HCF, LCM, Prime Factorisation, and Discrete Frequency Distributions Study Guide

Prime Factorisation Definitions and Given Values

• Prime factorisation expresses a composite number as the product of its prime factors raised to specific integer powers.
• Standard reference numbers expressed in prime factor form:
  • $A = 2^3 \times 5^4$
  • $B = 2^4 \times 3^3 \times 7$
  • $C = 2^2 \times 3 \times 5^3$
  • $D = 2 \times 3^5 \times 5$
  • $E = 3^2 \times 5 \times 7^3$
  • $F = 2^2 \times 7^3 \times 11$
  • $G = 2^4 \times 3 \times 5^2 \times 13$

Highest Common Factor (HCF) Principles and Calculations

• Fundamental Rule for HCF: The Highest Common Factor of two or more numbers expressed in prime factor form is found by identifying all prime factors common to all terms and taking the lowest exponent for each common prime factor.

Two-Factor HCF Calculations

• (a) Find the HCF of $A$ and $D$:

  • Given prime decompositions:
  • $A = 2^3 \times 5^4$
  • $D = 2^1 \times 3^5 \times 5^1$
  • Identify common prime bases:
  • Base $2$ is present in both terms; the minimum exponent is $\min(3, 1) = 1$.
  • Base $5$ is present in both terms; the minimum exponent is $\min(4, 1) = 1$.
  • Base $3$ is not present in $A$.
  • Answer as product of prime factors:
  • $\text{HCF}(A, D) = 2 \times 5$

• (b) Find the HCF of $B$ and $C$:

  • Given prime decompositions:
  • $B = 2^4 \times 3^3 \times 7$
  • $C = 2^2 \times 3^1 \times 5^3$
  • Identify common prime bases:
  • Base $2$ is present in both terms; the minimum exponent is $\min(4, 2) = 2$.
  • Base $3$ is present in both terms; the minimum exponent is $\min(3, 1) = 1$.
  • Answer as product of prime factors:
  • $\text{HCF}(B, C) = 2^2 \times 3$

• (c) Find the HCF of $D$ and $E$:

  • Given prime decompositions:
  • $D = 2^1 \times 3^5 \times 5^1$
  • $E = 3^2 \times 5^1 \times 7^3$
  • Identify common prime bases:
  • Base $3$ is present in both terms; the minimum exponent is $\min(5, 2) = 2$.
  • Base $5$ is present in both terms; the minimum exponent is $\min(1, 1) = 1$.
  • Answer as product of prime factors:
  • $\text{HCF}(D, E) = 3^2 \times 5$

• (d) Find the HCF of $C$ and $G$:

  • Given prime decompositions:
  • $C = 2^2 \times 3^1 \times 5^3$
  • $G = 2^4 \times 3^1 \times 5^2 \times 13$
  • Identify common prime bases:
  • Base $2$ is present in both terms; minimum exponent is $\min(2, 4) = 2$.
  • Base $3$ is present in both terms; minimum exponent is $\min(1, 1) = 1$.
  • Base $5$ is present in both terms; minimum exponent is $\min(3, 2) = 2$.
  • Answer as product of prime factors:
  • $\text{HCF}(C, G) = 2^2 \times 3 \times 5^2$

Three-Factor HCF Calculations

• (a) Find the HCF of $A$, $C$, and $D$:

  • Given prime decompositions:
  • $A = 2^3 \times 5^4$
  • $C = 2^2 \times 3^1 \times 5^3$
  • $D = 2^1 \times 3^5 \times 5^1$
  • Determine common factors present across ALL three numbers:
  • Factor $2$ appears in $A$ ($2^3$), $C$ ($2^2$), and $D$ ($2^1$). Minimum power is $2^1$.
  • Factor $5$ appears in $A$ ($5^4$), $C$ ($5^3$), and $D$ ($5^1$). Minimum power is $5^1$.
  • Factor $3$ appears in $C$ and $D$, but NOT in $A$.
  • Answer as product of prime factors:
  • $\text{HCF}(A, C, D) = 2 \times 5$

• (b) Find the HCF of $C$, $E$, and $G$:

