Solved Exercises on Extended Form Numbers in Grade 4

Master extended form numbers: place value decomposition, digit representation, and number composition through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Basic extended form decomposition
Exercise 1
Write the extended form of 6,842.
Definition:

Extended form: Writing a number to show the value of each digit based on its place value, expressed as the sum of each digit multiplied by its place value.

Extended Form Method:
  1. Identify the place value of each digit
  2. Write each digit multiplied by its place value
  3. Join all parts with addition signs
Number
6,842
Extended Form
(6×1,000) + (8×100) + (4×10) + (2×1)
Step 1: Identify place values

6 is in thousands place: 6 × 1,000 = 6,000

8 is in hundreds place: 8 × 100 = 800

4 is in tens place: 4 × 10 = 40

2 is in ones place: 2 × 1 = 2

Step 2: Write in extended form

6,842 = (6×1,000) + (8×100) + (4×10) + (2×1)

Step 3: Verify the calculation

(6×1,000) + (8×100) + (4×10) + (2×1) = 6,000 + 800 + 40 + 2 = 6,842 ✓

6,842 = (6×1,000) + (8×100) + (4×10) + (2×1)
Final answer:

6,842 = (6×1,000) + (8×100) + (4×10) + (2×1)

Applied rules:

• Place Value Multiplication: Each digit multiplied by its place value

• Extended Form Format: (digit × place value) + (digit × place value)...

• Verification: Sum of parts equals original number

2 Extended form with zeros
Exercise 2
Write 7,053 in extended form.
Definition:

Zero in extended form: Even zeros are included as (0 × place value) = 0

Number
7,053
Extended Form
(7×1,000) + (0×100) + (5×10) + (3×1)
Simplified
(7×1,000) + (5×10) + (3×1)
Step 1: Identify place values

7 is in thousands place: 7 × 1,000 = 7,000

0 is in hundreds place: 0 × 100 = 0

5 is in tens place: 5 × 10 = 50

3 is in ones place: 3 × 1 = 3

Step 2: Write in extended form

7,053 = (7×1,000) + (0×100) + (5×10) + (3×1)

Step 3: Simplify by omitting zero terms

7,053 = (7×1,000) + (5×10) + (3×1)

Step 4: Verify the calculation

(7×1,000) + (5×10) + (3×1) = 7,000 + 50 + 3 = 7,053 ✓

7,053 = (7×1,000) + (5×10) + (3×1)
Final answer:

7,053 = (7×1,000) + (5×10) + (3×1)

Applied rules:

• Zero Inclusion: Zeros are written as (0 × place value)

• Simplification: Terms with (0 × place value) can be omitted

• Verification: Sum of parts equals original number

3 Converting extended form to standard form
Exercise 3
What number is represented by (5×10,000) + (2×1,000) + (0×100) + (6×10) + (4×1)?
Definition:

Standard form: The regular way we write numbers without showing place values

Extended Form
(5×10,000) + (2×1,000) + (0×100) + (6×10) + (4×1)
Standard Form
52,064
Step 1: Calculate each part of the extended form

(5×10,000) = 50,000

(2×1,000) = 2,000

(0×100) = 0

(6×10) = 60

(4×1) = 4

Step 2: Add all parts together

50,000 + 2,000 + 0 + 60 + 4 = 52,064

Step 3: Verify by converting back to extended form

52,064 = (5×10,000) + (2×1,000) + (0×100) + (6×10) + (4×1) ✓

(5×10,000) + (2×1,000) + (0×100) + (6×10) + (4×1) = 52,064
Final answer:

The number is 52,064

Applied rules:

• Place Value Addition: Sum all extended form parts

• Digit Placement: Each part corresponds to a specific place

• Verification: Convert back to ensure accuracy

Extended Form Rules and Methods
\(6,842 = (6×1,000) + (8×100) + (4×10) + (2×1)\)
Extended Form Example
Ones Place
\(d \times 1\)
Digit × 1
Tens Place
\(d \times 10\)
Digit × 10
Hundreds Place
\(d \times 100\)
Digit × 100
Thousands Place
\(d \times 1,000\)
Digit × 1,000
Ten-Thousands Place
\(d \times 10,000\)
Digit × 10,000
Key definitions:

Extended Form: A number expressed as the sum of each digit multiplied by its place value

Place Value: The value of a digit based on its position in a number

Standard Form: The regular way we write numbers (e.g., 6,842)

Expanded Form: Writing a number as the sum of place values (e.g., 6,000 + 800 + 40 + 2)

