Place Value Worksheets with Addition and Subtraction - 4th Grade ... - Free Printable

Let's solve each of the problems on the worksheet step by step. The task is to write a number that matches each description, based on adjacent place value relationships.

We are given six numbers at the top:
- 255,392
- 34,482
- 993,271
- 1,368,339
- 136,077
- 1,193,167

But these appear to be examples or reference numbers — not necessarily required for solving the problems. The actual task is to create a number that satisfies each condition.

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Understanding the Concept:

"Adjacent place value" means comparing digits in neighboring places (e.g., hundreds and tens, thousands and hundreds, etc.).

A digit in a higher place value is ten times greater than the same digit in the next lower place.

For example:
- The digit in the hundreds place is 10 times the digit in the tens place.
- So if the tens digit is 3, then the hundreds digit must be 30? No — wait! That’s not possible because digits are from 0–9.

Ah! Here’s the key: We’re looking for a number where the digit in one place is ten times the digit in the adjacent lower place.

But since digits are only 0–9, the only way a digit can be ten times another digit is:

> If the lower digit is 0, then the higher digit could be anything (since 0 × 10 = 0), but that doesn't help.

Wait — actually, no digit from 1 to 9 multiplied by 10 gives a single digit.

So how can a digit be ten times greater?

Let’s re-analyze.

The phrase "the digit in the hundreds place is ten times greater than the digit in the tens place" means:

> Let the digit in the tens place be $ x $. Then the digit in the hundreds place is $ 10x $.

But $ x $ must be a digit (0–9), and $ 10x $ must also be a digit (0–9). That only works if $ x = 0 $, then $ 10x = 0 $.

So the only possibility is both digits are 0.

But that seems trivial. Maybe we're misunderstanding.

Wait — perhaps it's not that the digit itself is ten times greater, but that the value of the digit in the higher place is ten times the value of the digit in the lower place.

That makes more sense.

Let’s clarify:

- The value of a digit depends on its place.
- For example, in 255,392:
- The digit 3 is in the hundreds place → value = 300
- The digit 9 is in the tens place → value = 90
- But 300 ≠ 10 × 90 → no.

But suppose the digit in the hundreds place is 3, and the digit in the tens place is 3. Then:
- Hundreds place value = 300
- Tens place value = 30
→ 300 = 10 × 30 → yes!

So the value of the digit in the hundreds place is 10 times the value of the digit in the tens place — if the digits are the same.

But the problem says: “the digit in the hundreds place is ten times greater than the digit in the tens place.”

This wording is ambiguous.

But in math, when we say "the digit in the hundreds place is ten times greater than the digit in the tens place", it usually means:

> The value contributed by the hundreds digit is ten times the value contributed by the tens digit.

But that would mean:

> Value in hundreds = 10 × value in tens

But value in hundreds = (digit) × 100
Value in tens = (digit) × 10

So:
> (H_digit × 100) = 10 × (T_digit × 10)
> H_digit × 100 = 10 × T_digit × 10 = 100 × T_digit
> So H_digit = T_digit

So again, the digits must be equal.

Wait — that means the digit values are the same.

But the place value is already 10 times.

So this suggests that any number where the digits in adjacent places are equal will satisfy that the higher place value is ten times the lower one.

But the question says: “a number in which the digit in the hundreds place is ten times greater than the digit in the tens place.”

This is likely a misphrasing.

In standard interpretation, “the digit” refers to the numerical value of the digit, not its place value.

So if the digit in the hundreds place is d_h, and in the tens place is d_t, then:

> d_h = 10 × d_t

But d_h and d_t are digits (0–9), so the only solution is:

- d_t = 0 → d_h = 0

So both digits must be zero.

Thus, the only number satisfying this would have 0 in both hundreds and tens places.

But that feels too restrictive.

Alternatively, maybe the problem means:

> The value of the digit in the hundreds place is ten times the value of the digit in the tens place.

Then:

> (hundreds digit) × 100 = 10 × (tens digit) × 10
> → h × 100 = 10 × t × 10 = 100 × t
> → h = t

So again, the digits must be equal.

But that contradicts the wording.

Wait — let’s look at an example.

Suppose we have a number like 300:
- Hundreds digit: 3 → value = 300
- Tens digit: 0 → value = 0
→ 300 ≠ 10 × 0

Not helpful.

What about 330?
- Hundreds digit: 3 → value = 300
- Tens digit: 3 → value = 30
→ 300 = 10 × 30 → YES!

So the value in the hundreds place is 10 times the value in the tens place.

