2.3 HCF and LCM Properties, Prime Factorization & Practical Word Problems

Definitions and Fundamental Methods

Highest Common Factor (HCF) / Greatest Common Divisor (GCD)

The HCF of two or more positive integers is the largest positive integer that divides each of the numbers without leaving a remainder.

Least Common Multiple (LCM)

The LCM of two or more positive integers is the smallest positive integer that is divisible by each of the given numbers.

+-----------------------------------------------------------------------------+
|                        COMPUTATIONAL METHODS COMPARISON                     |
+-----------------------------------------------------------------------------+
| 1. PRIME FACTORIZATION METHOD                                               |
|    Write each number in canonical prime powers:                             |
|       A = p_1^(a_1) * p_2^(a_2) * ...                                       |
|       B = p_1^(b_1) * p_2^(b_2) * ...                                       |
|    HCF = Product of MINIMUM powers:  Prod p_i^min(a_i, b_i)                 |
|    LCM = Product of MAXIMUM powers:  Prod p_i^max(a_i, b_i)                 |
+-----------------------------------------------------------------------------+
| 2. EUCLID'S DIVISION ALGORITHM (FOR HCF)                                    |
|    Based on the identity: gcd(A, B) = gcd(B, A mod B)                       |
|    Divide larger by smaller until remainder is 0. Last non-zero divisor is  |
|    the exact HCF. Highly efficient for large, multi-digit numbers.          |
+-----------------------------------------------------------------------------+

---

Fundamental Product Rules & Multi-Variable Relations

The Two-Number Product Theorem

For any two positive integers $a$ and $b$:

$$\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$$

For three positive integers $a, b, c$, the mathematically rigorous identity is:

$$\text{LCM}(a, b, c) = \frac{a \cdot b \cdot c \cdot \text{HCF}(a, b, c)}{\text{HCF}(a, b) \cdot \text{HCF}(b, c) \cdot \text{HCF}(c, a)}$$

---

HCF and LCM of Fractions, Decimals & Powers

Fractions

Before applying formulas, all fractions must be reduced to their simplest irreducible form $\frac{p}{q}$ where $\gcd(p, q) = 1$:

$$\text{HCF}\left(\frac{a}{b}, \frac{c}{d}, \frac{e}{f}\right) = \frac{\text{HCF}(a, c, e)}{\text{LCM}(b, d, f)}$$ $$\text{LCM}\left(\frac{a}{b}, \frac{c}{d}, \frac{e}{f}\right) = \frac{\text{LCM}(a, c, e)}{\text{HCF}(b, d, f)}$$

Decimals

To compute the HCF or LCM of decimal numbers:

1. Count the maximum number of decimal places across all terms (say, $k$ places).
2. Multiply every number by $10^k$ to convert them into integers.
3. Compute the HCF or LCM of the resulting integers.
4. Divide the final result by $10^k$.

Exponential Powers with Common Bases

For powers of the same integer base $x>1$ (using the rational-fraction convention for negative exponents):

• Positive exponents: $\text{HCF}(x^p, x^q, x^r) = x^{\min(p, q, r)}$ and $\text{LCM}(x^p, x^q, x^r) = x^{\max(p, q, r)}$.
• Negative exponents: $\text{HCF}(x^{-2}, x^{-5}, x^{-8}) = x^{-8}$ (smallest value) and $\text{LCM}(x^{-2}, x^{-5}, x^{-8}) = x^{-2}$ (largest value).

---

Operational Railway & Aptitude Word Problem Models

Model 1: Synchronized Fictional Indicators

In a fictional timing exercise, independent indicator lamps flash at fixed intervals of $t_1, t_2, t_3, \dots$ seconds. This is a periodic-event model, not a description of railway signalling.

• The events occur simultaneously at time intervals equal to $\text{LCM}(t_1, t_2, t_3, \dots)$.
• If the events start simultaneously at $t = 0$, the number of simultaneous occurrences within a total elapsed time $T$ is:

$$\text{Total Simultaneous Occurrences} = \left\lfloor \frac{T}{\text{LCM}} \right\rfloor + 1$$

(If the question asks how many times they toll after the start, do not add 1.)

Model 2: Circular Track Rendezvous

Two or more locomotives or athletes circulate around a closed circular loop of length $L$ with constant speeds $v_1, v_2, v_3$.

• Time taken by each to complete one full revolution: $T_1 = \frac{L}{v_1}, T_2 = \frac{L}{v_2}, T_3 = \frac{L}{v_3}$.
• Rendezvous at Starting Point: They will all meet simultaneously at the starting point after a time equal to:

$$T_{\text{start}} = \text{LCM}(T_1, T_2, T_3)$$

• Rendezvous Anywhere on Track (Two Bodies): They meet anywhere on the loop at intervals of $\frac{L}{|v_1 - v_2|}$ (same direction, unequal speeds) or $\frac{L}{v_1 + v_2}$ (opposite direction).

---

For the fractional HCF/LCM convention, use positive rational quantities: a common divisor divides each quantity by an integer, and a common multiple is an integer multiple of each. Circular return formulas assume simultaneous departure and commensurable lap times.