  • Given prime decompositions:
  • $C = 2^2 \times 3^1 \times 5^3$
  • $E = 3^2 \times 5^1 \times 7^3$
  • $G = 2^4 \times 3^1 \times 5^2 \times 13$
  • Determine common factors present across ALL three numbers:
  • Factor $3$ appears in $C$ ($3^1$), $E$ ($3^2$), and $G$ ($3^1$). Minimum power is $3^1$.
  • Factor $5$ appears in $C$ ($5^3$), $E$ ($5^1$), and $G$ ($5^2$). Minimum power is $5^1$.
  • Factor $2$ appears in $C$ and $G$, but NOT in $E$.
  • Factor $7$ appears only in $E$; Factor $13$ appears only in G$.\n - Answer as product of prime factors:\n - \text{HCF}(C, E, G) = 3 \times 5\n\n## Modified Multiples HCF Calculations\n\n- **(a) Find the HCF of 8A$and$B:**\n - First express 8A in prime factor form:\n - 8 = 2^3\n - 8A = 2^3 \times A = 2^3 \times (2^3 \times 5^4) = 2^6 \times 5^4\n - Compare with B = 2^4 \times 3^3 \times 7:\n - Common prime factor is 2.\n - Lowest power between 2^6$and$2^4$is$2^4.\n - Answer as product of prime factors:\n - \text{HCF}(8A, B) = 2^4\n\n- **(b) Find the HCF of 5B$and$E:**\n - First express 5B in prime factor form:\n - 5B = 5 \times (2^4 \times 3^3 \times 7) = 2^4 \times 3^3 \times 5^1 \times 7^1\n - Compare with E = 3^2 \times 5^1 \times 7^3:\n - Prime factor 3$: minimum exponent is$\min(3, 2) = 2.\n - Prime factor 5$: minimum exponent is$\min(1, 1) = 1.\n - Prime factor 7$: minimum exponent is$\min(1, 3) = 1.\n - Answer as product of prime factors:\n - \text{HCF}(5B, E) = 3^2 \times 5 \times 7\n\n# Lowest Common Multiple (LCM) Principles and Calculations\n\n- **Fundamental Rule for LCM:** The Lowest Common Multiple of two or more numbers expressed in prime factor form is found by identifying all prime factors present in ANY of the numbers and taking the highest exponent for each prime factor.\n\n## Two-Factor LCM Calculations\n\n- **(a) Find the LCM of A$and$C:**\n - Given prime decompositions:\n - A = 2^3 \times 5^4\n - C = 2^2 \times 3^1 \times 5^3\n - Select maximum exponent for every prime base:\n - Base 2$:$\max(3, 2) = 3\n - Base 3$:$\max(0, 1) = 1\n - Base 5$:$\max(4, 3) = 4\n - Answer as product of prime factors:\n - \text{LCM}(A, C) = 2^3 \times 3 \times 5^4\n\n- **(b) Find the LCM of C$and$D:**\n - Given prime decompositions:\n - C = 2^2 \times 3^1 \times 5^3\n - D = 2^1 \times 3^5 \times 5^1\n - Select maximum exponent for every prime base:\n - Base 2$:$\max(2, 1) = 2\n - Base 3$:$\max(1, 5) = 5\n - Base 5$:$\max(3, 1) = 3\n - Answer as product of prime factors:\n - \text{LCM}(C, D) = 2^2 \times 3^5 \times 5^3\n\n- **(c) Find the LCM of B$and$F:**\n - Given prime decompositions:\n - B = 2^4 \times 3^3 \times 7^1\n - F = 2^2 \times 7^3 \times 11^1\n - Select maximum exponent for every prime base:\n - Base 2$:$\max(4, 2) = 4\n - Base 3$:$\max(3, 0) = 3\n - Base 7$:$\max(1, 3) = 3\n - Base 11$:$\max(0, 1) = 1\n - Answer as product of prime factors:\n - \text{LCM}(B, F) = 2^4 \times 3^3 \times 7^3 \times 11\n\n- **(d) Find the LCM of E$and$G:**\n - Given prime decompositions:\n - E = 3^2 \times 5^1 \times 7^3\n - G = 2^4 \times 3^1 \times 5^2 \times 13^1\n - Select maximum exponent for every prime base:\n - Base 2$:$\max(0, 4) = 4\n - Base 3$:$\max(2, 1) = 2\n - Base 5$:$\max(1, 2) = 2\n - Base 7$:$\max(3, 0) = 3\n - Base 13$:$\max(0, 1) = 1\n - Answer as product of prime factors:\n - \text{LCM}(E, G) = 2^4 \times 3^2 \times 5^2 \times 7^3 \times 13\n\n## Three-Factor LCM Calculations\n\n- **(c) Find the LCM of A$,$C$, and$D:**\n - Given prime decompositions:\n - A = 2^3 \times 5^4\n - C = 2^2 \times 3^1 \times 5^3\n - D = 2^1 \times 3^5 \times 5^1\n - Select maximum exponent for every prime base present:\n - Prime 2$:$\max(3, 2, 1) = 3\n - Prime 3$:$\max(0, 1, 5) = 5\n - Prime 5$:$\max(4, 3, 1) = 4\n - Answer as product of prime factors:\n - \text{LCM}(A, C, D) = 