SSC GD LCM And HCF

• Introduction to SSC GD LCM and HCF
• Understanding Factors and Multiples
• Finding the Highest Common Factor (HCF)
• Finding the Least Common Multiple (LCM)
• The Special Relationship between LCM and HCF
• Solving SSC GD Word Problems using LCM and HCF

Introduction to SSC GD LCM and HCF

Welcome to your study notes for the SSC GD LCM and HCF topic! This is a very important part of the math section. If you understand these concepts well, you can easily solve many lcm and hcf ssc gd questions. We will break down every step so you can master these skills for your SSC GD Constable 2026 exam.

Understanding Factors and Multiples

What are Factors and Multiples?

• Factors: Factors are numbers that divide another number exactly, leaving no remainder. Think of them as the building blocks of a number.
• Example: The factors of 10 are 1, 2, 5, and 10. Why? Because $10÷2=5$ (exact division).
• Multiples: Multiples are the results you get when you multiply a number by 1, 2, 3, 4, and so on. They are like the counting table for that number.
• Example: The multiples of 3 are 3, 6, 9, 12, 15, and it goes on forever.
• Understanding factors and multiples is the first step to solving any ssc gd lcm hcf question.

Solved Examples

Example 1: Find all the factors of 24.

• Solution:
  1. We look for pairs of numbers that multiply to 24: $1\times 24$, $2\times 12$, $3\times 8$, $4\times 6$.
  2. We check if 5 divides 24 (it does not). We check if 6 divides 24 (it does, $6\times 4$).
  3. So, the factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.

Example 2: List the first five multiples of 7.

• Solution:
  1. We multiply 7 by 1, 2, 3, 4, and 5.
  2. $7\times 1=7$
  3. $7\times 2=14$
  4. $7\times 3=21$
  5. $7\times 4=28$
  6. $7\times 5=35$.
  7. The first five multiples are 7, 14, 21, 28, 35.

Example 3: Is 9 a factor of 72?

• Solution:
  1. To check if 9 is a factor of 72, we divide 72 by 9.
  2. $72÷9=8$.
  3. Since the division is exact (remainder is 0), yes, 9 is a factor of 72.

Example 4: Is 56 a multiple of 6?

• Solution:
  1. To check if 56 is a multiple of 6, we divide 56 by 6.
  2. $56÷6$. We know $6\times 9=54$ and $6\times 10=60$.
  3. $56÷6$ leaves a remainder of 2.
  4. Since the division is not exact, 56 is NOT a multiple of 6.

Finding the Highest Common Factor (HCF)

What is HCF?

• HCF stands for Highest Common Factor. It is also sometimes called the Greatest Common Divisor (GCD).
• The HCF is the biggest number that can divide two or more numbers exactly.
• Think of it as finding the largest possible group size when you need to divide things equally. This is very common in ssc gd lcm and hcf questions.
• We usually use the Prime Factorization Method to find HCF quickly in the SSC GD exam.

Prime Factorization Method for HCF

1. Step 1: Find the Prime Factors of each number. (Prime factors are numbers like 2, 3, 5, 7, 11, etc.)
2. Step 2: Look for the factors that are Common (shared) between all the numbers.
3. Step 3: Multiply these common factors, using the smallest power (exponent) they appear with.

Solved Examples

Example 1: Find the HCF of 12 and 18.

• Solution:
  1. Prime Factors of 12: $12=2\times 6=2\times 2\times 3=2^2\times 3^1$.
  2. Prime Factors of 18: $18=2\times 9=2\times 3\times 3=2^1\times 3^2$.
  3. Common Factors: 2 and 3. We take the smallest power of each: $2^1$ and $3^1$.
  4. HCF $=2^1\times 3^1=6$. The HCF of 12 and 18 is 6.

Example 2: What is the HCF of 30, 45, and 75? This is a typical lcm hcf ssc gd question.

• Solution:
  1. Prime Factors:
     • $30=2\times 3\times 5$
     • $45=3\times 3\times 5=3^2\times 5$
     • $75=3\times 5\times 5=3\times 5^2$
  2. Common Factors: 3 and 5 appear in all three numbers.
  3. Smallest Powers: $3^1$ (from 30 and 75) and $5^1$ (from 30 and 45).
  4. HCF $=3\times 5=15$.

Example 3: Find the HCF of $2^3\times 3^4\times 5^1$ and $2^2\times 3^5\times 7^1$.