But the digits are both 3.

So the digit in the hundreds place is not ten times the digit in the tens place — they are equal.

So the statement "the digit in the hundreds place is ten times greater than the digit in the tens place" is false — the values are ten times, but the digits are equal.

Therefore, the intended meaning is likely:

> The value of the digit in the higher place is ten times the value of the digit in the lower place.

But since the place value is already 10 times, this happens whenever the digits are equal.

But the problem says “the digit in the hundreds place is ten times greater than the digit in the tens place”.

This suggests that the digit itself is 10 times larger.

But digits are 0–9, so only possible if tens digit is 0 and hundreds digit is 0.

So only numbers with 0 in both places work.

But that seems pointless.

Alternative interpretation: Perhaps it's a typo, and it should be:

> “the value of the digit in the hundreds place is ten times the value of the digit in the tens place”

Which is true whenever the digits are equal.

But let’s check the examples at the top.

Look at 255,392:
- Hundreds digit: 3
- Tens digit: 9
→ 3 ≠ 10×9 → no

34,482:
- Hundreds: 4
- Tens: 8 → 4 ≠ 10×8

993,271:
- Hundreds: 2
- Tens: 7 → no

1,368,339:
- Hundreds: 3
- Tens: 3 → 3 = 10×3? No.

Wait — unless it's saying that the value is ten times.

In 1,368,339:
- Hundreds place: 3 → value = 300
- Tens place: 3 → value = 30
→ 300 = 10 × 30 → YES!

So the value in the hundreds place is 10 times the value in the tens place.

And the digits are equal.

So likely, the problem means:

> A number in which the value of the digit in the higher place is ten times the value of the digit in the lower place.

But the wording says “the digit... is ten times greater than the digit...”, which is incorrect.

But given the context, we can assume it means:

> The digit in the higher place is such that its value is ten times the value of the digit in the lower place.

Which implies the digits are equal.

But let’s try a different approach.

Maybe it means:

> The digit in the hundreds place is ten times the digit in the tens place.

But as digits are 0–9, the only way is:

- tens digit = 0 → hundreds digit = 0

So any number with 0 in both hundreds and tens places.

Example: 100, 200, 300, ..., 900, 1000, etc.

But let’s see if that fits.

But then for #14: digit in thousands place is ten times digit in hundreds place.

So thousands digit = 10 × hundreds digit

Only possible if hundreds digit = 0, thousands digit = 0.

So all such digits must be 0.

That leads to trivial answers.

But that seems unlikely.

Another idea: Perhaps the problem is asking for a number where the digit in the higher place is one-tenth of the lower place? No.

Wait — maybe it's a trick.

Let’s consider that the place value is always 10 times, so the value of a digit in a higher place is 10 times the value of the same digit in the lower place.

So if the digit is the same, the value is 10 times.

So the only way the value is 10 times is if the digits are equal.

But the problem says “the digit in the hundreds place is ten times greater than the digit in the tens place”.

This is a common confusion.

“Ten times greater” means: original + 10×original = 11×original.

But in common usage, people often mean “ten times as great”.

But even then, a digit cannot be ten times another digit unless it’s 0.

So let’s go back to the examples.

Look at 1,368,339

- Thousands place: 8
- Hundreds place: 3
→ 8 vs 3 → not 10×3=30

No.

136,077
- Thousands: 6
- Hundreds: 0 → 6 vs 0 → 6 ≠ 10×0

No.

1,193,167
- Ten thousands: 9
- Thousands: 3 → 9 vs 3 → 9 ≠ 10×3

No.

But let’s try to find a number where the value in the hundreds place is 10 times the value in the tens place.

As before, this happens when the digits are equal.

For example, in 330, hundreds digit = 3, tens digit = 3 → value 300 = 10×30 → yes.

Similarly, 550: 500 = 10×50 → yes.

So for #13: “A number in which the digit in the hundreds place is ten times greater than the digit in the tens place”

If we interpret “digit” as the value, then it’s impossible unless both are 0.

But if we interpret it as “the value of the digit in the hundreds place is ten times the value of the digit in the tens place”, then it’s possible when digits are equal.

But the word “digit” refers to the symbol, not the value.

So the only logical conclusion is that the problem has a wording issue, and it means:

> The value of the digit in the higher place is ten times the value of the digit in the lower place.

Which is true when the digits are equal.

So for each problem, we need a number where the digits in adjacent places are equal.

But let’s read carefully.

#13: “A number in which the digit in the hundreds place is ten times greater than the digit in the tens place.”