Remainder Classification Models

Remainder problems in HCF and LCM fall into distinct mathematical patterns that frequently appear on the CBT:

+-----------------------------------------------------------------------------+
|                        REMAINDER MODELS TAXONOMY                            |
+-----------------------------------------------------------------------------+
| HCF REMAINDER MODELS (Finding the GREATEST DIVISOR)                         |
|                                                                             |
| Case 1: Leaves same known remainder r when dividing x, y, z                 |
|         Required Number = HCF(x - r, y - r, z - r)                          |
|                                                                             |
| Case 2: Leaves same UNKNOWN remainder when dividing x, y, z                 |
|         Required Number = HCF(|x - y|, |y - z|, |z - x|)                    |
|                                                                             |
| Case 3: Leaves different remainders a, b, c when dividing x, y, z           |
|         Required Number = HCF(x - a, y - b, z - c)                          |
+-----------------------------------------------------------------------------+
| LCM REMAINDER MODELS (Finding the SMALLEST DIVIDEND)                        |
|                                                                             |
| Case 1: Leaves same remainder r when divided by x, y, z                     |
|         Required Number = k * LCM(x, y, z) + r                              |
|                                                                             |
| Case 2: Leaves different remainders a, b, c such that:                      |
|         (x - a) = (y - b) = (z - c) = d  (Constant Difference)              |
|         Required Number = k * LCM(x, y, z) - d                              |
+-----------------------------------------------------------------------------+

---

For greatest-divisor problems, verify that the proposed divisor exceeds every required remainder. Otherwise no such divisor exists. For LCM-offset models, choose an integer $k$ giving a positive dividend and valid remainders. If no lower bound on the dividend is imposed, $k=0$ can be valid in the common-positive-remainder model; do not automatically select $k=1$.

Worked Examples

Worked Example 1: Periodic Indicator Synchronization

Problem: Four fictional indicator lamps flash every 24 seconds, 36 seconds, 54 seconds, and 72 seconds respectively. If all four lamps flash simultaneously at exactly 08:00:00 AM, how many times will they flash simultaneously between 08:00:00 AM and 09:12:00 AM (inclusive of both endpoints)?

• Step 1: Compute the LCM of the signal intervals: Express each interval in canonical prime factor form:

$$24 = 2^3 \times 3^1$$ $$36 = 2^2 \times 3^2$$ $$54 = 2^1 \times 3^3$$ $$72 = 2^3 \times 3^2$$ $$\text{LCM} = 2^{\max(3, 2, 1, 3)} \times 3^{\max(1, 2, 3, 2)} = 2^3 \times 3^3 = 8 \times 27 = 216 \text{ seconds}$$

The lamps synchronize every 216 seconds ($3 \text{ minutes and } 36 \text{ seconds}$).

• Step 2: Determine total elapsed duration in seconds: From 08:00:00 AM to 09:12:00 AM is 1 hour and 12 minutes $= 72 \text{ minutes}$.

$$T = 72 \times 60 = 4,320 \text{ seconds}$$

• Step 3: Compute simultaneous occurrences:

$$\frac{T}{\text{LCM}} = \frac{4320}{216} = 20$$

Since both endpoints (including the start at 08:00:00 AM) are included:

$$\text{Total Changes} = 20 + 1 = 21$$

• Conclusion: The lamps flash simultaneously 21 times.

---

Worked Example 2: LCM Model with Constant Difference

Problem: Find the least number of wagons in a freight rake inventory such that when partitioned into sidings containing 12, 16, 18, or 21 wagons each, there remain 7, 11, 13, and 16 wagons unaccommodated respectively.

• Step 1: Check for a constant difference between divisor and remainder:

$$12 - 7 = 5$$ $$16 - 11 = 5$$ $$18 - 13 = 5$$ $$21 - 16 = 5$$

A constant difference exists: $d = 5$.

• Step 2: Compute $\text{LCM}(12, 16, 18, 21)$:

$$12 = 2^2 \times 3$$ $$16 = 2^4$$ $$18 = 2 \times 3^2$$ $$21 = 3 \times 7$$ $$\text{LCM} = 2^4 \times 3^2 \times 7 = 16 \times 9 \times 7 = 144 \times 7 = 1,008$$

• Step 3: Apply the constant difference formula:

$$N = \text{LCM} - d = 1,008 - 5 = 1,003$$

• Step 4: Verification:

$$1003 \div 12 = 83 \text{ with remainder } 7 \quad (12 \times 83 = 996)$$ $$1003 \div 16 = 62 \text{ with remainder } 11 \quad (16 \times 62 = 992)$$ $$1003 \div 18 = 55 \text{ with remainder } 13 \quad (18 \times 55 = 990)$$ $$1003 \div 21 = 47 \text{ with remainder } 16 \quad (21 \times 47 = 987)$$

• Conclusion: The least number of wagons is 1,003.

---

Worked Example 3: HCF Unknown Same Remainder Model

Problem: Find the greatest positive integer that divides 1354, 1866, and 2762 leaving the exact same remainder $R$ in each case, and determine the value of that remainder $R$.

• Step 1: Identify model type: Because the remainder $r$ is identical but unknown, compute the HCF of the pairwise absolute differences:

$$D_1 = |1866 - 1354| = 512$$ $$D_2 = |2762 - 1866| = 896$$ $$D_3 = |2762 - 1354| = 1408$$

• Step 2: Factorize the differences to find HCF:

$$512 = 2^9$$ $$896 = 2^7 \times 7 = 128 \times 7$$ $$1408 = 2^7 \times 11 = 128 \times 11$$ $$\text{HCF}(512, 896, 1408) = 2^7 = 128$$

The greatest dividing number is 128.

• Step 3: Calculate the common remainder $R$: Divide any of the original numbers by 128:

$$1354 \div 128: \quad 128 \times 10 = 1280$$ $$R = 1354 - 1280 = 74$$

(Verification: $1866 = 128 \times 14 + 74 = 1792 + 74 = 1866$; $2762 = 128 \times 21 + 74 = 2688 + 74 = 2762$.)

• Conclusion: The greatest divisor is 128, leaving a constant remainder of 74.

The periodic bells or signal names in these arithmetic examples are fictional timing devices. Real railway signal aspects are not set by a fixed LCM timer; use the numbers only to practise divisibility and periodic-event models.