Extended Form: Writing a number as the sum of digits multiplied by their place values (e.g., (6×1,000) + (8×100) + (4×10) + (2×1))

Complete methodology:
  1. Identify positions: Recognize place values (ones, tens, hundreds, thousands, etc.)
  2. Write multiplication: Express each digit as (digit × place value)
  3. Join with addition: Combine all parts with plus signs
  4. Verify: Check that the sum equals the original number
Tip 1: Remember the place value pattern: 1, 10, 100, 1,000, 10,000...
Tip 2: Zeros in extended form can be written as (0 × place value) or omitted.
Tip 3: Always verify by adding the extended form parts to get the original number.
Tip 4: Use base-10 blocks to visualize place values for better understanding.
Common errors: Forgetting place values, misplacing digits, confusing expanded form with extended form.
Memory aid: Think of extended form as "digit times place value".
Formulas to know by heart:

• Extended Form: Sum of (digit × place value)

• Place value pattern: 1, 10, 100, 1,000, 10,000...

• Verification: Sum of extended parts = Original number

• Zero rule: (0 × any place value) = 0

Solution: Exercises 4 to 5
4 Comparing extended forms
Exercise 4
Which is greater: (4×1,000) + (6×100) + (3×10) + (8×1) or (4×1,000) + (5×100) + (9×10) + (2×1)? Show your work.
Definition:

Comparison in extended form: Convert both expressions to standard form to compare

First Expression
(4×1,000) + (6×100) + (3×10) + (8×1)
Equals
4,000 + 600 + 30 + 8 = 4,638
Second Expression
(4×1,000) + (5×100) + (9×10) + (2×1)
Equals
4,000 + 500 + 90 + 2 = 4,592
Step 1: Calculate first expression

(4×1,000) = 4,000

(6×100) = 600

(3×10) = 30

(8×1) = 8

Total: 4,000 + 600 + 30 + 8 = 4,638

Step 2: Calculate second expression

(4×1,000) = 4,000

(5×100) = 500

(9×10) = 90

(2×1) = 2

Total: 4,000 + 500 + 90 + 2 = 4,592

Step 3: Compare the results

4,638 vs 4,592

Since 4,638 > 4,592, the first expression is greater.

Step 4: Alternative method - compare by place value

Thousands: Both have 4,000

Hundreds: 600 > 500, so the first is greater

(4×1,000) + (6×100) + (3×10) + (8×1) = 4,638
(4×1,000) + (5×100) + (9×10) + (2×1) = 4,592
4,638 > 4,592
Final answer:

(4×1,000) + (6×100) + (3×10) + (8×1) is greater than (4×1,000) + (5×100) + (9×10) + (2×1)

Because 4,638 > 4,592

Applied rules:

• Comparison method: Convert to standard form to compare

• Place value comparison: Compare digits from left to right

• Verification: Confirm the comparison is correct

5 Missing digit in extended form
Exercise 5
If (7×1,000) + (2×100) + (n×10) + (5×1) = 7,275, what is the value of n?
Definition:

Missing digit problem: Finding an unknown digit in extended form

Given
(7×1,000) + (2×100) + (n×10) + (5×1) = 7,275
Calculate known parts
7,000 + 200 + (n×10) + 5 = 7,275
Step 1: Calculate the known parts

(7×1,000) = 7,000

(2×100) = 200

(5×1) = 5

So: 7,000 + 200 + (n×10) + 5 = 7,275

Step 2: Simplify the equation

7,000 + 200 + 5 = 7,205

So: 7,205 + (n×10) = 7,275

Step 3: Solve for n

(n×10) = 7,275 - 7,205 = 70

n×10 = 70

n = 70 ÷ 10 = 7

Step 4: Verify the answer

(7×1,000) + (2×100) + (7×10) + (5×1) = 7,000 + 200 + 70 + 5 = 7,275 ✓

n = 7
(7×1,000) + (2×100) + (7×10) + (5×1) = 7,275
Final answer:

The value of n is 7

Because (7×1,000) + (2×100) + (7×10) + (5×1) = 7,000 + 200 + 70 + 5 = 7,275

Applied rules:

• Algebraic reasoning: Treat unknown as variable to solve

• Place value understanding: Each part contributes to total value

• Verification: Substitute back to confirm answer

Complete Summary: Extended Form Numbers
\(d_n \times 10^n + d_{n-1} \times 10^{n-1} + ... + d_1 \times 10^1 + d_0 \times 10^0\)
General Extended Form Formula
Key definitions:

Extended Form: A number expressed as the sum of each digit multiplied by its place value

Place Value: The value of a digit based on its position in a number

Standard Form: The usual way we write numbers (e.g., 6,842)

Expanded Form: Writing a number as the sum of place values (e.g., 6,000 + 800 + 40 + 2)

Extended Form: Writing a number as the sum of digits multiplied by their place values (e.g., (6×1,000) + (8×100) + (4×10) + (2×1))

Digit: Any of the symbols 0-9 used to make numbers

Complete methodology:
  1. Identify positions: Recognize the place value of each digit
  2. Multiplication: Express each digit as (digit × place value)
  3. Addition: Join all parts with plus signs
  4. Verification: Add the extended form parts to confirm they equal the original number
Tip 1: Use the place value pattern: 1, 10, 100, 1,000, 10,000...
Tip 2: Draw a place value chart to organize digits before extending.
Tip 3: Always verify by calculating the sum of extended form parts.
Tip 4: Practice reading numbers aloud to reinforce place value understanding.
Common errors: Confusing extended form with expanded form, misplacing digits, forgetting zeros.
Memory aids: Use base-10 blocks, draw charts, remember "digit times place value".
Formulas and Rules to Know:

• Place Value Pattern: ..., 10,000, 1,000, 100, 10, 1

• Extended Form: Sum of (digit × place value)

• Zero Rule: (0 × any place value) = 0

• Verification: Sum of extended parts = Original number

• Format: (digit × place value) + (digit × place value) + ...

Extended Form Visualization: Number Breakdown
Exercise 6: Understanding Extended Form
Consider the following numbers and their extended form breakdowns:
4,567 = (4×1,000) + (5×100) + (6×10) + (7×1)
8,092 = (8×1,000) + (0×100) + (9×10) + (2×1)
6,105 = (6×1,000) + (1×100) + (0×10) + (5×1)

Analysis: The chart shows how numbers decompose into extended form components.

  • 4,567 = (4×1,000) + (5×100) + (6×10) + (7×1) (Four thousand five hundred sixty-seven)
  • 8,092 = (8×1,000) + (0×100) + (9×10) + (2×1) (Eight thousand ninety-two)
  • 6,105 = (6×1,000) + (1×100) + (0×10) + (5×1) (Six thousand one hundred five)

Questions & Answers

Question: I don't understand why we need to learn extended form. Why can't we just write numbers normally?

Answer: Great question! Extended form is crucial for understanding how our number system works:

  • It helps you understand what each digit means in a number
  • It builds a foundation for addition and subtraction with regrouping
  • It prepares you for working with larger numbers and decimals later
  • It makes it easier to compare numbers and understand their magnitude
  • It shows the relationship between digits and their place values

For example, in extended form (6×1,000) + (8×100) + (4×10) + (2×1), you can see that the 6 represents 6 thousands (6,000), not just 6. This helps you understand the true value of each digit!

Question: My child gets confused when there are zeros in the extended form like (8×1,000) + (0×100) + (5×10) + (3×1). How do I explain this?

Answer: Here's a helpful explanation for zeros in extended form:

  • In (8×1,000) + (0×100) + (5×10) + (3×1), the (0×100) means there are 0 hundreds
  • Even though 0×100 = 0, we include it to show that there's a place for hundreds (even if it's empty)
  • This helps us see that the number is 8,053: 8 thousands, 0 hundreds, 5 tens, 3 ones
  • The zero keeps the 5 in the tens place and the 3 in the ones place in the right positions

You can think of it like having an empty box in a row of boxes. Even though the box is empty, it still takes up space and keeps the other items in their correct positions!

Sometimes we simplify by writing (8×1,000) + (5×10) + (3×1), but including the zero version shows the complete place value structure.

Question: What's the difference between expanded form and extended form? They seem the same to me.

Answer: Actually, extended form and expanded form refer to the same concept in 4th grade mathematics. Both terms describe writing a number as the sum of each digit multiplied by its place value:

  • Extended Form: Shows each digit multiplied by its place value (e.g., (6×1,000) + (8×100) + (4×10) + (2×1))
  • Expanded Form: Shows each digit multiplied by its place value (e.g., (6×1,000) + (8×100) + (4×10) + (2×1))

Extended Form: 6,842 = (6×1,000) + (8×100) + (4×10) + (2×1)

Expanded Form: 6,842 = (6×1,000) + (8×100) + (4×10) + (2×1)

In 4th grade, these terms are often used interchangeably. Both show the "work" behind how we get each part of the number by multiplying digits by their place values!