This is only possible if the digit in the hundreds place is 0 and the digit in the tens place is 0.

Because 0 = 10 × 0.

Any other digit: 10×1 = 10, not a digit.

So only possible if both are 0.

So answer: any number with 0 in hundreds and tens places, e.g., 100, 200, 300, ..., 900, 1000, etc.

But let’s see #14: “the digit in the thousands place is ten times greater than the digit in the hundreds place”

Same logic: only possible if both are 0.

So the only number that satisfies all conditions is one with many zeros.

But that seems too restrictive.

Perhaps the problem means:

> The digit in the higher place is equal to the digit in the lower place.

But it says “ten times greater”.

Another possibility: maybe “ten times greater” is meant to emphasize that the place value is ten times, and the digit is the same.

But the phrasing is poor.

Given the context of the worksheet, and the fact that the numbers provided include ones like 1,368,339 where the hundreds and tens digits are both 3, suggesting that the values are related, I think the intended meaning is:

> A number in which the value of the digit in the higher place is ten times the value of the digit in the lower place.

Which occurs when the digits are equal.

So for each problem, we need to create a number where the two adjacent digits are the same.

Let’s proceed with that interpretation.

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Solving Each Problem

#### 13) A number in which the digit in the hundreds place is ten times greater than the digit in the tens place.

We want the value in hundreds place to be 10 times the value in tens place.

Since place value is already 10 times, this happens when the digits are equal.

So pick a digit, say 5, and make both hundreds and tens digits 5.

Example: 1550 → hundreds = 5, tens = 5 → value: 500 and 50 → 500 = 10×50 → YES.

But the problem says “the digit in the hundreds place is ten times greater than the digit in the tens place”.

Digit 5 is not ten times digit 5.

So contradiction.

Unless the problem means: the digit in the hundreds place is the same as the digit in the tens place, and the place value is ten times.

But it says “ten times greater”.

Perhaps it's a mistake, and it should be:

> “the value of the digit in the hundreds place is ten times the value of the digit in the tens place”

Then answer: any number where hundreds and tens digits are equal.

Example: 1550, 2660, 3770, etc.

Or even 550.

Let’s assume that.

So for #13: 550 (hundreds digit 5, tens digit 5)

Even though the digit is not ten times greater, the value is.

But the problem says “digit”.

Perhaps the problem is asking for a number where the digit in the higher place is 10 times the digit in the lower place, but that's impossible unless both are 0.

So the only possibility is both digits are 0.

So answer: 100 (hundreds digit 1, tens digit 0) → 1 ≠ 10×0

No.

100: hundreds digit = 1, tens digit = 0 → 1 = 10×0? No.

Only if hundreds digit = 0, tens digit = 0.

So number like 1000 → hundreds digit = 0, tens digit = 0 → 0 = 10×0 → YES.

So 1000 works.

Similarly, 10000 works.

So for #13: 1000

For #14: “digit in thousands place is ten times greater than digit in hundreds place”

So thousands digit = 10 × hundreds digit

Only possible if hundreds digit = 0, thousands digit = 0.

So number like 10000 → thousands digit = 0, hundreds digit = 0 → 0 = 10×0 → YES.

So 10000 works.

For #15: “digit in millions place is ten times greater than digit in hundred thousands place”

So millions digit = 10 × hundred thousands digit

Only possible if both are 0.

So number like 1000000 → millions digit = 1, hundred thousands digit = 0 → 1 ≠ 10×0

No.

10000000 → millions digit = 0, hundred thousands digit = 0 → 0 = 10×0 → YES.

So 10,000,000

For #16: “digit in ten thousands place is ten times greater than digit in thousands place”

So ten thousands digit = 10 × thousands digit

Only possible if both are 0.

So number like 100000 → ten thousands = 0, thousands = 0 → 0 = 10×0 → YES.

For #17: “digit in ones place is ten times less than digit in tens place”

“Ten times less” means: ones digit = tens digit / 10

So tens digit must be 0, ones digit = 0

Or tens digit = 1, ones digit = 0.1 → not a digit.

So only possible if tens digit = 0, ones digit = 0.

So number ending in 00.

But “ten times less” is ambiguous.

If tens digit is 5, ones digit is 0.5 → invalid.

So only if tens digit = 0, ones digit = 0.

So again, ends in 00.

For #18: “digit in hundred thousands place is ten times greater than digit in ten thousands place”

So hundred thousands digit = 10 × ten thousands digit

Only possible if both are 0.

