
The stated counts are not all the standard digits, so the $\blacksquare$ may exceed 9.

Unknowns

How many tens make the parts add up to 6479.

Constraints

All parts together must equal 6479.
Ten of any place regroup into one of the next higher place.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #11 Work Backwards#1 Draw a Diagram

The counts are non-standard, so I cannot just read a digit. I add the known parts, subtract from 6479 to see what the tens must supply, then convert that amount into tens using the rule that 10 of one place equal 1 of the next.

3 · Execute3 carry out the plan

1Add up the parts you already know

#11 Work Backwards 4.NBT.A.2

The stated places give 6 thousands, 0 hundreds, 9 ones. In value that is 6000 + 0 + 9 = 6009.

6000 + 0 + 9 = 6009

Pin down everything that is fixed first; whatever is left over has to come from the blank.

2Find what the tens must supply

#11 Work Backwards 4.NBT.A.2

Subtract the known part from the whole number: 6479 - 6009 = 470. So the tens place must account for 470.

6479 - 6009 = 470

Working backwards from the total isolates exactly the blank's job.

3Regroup 470 into tens

#8 Analyze the Units 4.NBT.A.1

Each ten is worth 10, so 470 divided by 10 is 47 tens. That is more than the 7 you would see in the tens column of 6479: the hundreds are stated below their usual amounts, so those higher units are traded down — ten of each becoming the next place — adding 40 extra tens. So

\blacksquare = 47

470 \div 10 = 47

Because ten of a smaller place make one of the next, a place can legally hold a count bigger than 9.

Answer: 47

4 · Reviewdoes it hold up?

Rebuild the number: 6009 + 470 = 6479, which matches, so $\blacksquare = 47$ .

Another way: Start from the standard 7 tens in 6479 and add the 40 tens regrouped from the missing higher places, again giving 47.

Standardsmin grade 4

4.NBT.A.1 Recognize that a digit represents ten times what it represents in place to its right — Regrouping between places — trading 1 of a higher place for 10 of the next — so a place can hold more than 9 units.
4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Composing the four-digit number from non-standard place-value parts.

💡Takeaway. Ten of one place make one of the next, so you can trade higher units down — that is why the

\blacksquare

can be bigger than a single digit!

Variant 9 hard

Find the number $\blacksquare$ that makes the statement true. $\blacksquare$ may be larger than $9$ .

$7545$ is the same as $7$ thousands, $4$ hundreds, $\blacksquare$ tens, and $5$ ones.

Your answer

Show solution

1 · Understandwhat's really being asked

7545 is re-expressed as 7 thousands, 4 hundreds, 5 ones, and an unknown number of tens. Because the other places do not use their standard digits, find how many tens are needed to rebuild 7545.

Givens

The number is 7545.
It is described as 7 thousands, 4 hundreds, 5 ones, and $\blacksquare$ tens.
The stated counts are not all the standard digits, so the $\blacksquare$ may exceed 9.

Unknowns

How many tens make the parts add up to 7545.

Constraints

All parts together must equal 7545.
Ten of any place regroup into one of the next higher place.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #11 Work Backwards#1 Draw a Diagram

The counts are non-standard, so I cannot just read a digit. I add the known parts, subtract from 7545 to see what the tens must supply, then convert that amount into tens using the rule that 10 of one place equal 1 of the next.

3 · Execute3 carry out the plan

1Add up the parts you already know

#11 Work Backwards 4.NBT.A.2

The stated places give 7 thousands, 4 hundreds, 5 ones. In value that is 7000 + 400 + 5 = 7405.

7000 + 400 + 5 = 7405

Pin down everything that is fixed first; whatever is left over has to come from the blank.

2Find what the tens must supply

#11 Work Backwards 4.NBT.A.2

Subtract the known part from the whole number: 7545 - 7405 = 140. So the tens place must account for 140.

7545 - 7405 = 140

Working backwards from the total isolates exactly the blank's job.

3Regroup 140 into tens

#8 Analyze the Units 4.NBT.A.1

Each ten is worth 10, so 140 divided by 10 is 14 tens. That is more than the 4 you would see in the tens column of 7545: the hundreds are stated below their usual amounts, so those higher units are traded down — ten of each becoming the next place — adding 10 extra tens. So

\blacksquare = 14

140 \div 10 = 14

Because ten of a smaller place make one of the next, a place can legally hold a count bigger than 9.

Answer: 14

4 · Reviewdoes it hold up?

Rebuild the number: 7405 + 140 = 7545, which matches, so $\blacksquare = 14$ .

