3.1 Place Value, Powers of 10, Factors & Multiples

Place Value, Powers of 10, Factors & Multiples

The Hindu-Arabic numeral system is a positional base-10 system. In this system, the value of any digit depends entirely on its position, or place, within the numeral. Each move one position to the left increases a digit's value tenfold ($10^1$), while each move one position to the right decreases its value by a factor of 10 ($\frac{1}{10}$ or $10^{-1}$).

Understanding place value is the essential foundation for arithmetic operations, estimation, and algebraic reasoning tested on national primary assessments in Jamaica.

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Face Value, Place Value, and Total Value

Every digit in a number possesses three distinct attributes:

1. Face Value: The intrinsic, unchanging value of the digit itself regardless of position (e.g., the face value of 7 is always 7).
2. Place Value: The positional name or column value assigned to the location the digit occupies (such as tens, thousands, or ten thousands).
3. Total Value (or Value): The product of the digit's face value and its place value ($\text{Total Value} = \text{Face Value} \times \text{Place Value}$).

Consider the numeral $4,725,318$:

• The digit 7 has a face value of 7.
• It sits in the hundred thousands place ($10^5 = 100,000$).
• Its total value is $7 \times 100,000 = 700,000$.

The Place Value Chart from Units to Millions

Numbers are divided into groups of three digits called periods (Units period, Thousands period, Millions period), separated by commas for clarity:

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Expanded Notation: Additive and Exponential Forms

Expanded notation expresses a number as the sum of the total values of each of its constituent non-zero digits. Grade 6 students must be comfortable switching between two formats:

1. Additive Expanded Form

In additive expanded form, each digit is written as its complete numerical value:

$4,725,318 = 4,000,000 + 700,000 + 20,000 + 5,000 + 300 + 10 + 8$

2. Exponential Form (Powers of 10)

In exponential form, each digit is multiplied by 10 raised to the exponent corresponding to its place value column:

$4,725,318 = (4 \times 10^6) + (7 \times 10^5) + (2 \times 10^4) + (5 \times 10^3) + (3 \times 10^2) + (1 \times 10^1) + (8 \times 10^0)$

Worked Example: Decomposing a Number with Zeros

Problem: Express $6,043,050$ in both additive and exponential expanded notation.

Solution Walkthrough:

1. Identify the non-zero digits and their column positions:
   • 6 is in the millions place ($10^6$)
   • 0 is in the hundred thousands place (omitted from the sum)
   • 4 is in the ten thousands place ($10^4$)
   • 3 is in the thousands place ($10^3$)
   • 0 is in the hundreds place (omitted)
   • 5 is in the tens place ($10^1$)
   • 0 is in the ones place (omitted)
2. Additive form: $6,000,000 + 40,000 + 3,000 + 50$
3. Exponential form: $(6 \times 10^6) + (4 \times 10^4) + (3 \times 10^3) + (5 \times 10^1)$

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Powers of 10: Multiplying and Dividing

Multiplying and dividing whole numbers and decimals by powers of 10 ($10, 100, 1,000, 10,000$) relies on shifting columns:

• Multiplying by $10^n$: Each digit shifts $n$ places to the left, making the number larger. In whole numbers, this appends $n$ trailing zeros.
  • $54 \times 10 = 540$
  • $54 \times 100 = 5,400$
  • $54 \times 1,000 = 54,000$
• Dividing by $10^n$: Each digit shifts $n$ places to the right, making the number smaller by a factor of $10^n$.
  • $68,000 \div 10 = 6,800$
  • $68,000 \div 100 = 680$
  • $68,000 \div 1,000 = 68$

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Rounding Whole Numbers

Rounding replaces an exact number with an approximate value that is simpler to manage in mental calculation and estimation.

The Systematic Rounding Procedure

1. Locate the Target Place: Underline or highlight the digit in the place you are rounding to.
2. Inspect the Decision Digit: Look at the single digit immediately to its right (one place value column lower).
3. Apply the Rule of 5:
   • If the decision digit is 5, 6, 7, 8, or 9: Round up (add 1 to the target digit).
   • If the decision digit is 0, 1, 2, 3, or 4: Round down (leave the target digit unchanged).
4. Replace Following Digits with Zeros: Change all digits to the right of the target place into zeros.

Worked Example: Rounding with Cascading Nines

Problem: Round $49,962$ to the nearest hundred.

Solution Walkthrough:

1. Target place is the hundreds digit: $49,\underline{9}62$.
2. Decision digit is immediately to the right in the tens column: 6.
3. Because $6 \ge 5$, round up: add 1 to 9 hundreds.
4. $9 + 1 = 10$ hundreds. Write 0 in the hundreds place and carry 1 to the thousands place ($9 + 1 = 10$ thousands, carry 1 to the ten-thousands place: $4 + 1 = 5$).
5. Replace the tens and ones digits with zeros.
6. Result: $50,000$.

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Classifying Properties of Numbers

1. Even and Odd Numbers

• Even numbers: Whole numbers divisible by 2 with no remainder. Their ones digit is always $0, 2, 4, 6,$ or $8$.
• Odd numbers: Whole numbers not divisible by 2 (leaving a remainder of 1). Their ones digit is always $1, 3, 5, 7,$ or $9$.

2. Prime and Composite Numbers

• Prime Number: A whole number strictly greater than 1 that has exactly two distinct positive factors: 1 and itself.
  • The first ten prime numbers are: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29$.
  • The number 2 is the smallest prime and the only even prime number.
• Composite Number: A whole number greater than 1 that has three or more factors (e.g., $4, 6, 8, 9, 10, 12, 14, 15$).
• The Numbers 0 and 1: Neither prime nor composite. The number 1 has only one positive factor (itself), failing the definition of prime which requires exactly two factors.

