Instructions
This worksheet explores two fundamental ways to represent large natural numbers: using place value (Powers of 10) and using prime factors (Exponents). Follow the instructions for each section carefully.
- Complete all tables by showing the required notation.
- Use a calculator only for checking your final answers in Section 4, but show all factorization steps.
- The goal is to practice precision in using exponents to simplify large products.
Section 1: Expanded Notation (Powers of 10)
Expanded notation using powers of 10 clearly shows the place value of each digit. Remember that $10^0 = 1$.
Example: $7,130 = (7 \times 10^3) + (1 \times 10^2) + (3 \times 10^1) + (0 \times 10^0)$
Complete the table below.
| Natural Number | Expanded Notation using Powers of 10 |
|---|---|
| 1,945 | $(1 \times 10^3) + (9 \times 10^2) + (4 \times 10^1) + (5 \times 10^0)$ |
| 821 | |
| 5,006 | |
| 32,890 | |
| 100,000 | |
| 403,117 |
Section 2: Understanding Prime Factorization
Prime factorization is the process of finding which prime numbers multiply together to make the original number. When we use exponents, we simplify the notation.
Quick Check: Identify which of the following expressions are not fully factored using only prime numbers. Circle the letter(s) of the invalid expression(s).
A. $2^3 \times 5$ B. $4 \times 3^2$ C. $11^1 \times 2^2$ D. $6 \times 5 \times 3$
Section 3: Prime Factorization using Exponents
Find the prime factors for the numbers below and express the final result using exponents for repeated factors.
Example: $24 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3^1$
Complete the table.
| Natural Number (N) | Prime Factors (Show work conceptually) | Prime Factorization (Exponent Form) |
|---|---|---|
| 18 | $18 = 2 \times 9 = 2 \times 3 \times 3$ | $2^1 \times 3^2$ |
| 100 | ||
| 75 | ||
| 144 | ||
| 210 | ||
| 6,000 |
Section 4: Application and Challenge
Part A: Decoding the Number
Sometimes, numbers are stored or transmitted in their exponent form to save space. Decode the following numbers by calculating their values.
-
$N_A = 2^2 \times 3^1 \times 5^1$ Calculation: $N_A = $
-
$N_B = 3^3 \times 5^2 \times 2^1$ Calculation: $N_B = $
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Challenge Question: A very large computer inventory code is represented by $C = 2^4 \times 7^2$. What is the total numerical value of this code? Calculation: $C = $
Part B: Real-World Place Value
Imagine you are an astronomer counting the approximate number of stars in a distant galaxy cluster. The count is $4,000,000,000$.
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Represent this star count using scientific notation (a single digit times a power of 10). Star Count = $\times 10^?$
-
Write the expanded notation for the number $4,000,000,000$. Expanded Notation =
Answer Key
Section 1: Expanded Notation (Powers of 10)
| Natural Number | Expanded Notation using Powers of 10 |
|---|---|
| 1,945 | $(1 \times 10^3) + (9 \times 10^2) + (4 \times 10^1) + (5 \times 10^0)$ |
| 821 | $(8 \times 10^2) + (2 \times 10^1) + (1 \times 10^0)$ |
| 5,006 | $(5 \times 10^3) + (0 \times 10^2) + (0 \times 10^1) + (6 \times 10^0)$ |
| 32,890 | $(3 \times 10^4) + (2 \times 10^3) + (8 \times 10^2) + (9 \times 10^1) + (0 \times 10^0)$ |
| 100,000 | $(1 \times 10^5) + (0 \times 10^4) + \dots + (0 \times 10^0)$ or simply $(1 \times 10^5)$ |
| 403,117 | $(4 \times 10^5) + (0 \times 10^4) + (3 \times 10^3) + (1 \times 10^2) + (1 \times 10^1) + (7 \times 10^0)$ |
Section 2: Understanding Prime Factorization
Invalid expressions (contain composite factors):
B. $4 \times 3^2$ (4 is composite) D. $6 \times 5 \times 3$ (6 is composite)
Section 3: Prime Factorization using Exponents
| Natural Number (N) | Prime Factors (Show work conceptually) | Prime Factorization (Exponent Form) |
|---|---|---|
| 18 | $18 = 2 \times 9 = 2 \times 3 \times 3$ | $2^1 \times 3^2$ |
| 100 | $100 = 10 \times 10 = (2 \times 5) \times (2 \times 5)$ | $2^2 \times 5^2$ |
| 75 | $75 = 3 \times 25 = 3 \times 5 \times 5$ | $3^1 \times 5^2$ |
| 144 | $144 = 12 \times 12 = (2^2 \times 3) \times (2^2 \times 3)$ | $2^4 \times 3^2$ |
| 210 | $210 = 10 \times 21 = (2 \times 5) \times (3 \times 7)$ | $2^1 \times 3^1 \times 5^1 \times 7^1$ |
| 6,000 | $6,000 = 6 \times 1000 = (2 \times 3) \times (10^3) = (2 \times 3) \times (2^3 \times 5^3)$ | $2^4 \times 3^1 \times 5^3$ |
Section 4: Application and Challenge
Part A: Decoding the Number
-
$N_A = 2^2 \times 3^1 \times 5^1$ Calculation: $4 \times 3 \times 5 = 12 \times 5$ $N_A = 60$
-
$N_B = 3^3 \times 5^2 \times 2^1$ Calculation: $27 \times 25 \times 2 = 27 \times 50$ $N_B = 1,350$
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Challenge Question: $C = 2^4 \times 7^2$ Calculation: $16 \times 49$ $C = 784$
Part B: Real-World Place Value
-
Represent this star count using scientific notation. Star Count = $4 \times 10^9$
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Write the expanded notation for the number $4,000,000,000$. Expanded Notation = $(4 \times 10^9) + \dots + (0 \times 10^0)$ (or just $4 \times 10^9$)