• Solution:
  1. We look for common bases: 2 and 3 are common. 5 and 7 are not common.
  2. For base 2, the powers are 3 and 2. We take the smallest: $2^2$.
  3. For base 3, the powers are 4 and 5. We take the smallest: $3^4$.
  4. HCF $=2^2\times 3^4=4\times 81=324$.

Example 4: Find the HCF of 17 and 23.

• Solution:
  1. 17 is a prime number, so its factors are 1 and 17.
  2. 23 is a prime number, so its factors are 1 and 23.
  3. The only common factor is 1.
  4. HCF $=1$. (If two numbers have an HCF of 1, they are called co-prime numbers.)

Finding the Least Common Multiple (LCM)

What is LCM?

• LCM stands for Least Common Multiple. It is the smallest number that is a multiple of two or more given numbers.
• Think of it as the first time two cycles or events will happen at the same time.
• The LCM is always greater than or equal to the largest of the given numbers.
• Mastering the LCM is essential for solving ssc gd lcm and hcf questions pdf problems involving time and distance.

Prime Factorization Method for LCM

1. Step 1: Find the Prime Factors of each number.
2. Step 2: List ALL the prime factors that appear in any of the numbers (common or not).
3. Step 3: Multiply these factors, using the highest power (exponent) they appear with.

Solved Examples

Example 1: Find the LCM of 12 and 18.

• Solution:
  1. Prime Factors:
     • $12=2^2\times 3^1$
     • $18=2^1\times 3^2$
  2. List all factors: 2 and 3.
  3. Highest Powers: For 2, the highest power is $2^2$. For 3, the highest power is $3^2$.
  4. LCM $=2^2\times 3^2=4\times 9=36$.

Example 2: Calculate the LCM of 8, 15, and 20. This is a common ssc gd lcm hcf mock test question.

• Solution:
  1. Prime Factors:
     • $8=2\times 2\times 2=2^3$
     • $15=3\times 5$
     • $20=2\times 2\times 5=2^2\times 5$
  2. List all factors: 2, 3, and 5.
  3. Highest Powers: $2^3$ (from 8), $3^1$ (from 15), $5^1$ (from 15 and 20).
  4. LCM $=2^3\times 3^1\times 5^1=8\times 3\times 5=120$.

Example 3: Find the LCM of $2^3\times 3^4\times 5^1$ and $2^2\times 3^5\times 7^1$.

• Solution:
  1. List all bases: 2, 3, 5, and 7.
  2. Highest Power of 2: $2^3$.
  3. Highest Power of 3: $3^5$.
  4. Highest Power of 5: $5^1$.
  5. Highest Power of 7: $7^1$.
  6. LCM $=2^3\times 3^5\times 5^1\times 7^1=8\times 243\times 5\times 7=68,040$.

Example 4: Find the LCM of 6, 9, and 10 using the Division Method.

• Solution:
  1. We write the numbers and divide by the smallest prime number (2): $$2|6,9,10$$ $$3,9,5$$
  2. Divide by the next prime number (3): $$3|3,9,5$$ $$1,3,5$$
  3. Divide by 3 again: $$3|1,3,5$$ $$1,1,5$$
  4. Divide by 5: $$5|1,1,5$$ $$1,1,1$$
  5. LCM = Multiply all the divisors: $2\times 3\times 3\times 5=90$.

The Special Relationship between LCM and HCF

The Core Formula for SSC GD LCM and HCF

• There is a very important rule that connects the HCF and LCM of two numbers. You must memorize this for the SSC GD LCM and HCF exam.
• If A and B are two numbers, the rule is: $$\text{Product of the two numbers}=\text{LCM}\times \text{HCF}$$ $$A\times B=\text{LCM}(A,B)\times \text{HCF}(A,B)$$
• This formula only works for two numbers.
• If you know any three of these values (A, B, LCM, HCF), you can always find the fourth. This helps solve many lcm and hcf ssc gd questions.

Solved Examples

Example 1: The product of two numbers is 1200. If their HCF is 10, find their LCM.

• Solution:
  1. We use the formula: $A\times B=\text{LCM}\times \text{HCF}$.
  2. We know $A\times B=1200$ and $\text{HCF}=10$.
  3. $1200=\text{LCM}\times 10$.
  4. To find LCM, we divide 1200 by 10: $\text{LCM}=1200/10=120$.