3. Square Numbers

A square number is the product of an integer multiplied by itself ($n \times n = n^2$). Geometrically, it represents the area of a square whose side length is $n$.

$1^2 = 1, \quad 2^2 = 4, \quad 3^2 = 9, \quad 4^2 = 16, \quad 5^2 = 25, \quad 6^2 = 36, \quad 7^2 = 49, \quad 8^2 = 64, \quad 9^2 = 81, \quad 10^2 = 100, \quad 11^2 = 121, \quad 12^2 = 144$

4. Triangular Numbers

A triangular number can be arranged as dots in an equilateral triangle. Each consecutive triangular number adds another row with one more dot than the previous row:

$T_1 = 1$ $T_2 = 1 + 2 = 3$ $T_3 = 1 + 2 + 3 = 6$ $T_4 = 1 + 2 + 3 + 4 = 10$ $T_5 = 1 + 2 + 3 + 4 + 5 = 15$ $T_6 = 1 + 2 + 3 + 4 + 5 + 6 = 21$ $T_7 = 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28$

The general formula for the $n$-th triangular number is:

$T_n = \frac{n(n + 1)}{2}$

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Prime Factorization, HCF, and LCM

According to the Fundamental Theorem of Arithmetic, every composite number can be uniquely factored into a product of prime numbers (apart from the order of factors).

Finding Prime Factors Using the Factor Tree

To construct a factor tree for 72:

1. Choose any pair of factors: $72 = 8 \times 9$.
2. Break each factor into smaller factors: $8 = 2 \times 4$; $9 = 3 \times 3$.
3. Break down any composite factors until only primes remain: $4 = 2 \times 2$.
4. Collect the prime leaves: $72 = 2 \times 2 \times 2 \times 3 \times 3 = 2^3 \times 3^2$.

Similarly, for 120:

• $120 = 10 \times 12$
• $10 = 2 \times 5$
• $12 = 3 \times 4 = 3 \times 2 \times 2$
• Prime factorization: $120 = 2 \times 2 \times 2 \times 3 \times 5 = 2^3 \times 3^1 \times 5^1$.

Highest Common Factor (HCF / GCF)

The Highest Common Factor is the largest whole number that divides evenly into two or more given numbers.

• Method: Write out prime factorizations in exponential form. For every prime factor common to both numbers, take the factor with the smallest exponent and multiply them together.
• For 72 and 120:
  • Common prime factor 2: smallest power is $2^3 = 8$.
  • Common prime factor 3: smallest power is $3^1 = 3$.
  • Prime factor 5 is not common to both.
  • $\text{HCF}(72, 120) = 2^3 \times 3^1 = 8 \times 3 = 24$.

Lowest Common Multiple (LCM)

The Lowest Common Multiple is the smallest non-zero whole number that is a multiple of two or more given numbers.

• Method: Take each prime factor that appears in either number with its highest exponent and multiply them together.
• For 72 and 120:
  • Prime 2: highest power is $2^3 = 8$.
  • Prime 3: highest power is $3^2 = 9$.
  • Prime 5: highest power is $5^1 = 5$.
  • $\text{LCM}(72, 120) = 2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = 72 \times 5 = 360$.

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Divisibility Rules

Divisibility rules allow quick mental testing of whether a large number can be divided by an integer without leaving a remainder:

Worked Example: Applying Divisibility

Problem: Is the number $85,416$ divisible by 6 and by 9?

1. Test for 2: The final digit is 6 (even), so $85,416$ is divisible by 2.
2. Test for 3: Sum of digits $= 8 + 5 + 4 + 1 + 6 = 24$. Since 24 is divisible by 3 ($24 \div 3 = 8$), $85,416$ is divisible by 3.
3. Since it is divisible by both 2 and 3, it is divisible by 6.
4. Test for 9: The sum of digits is 24. Since 24 is not divisible by 9 ($24 \div 9 = 2 \text{ R } 6$), $85,416$ is not divisible by 9.

Sets: Members, Union and Intersection

Grade 6 Mathematics in the National Standards Curriculum begins with sets, and the Ministry's sample Curriculum-Based Test items ask students to list the members of an intersection and to describe the elements of a set.

Key Set Language

• A set is a well-defined collection of objects. Each object is a member or element of the set.
• Sets are named with capital letters and their members are listed inside curly brackets: $A = \{2, 4, 6, 8\}$.
• The symbol $\in$ means "is a member of" ($6 \in A$) and $\notin$ means "is not a member of" ($5 \notin A$).
• A finite set has a countable number of members, such as the set of parishes in Jamaica (14 members).
• An infinite set goes on forever, such as the set of multiples of 5: $\{5, 10, 15, 20, \dots\}$. The three dots show that the pattern continues.
• The empty set has no members and is written $\{\ \}$ or $\varnothing$, for example the set of months with 32 days.
• $n(A)$ means the number of members in set $A$.

Union and Intersection

Worked Example: Factors of 12 and 18

Let $A$ be the set of factors of 12 and $B$ the set of factors of 18.

• $A = \{1, 2, 3, 4, 6, 12\}$ and $B = \{1, 2, 3, 6, 9, 18\}$.
• $A \cap B = \{1, 2, 3, 6\}$: these are the common factors, and the largest one, 6, is the HCF of 12 and 18.
• $A \cup B = \{1, 2, 3, 4, 6, 9, 12, 18\}$, so $n(A \cup B) = 8$.
• In a Venn diagram, 1, 2, 3 and 6 go in the overlap; 4 and 12 go in the part of circle $A$ only; 9 and 18 go in the part of circle $B$ only.