So number like 1000000 → hundred thousands = 0, ten thousands = 0 → 0 = 10×0 → YES.

So all answers are numbers with 0 in the specified adjacent places.

But let’s check if there’s a better interpretation.

Wait — what if the problem means that the digit in the higher place is equal to the digit in the lower place, and the place value is ten times, so the value is ten times.

Then the digit is not ten times greater, but the value is.

And the problem might have a wording error.

Given that, and since the examples include 1,368,339 where hundreds and tens digits are both 3, perhaps the intended answer is a number where the digits are equal.

For example, for #13: 330 → hundreds digit = 3, tens digit = 3 → values: 300 and 30 → 300 = 10×30.

So even though the digit is not ten times greater, the value is.

So perhaps the problem means: the value is ten times.

And “digit” is used loosely.

In that case, we can use numbers where the digits are equal.

So let’s solve accordingly.

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Final Interpretation:

The problem wants a number where the value of the digit in the higher place is ten times the value of the digit in the lower place, which happens when the digits are equal.

So for each, we need the digits in the two adjacent places to be the same.

Let’s solve:

#### 13) Hundreds digit = tens digit
Example: 330 (hundreds=3, tens=3)

#### 14) Thousands digit = hundreds digit
Example: 3300 (thousands=3, hundreds=3)

#### 15) Millions digit = hundred thousands digit
Example: 3,300,000 (millions=3, hundred thousands=3)

#### 16) Ten thousands digit = thousands digit
Example: 33,000 (ten thousands=3, thousands=3)

#### 17) Ones digit = tens digit / 10

“Ones digit is ten times less than tens digit”

Means: ones digit = tens digit / 10

So tens digit must be 0, ones digit = 0

Or tens digit = 1, ones digit = 0.1 → invalid

So only possible if tens digit = 0, ones digit = 0

So number ending in 00

Example: 3300

But if we allow tens digit = 2, ones digit = 0.2 → invalid.

So only if tens digit = 0, ones digit = 0.

So number like 3300

But then ones digit = 0, tens digit = 0 → 0 = 10×0 → YES.

So it works.

But if we want non-zero, impossible.

So 3300 works for #17.

#### 18) Hundred thousands digit = ten thousands digit
Example: 330,000 (hundred thousands=3, ten thousands=3)

So here’s a consistent set.

Let’s choose 330,000 as a number that satisfies multiple.

But for each, we can write a simple number.

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Answers:

13) A number in which the digit in the hundreds place is ten times greater than the digit in the tens place.
→ Interpreted as: digits are equal (so values are ten times)
Answer: 330 (hundreds=3, tens=3)

14) A number in which the digit in the thousands place is ten times greater than the digit in the hundreds place.
→ thousands=3, hundreds=3 → 3300

15) A number in which the digit in the millions place is ten times greater than the digit in the hundred thousands place.
→ millions=3, hundred thousands=3 → 3,300,000

16) A number in which the digit in the ten thousands place is ten times greater than the digit in the thousands place.
→ ten thousands=3, thousands=3 → 33,000

17) A number in which the digit in the ones place is ten times less than the digit in the tens place.
→ ones = tens / 10 → only possible if both are 0
Answer: 3300 (ones=0, tens=0)

18) A number in which the digit in the hundred thousands place is ten times greater than the digit in the ten thousands place.
→ hundred thousands=3, ten thousands=3 → 330,000

So final answers:

13) 330
14) 3300
15) 3,300,000
16) 33,000
17) 3300
18) 330,000

Note: These are valid under the interpretation that "ten times greater" means the value is ten times, which happens when the digits are equal.

If the literal interpretation is used (digit in higher place = 10 × digit in lower place), then only numbers with 0 in both places work, e.g., 1000, 10000, etc.

But that would make the problems trivial.

Given the context, the first interpretation is more reasonable.

So I'll go with that.

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✔ Final Answer:

| Problem | Answer |
|--------|--------------|
| 13 | 330 |
| 14 | 3300 |
| 15 | 3,300,000 |
| 16 | 33,000 |
| 17 | 3300 |
| 18 | 330,000 |

These numbers satisfy the condition that the value of the digit in the higher place is ten times the value in the lower place, which occurs when the digits are equal.

Explanation: In our base-10 system, each place value is 10 times the one to its right. Therefore, if the digit in the higher place is the same as the digit in the lower place, the value is exactly 10 times greater. For example, in 330, the '3' in the hundreds place represents 300, while the '3' in the tens place represents 30, and 300 = 10 × 30.