Another way: Start from the standard 4 tens in 7545 and add the 10 tens regrouped from the missing higher places, again giving 14.

Standardsmin grade 4

4.NBT.A.1 Recognize that a digit represents ten times what it represents in place to its right — Regrouping between places — trading 1 of a higher place for 10 of the next — so a place can hold more than 9 units.
4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Composing the four-digit number from non-standard place-value parts.

💡Takeaway. Ten of one place make one of the next, so you can trade higher units down — that is why the

\blacksquare

can be bigger than a single digit!

Variant 10 hard

Find the number $\blacksquare$ that makes the statement true. $\blacksquare$ may be larger than $9$ .

$7545$ is the same as $6$ thousands, $\blacksquare$ hundreds, $4$ tens, and $5$ ones.

Your answer

Show solution

1 · Understandwhat's really being asked

7545 is re-expressed as 6 thousands, 4 tens, 5 ones, and an unknown number of hundreds. Because the other places do not use their standard digits, find how many hundreds are needed to rebuild 7545.

Givens

The number is 7545.
It is described as 6 thousands, 4 tens, 5 ones, and $\blacksquare$ hundreds.
The stated counts are not all the standard digits, so the $\blacksquare$ may exceed 9.

Unknowns

How many hundreds make the parts add up to 7545.

Constraints

All parts together must equal 7545.
Ten of any place regroup into one of the next higher place.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #11 Work Backwards#1 Draw a Diagram

The counts are non-standard, so I cannot just read a digit. I add the known parts, subtract from 7545 to see what the hundreds must supply, then convert that amount into hundreds using the rule that 10 of one place equal 1 of the next.

3 · Execute3 carry out the plan

1Add up the parts you already know

#11 Work Backwards 4.NBT.A.2

The stated places give 6 thousands, 4 tens, 5 ones. In value that is 6000 + 40 + 5 = 6045.

6000 + 40 + 5 = 6045

Pin down everything that is fixed first; whatever is left over has to come from the blank.

2Find what the hundreds must supply

#11 Work Backwards 4.NBT.A.2

Subtract the known part from the whole number: 7545 - 6045 = 1500. So the hundreds place must account for 1500.

7545 - 6045 = 1500

Working backwards from the total isolates exactly the blank's job.

3Regroup 1500 into hundreds

#8 Analyze the Units 4.NBT.A.1

Each hundred is worth 100, so 1500 divided by 100 is 15 hundreds. That is more than the 5 you would see in the hundreds column of 7545: the thousands are stated below their usual amounts, so those higher units are traded down — ten of each becoming the next place — adding 10 extra hundreds. So

\blacksquare = 15

1500 \div 100 = 15

Because ten of a smaller place make one of the next, a place can legally hold a count bigger than 9.

Answer: 15

4 · Reviewdoes it hold up?

Rebuild the number: 6045 + 1500 = 7545, which matches, so $\blacksquare = 15$ .

Another way: Start from the standard 5 hundreds in 7545 and add the 10 hundreds regrouped from the missing higher places, again giving 15.

Standardsmin grade 4

4.NBT.A.1 Recognize that a digit represents ten times what it represents in place to its right — Regrouping between places — trading 1 of a higher place for 10 of the next — so a place can hold more than 9 units.
4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Composing the four-digit number from non-standard place-value parts.

💡Takeaway. Ten of one place make one of the next, so you can trade higher units down — that is why the

\blacksquare

can be bigger than a single digit!

Variant 11 hard

Find the number $\blacksquare$ that makes the statement true. $\blacksquare$ may be larger than $9$ .

$7545$ is the same as $5$ thousands, $\blacksquare$ hundreds, $4$ tens, and $5$ ones.

Your answer

Show solution

1 · Understandwhat's really being asked

7545 is re-expressed as 5 thousands, 4 tens, 5 ones, and an unknown number of hundreds. Because the other places do not use their standard digits, find how many hundreds are needed to rebuild 7545.

Givens

The number is 7545.
It is described as 5 thousands, 4 tens, 5 ones, and $\blacksquare$ hundreds.
The stated counts are not all the standard digits, so the $\blacksquare$ may exceed 9.

Unknowns

How many hundreds make the parts add up to 7545.

Constraints

All parts together must equal 7545.
Ten of any place regroup into one of the next higher place.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #11 Work Backwards#1 Draw a Diagram

3 · Execute3 carry out the plan

1Add up the parts you already know

#11 Work Backwards 4.NBT.A.2

The stated places give 5 thousands, 4 tens, 5 ones. In value that is 5000 + 40 + 5 = 5045.