2^3 \times 3^5 \times 5^4\n\n- **(d) Find the LCM of C$,$D$, and$G:**\n - Given prime decompositions:\n - C = 2^2 \times 3^1 \times 5^3\n - D = 2^1 \times 3^5 \times 5^1\n - G = 2^4 \times 3^1 \times 5^2 \times 13^1\n - Select maximum exponent for every prime base present:\n - Prime 2$:$\max(2, 1, 4) = 4\n - Prime 3$:$\max(1, 5, 1) = 5\n - Prime 5$:$\max(3, 1, 2) = 3\n - Prime 13$:$\max(0, 0, 1) = 1\n - Answer as product of prime factors:\n - \text{LCM}(C, D, G) = 2^4 \times 3^5 \times 5^3 \times 13\n\n## Modified Multiples LCM Calculations\n\n- **(c) Find the LCM of C$and$10D:**\n - Prime decomposition of C:\n - C = 2^2 \times 3^1 \times 5^3\n - Prime decomposition of 10D:\n - 10 = 2 \times 5\n - 10D = (2 \times 5) \times (2^1 \times 3^5 \times 5^1) = 2^2 \times 3^5 \times 5^2\n - Compare C = 2^2 \times 3^1 \times 5^3$and$10D = 2^2 \times 3^5 \times 5^2:\n - Prime 2$:$\max(2, 2) = 2\n - Prime 3$:$\max(1, 5) = 5\n - Prime 5$:$\max(3, 2) = 3\n - Answer as product of prime factors:\n - \text{LCM}(C, 10D) = 2^2 \times 3^5 \times 5^3\n\n- **(d) Find the LCM of 8D$and$G:**\n - Prime decomposition of 8D:\n - 8 = 2^3\n - 8D = 2^3 \times (2^1 \times 3^5 \times 5^1) = 2^4 \times 3^5 \times 5^1\n - Prime decomposition of G:\n - G = 2^4 \times 3^1 \times 5^2 \times 13^1\n - Select maximum powers:\n - Prime 2$:$\max(4, 4) = 4\n - Prime 3$:$\max(5, 1) = 5\n - Prime 5$:$\max(1, 2) = 2\n - Prime 13$:$\max(0, 1) = 1\n - Answer as product of prime factors:\n - \text{LCM}(8D, G) = 2^4 \times 3^5 \times 5^2 \times 13\n\n# Advanced Prime Factorisation: Solving for Unknown Exponents\n\n- Given expressions:\n - H = 2^x \times 5^y \times 7^2\n - F = 2^2 \times 7^3 \times 11^1\n - \text{HCF}(H, F) = 98\n - \text{LCM}(H, F) = 2^2 \times 5^3 \times 7^3 \times 11\n\n- **Step-by-step determination of exponent x:**\n - Express the given HCF in prime factor form:\n - 98 = 2 \times 49 = 2^1 \times 7^2\n - Apply the HCF rule for prime factor 2:\n - \text{HCF power of } 2 = \min(x, 2) = 1\n - Since \min(x, 2) = 1$,$x$must equal$1\n\n- **Step-by-step determination of exponent y:**\n - Examine the LCM expression:\n - \text{LCM}(H, F) = 2^2 \times 5^3 \times 7^3 \times 11\n - Apply the LCM rule for prime factor 5:\n - F$contains$5^0$(no factor of$5)\n - H$contains$5^y\n - \text{LCM power of } 5 = \max(y, 0) = 3\n - Therefore, y$must equal$3\n\n- **Final Answer:**\n - x = 1\n - y = 3\n\n# Discrete Frequency Distributions: Mean and Mode\n\n\n\n## Frequency Table Construction\n\n- In a discrete frequency table, x$represents the value of the variable,$f$represents the frequency (number of occurrences), and the 3rd column$f \times x represents the product of each value and its corresponding frequency.\n\n| Value (x$) | Frequency ($f$) | Product ($f \times x) |\n| :--- | :--- | :--- |\n| 1$|$5$|$1 \times 5 = 5 |\n| 2$|$8$|$2 \times 8 = 16 |\n| 3$|$10$|$3 \times 10 = 30 |\n| 4$|$11$|$4 \times 11 = 44 |\n| 5$|$6$|$5 \times 6 = 30 |\n| **Total** | **\sum f = 40$** | **$\sum f x = 125** |\n\n## Step-by-Step Mean Calculation\n\n- **Formula for the mean of a frequency distribution:**\n - \text{Mean} = \frac{\sum fx}{\sum f}\n- **Calculation:**\n - Calculate the sum of the frequencies:\n - \sum f = 5 + 8 + 10 + 11 + 6 = 40\n - Calculate the sum of the products (f \times x):\n - \sum fx = 5 + 16 + 30 + 44 + 30 = 125\n - Divide the total product by the total frequency:\n - \text{Mean} = \frac{125}{40} = 3.125\n\n## Determining the Modal Value\n\n- **Definition of Mode:** The mode is the value of the variable (x) that occurs with the highest frequency.\n- **Analysis of the table:**\n - Frequencies for values 1, 2, 3, 4, 5$are$5, 8, 10, 11, 6 respectively.\n - The maximum frequency is 11$, which corresponds to the variable value$x = 4\n - Therefore, the modal value is 4$$.