Example 2: The LCM of two numbers is 180 and their HCF is 6. If one number is 30, find the other number (B).

• Solution:
  1. Formula: $A\times B=\text{LCM}\times \text{HCF}$.
  2. We know $A=30$, $\text{LCM}=180$, $\text{HCF}=6$.
  3. $30\times B=180\times 6$.
  4. First, calculate the right side: $180\times 6=1080$.
  5. $30\times B=1080$.
  6. $B=1080/30=36$. The other number is 36.

Example 3: The HCF and LCM of two numbers are 8 and 48, respectively. If the ratio of the two numbers is 2:3, find the numbers. This is a tricky ssc gd lcm hcf question.

• Solution:
  1. Let the numbers be $A=2x$ and $B=3x$.
  2. Remember: The HCF is always a factor of the numbers. Since HCF is 8, the numbers must be $A=2\times 8=16$ and $B=3\times 8=24$. (This is a shortcut!)
  3. Let’s verify using the formula: $A\times B=\text{LCM}\times \text{HCF}$.
  4. $16\times 24=384$.
  5. $\text{LCM}\times \text{HCF}=48\times 8=384$.
  6. Since $384=384$, the numbers are 16 and 24.

Example 4: Can two numbers have an HCF of 5 and an LCM of 42? Explain why or why not.

• Solution:
  1. Rule: The HCF must always be a factor of the LCM.
  2. We check if 5 divides 42 exactly. $42÷5$.
  3. $42÷5=8$ with a remainder of 2.
  4. Since 5 does not divide 42 exactly, it is impossible for two numbers to have an HCF of 5 and an LCM of 42.

Solving SSC GD Word Problems using LCM and HCF

When to use HCF vs. LCM

When you see a word problem in your ssc gd lcm hcf mock test, you need to decide which tool to use.

Solved Examples

Example 1 (HCF): A shopkeeper has 48 apples and 60 oranges. He wants to pack them into baskets so that each basket has the same number of fruit, and no fruit is left over. What is the maximum number of fruits he can put in each basket?

• Solution:
  1. The word “maximum” tells us to find the HCF of 48 and 60.
  2. Prime Factors: $48=2^4\times 3^1$ and $60=2^2\times 3^1\times 5^1$.
  3. Common factors with smallest power: $2^2$ and $3^1$.
  4. HCF $=2^2\times 3=4\times 3=12$.
  5. The maximum number of fruits per basket is 12.

Example 2 (LCM): Three traffic lights change after every 30 seconds, 45 seconds, and 60 seconds. If they all change together at 8:00 AM, when will they change together again? This is a classic ssc gd lcm and hcf questions type.

• Solution:
  1. The phrase “change together again” means we need to find the smallest common meeting time, so we find the LCM of 30, 45, and 60.
  2. Using the Division Method: $$5|30,45,60$$ $$2|6,9,12$$ $$3|3,9,6$$ $$1,3,2$$
  3. LCM $=5\times 2\times 3\times 1\times 3\times 2=180$.
  4. The lights will change together after 180 seconds.
  5. Convert seconds to minutes: $180÷60=3$ minutes.
  6. They will change together again at 8:00 AM + 3 minutes = 8:03 AM.

Example 3 (HCF with Remainder): Find the greatest number that divides 130 and 170, leaving a remainder of 2 in each case.

• Solution:
  1. Since the remainder is 2, the number must divide $(130-2)$ and $(170-2)$ exactly.
  2. The new numbers are $130-2=128$ and $170-2=168$.
  3. We find the HCF of 128 and 168.
  4. Prime Factors: $128=2^7$ and $168=2^3\times 3\times 7$.
  5. Common factor with smallest power: $2^3$.
  6. HCF $=2^3=8$. The greatest number is 8.

Example 4 (LCM with Remainder): Find the smallest number which, when divided by 12, 15, and 20, leaves a remainder of 5 in each case. This is a key concept for ssc gd lcm and hcf questions pdf.

• Solution:
  1. First, find the LCM of 12, 15, and 20.
  2. LCM calculation (using division method): $$2|12,15,20$$ $$2|6,15,10$$ $$3|3,15,5$$ $$5|1,5,5$$ $$1,1,1$$
  3. LCM $=2\times 2\times 3\times 5=60$.
  4. The smallest number that leaves a remainder of 5 is found by adding the remainder to the LCM.
  5. Required Number = LCM + Remainder $=60+5=65$.