5000 + 40 + 5 = 5045

Pin down everything that is fixed first; whatever is left over has to come from the blank.

2Find what the hundreds must supply

#11 Work Backwards 4.NBT.A.2

Subtract the known part from the whole number: 7545 - 5045 = 2500. So the hundreds place must account for 2500.

7545 - 5045 = 2500

Working backwards from the total isolates exactly the blank's job.

3Regroup 2500 into hundreds

#8 Analyze the Units 4.NBT.A.1

Each hundred is worth 100, so 2500 divided by 100 is 25 hundreds. That is more than the 5 you would see in the hundreds column of 7545: the thousands are stated below their usual amounts, so those higher units are traded down — ten of each becoming the next place — adding 20 extra hundreds. So

\blacksquare = 25

2500 \div 100 = 25

Because ten of a smaller place make one of the next, a place can legally hold a count bigger than 9.

Answer: 25

4 · Reviewdoes it hold up?

Rebuild the number: 5045 + 2500 = 7545, which matches, so $\blacksquare = 25$ .

Another way: Start from the standard 5 hundreds in 7545 and add the 20 hundreds regrouped from the missing higher places, again giving 25.

Standardsmin grade 4

4.NBT.A.1 Recognize that a digit represents ten times what it represents in place to its right — Regrouping between places — trading 1 of a higher place for 10 of the next — so a place can hold more than 9 units.
4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Composing the four-digit number from non-standard place-value parts.

💡Takeaway. Ten of one place make one of the next, so you can trade higher units down — that is why the

\blacksquare

can be bigger than a single digit!

Variant 12 hard

Find the number $\blacksquare$ that makes the statement true. $\blacksquare$ may be larger than $9$ .

$8124$ is the same as $8$ thousands, $0$ hundreds, $\blacksquare$ tens, and $4$ ones.

Your answer

Show solution

1 · Understandwhat's really being asked

8124 is re-expressed as 8 thousands, 0 hundreds, 4 ones, and an unknown number of tens. Because the other places do not use their standard digits, find how many tens are needed to rebuild 8124.

Givens

The number is 8124.
It is described as 8 thousands, 0 hundreds, 4 ones, and $\blacksquare$ tens.
The stated counts are not all the standard digits, so the $\blacksquare$ may exceed 9.

Unknowns

How many tens make the parts add up to 8124.

Constraints

All parts together must equal 8124.
Ten of any place regroup into one of the next higher place.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #11 Work Backwards#1 Draw a Diagram

The counts are non-standard, so I cannot just read a digit. I add the known parts, subtract from 8124 to see what the tens must supply, then convert that amount into tens using the rule that 10 of one place equal 1 of the next.

3 · Execute3 carry out the plan

1Add up the parts you already know

#11 Work Backwards 4.NBT.A.2

The stated places give 8 thousands, 0 hundreds, 4 ones. In value that is 8000 + 0 + 4 = 8004.

8000 + 0 + 4 = 8004

Pin down everything that is fixed first; whatever is left over has to come from the blank.

2Find what the tens must supply

#11 Work Backwards 4.NBT.A.2

Subtract the known part from the whole number: 8124 - 8004 = 120. So the tens place must account for 120.

8124 - 8004 = 120

Working backwards from the total isolates exactly the blank's job.

3Regroup 120 into tens

#8 Analyze the Units 4.NBT.A.1

Each ten is worth 10, so 120 divided by 10 is 12 tens. That is more than the 2 you would see in the tens column of 8124: the hundreds are stated below their usual amounts, so those higher units are traded down — ten of each becoming the next place — adding 10 extra tens. So

\blacksquare = 12

120 \div 10 = 12

Because ten of a smaller place make one of the next, a place can legally hold a count bigger than 9.

Answer: 12

4 · Reviewdoes it hold up?

Rebuild the number: 8004 + 120 = 8124, which matches, so $\blacksquare = 12$ .

Another way: Start from the standard 2 tens in 8124 and add the 10 tens regrouped from the missing higher places, again giving 12.

Standardsmin grade 4

4.NBT.A.1 Recognize that a digit represents ten times what it represents in place to its right — Regrouping between places — trading 1 of a higher place for 10 of the next — so a place can hold more than 9 units.
4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols — Composing the four-digit number from non-standard place-value parts.

💡Takeaway. Ten of one place make one of the next, so you can trade higher units down — that is why the

\blacksquare

can be bigger than a single digit!

Where next?

← Back to the problem: Carry ten units to the next place