**Shortcuts for HCF:**

**There are two types of question usually asked in the examination based on H.C.F.**

**TYPE – 1**

When various number a, b, c is divided by number N and gives the same remainder r in each case. Then H.C.F. of a, b, c is given by the H.C.F. of (a – b) or (b – a) and (b–c) or (c – b) and (c –a) or (a –c). Let’s understand with help of example

**Example 1**: Find the largest possible value of N if all three numbers 59, 77,104 when divided by N leaves same remainder?

**Solution:** As per given

59 = N x a + r………………………..(1)

77 = N x b + r………………………..(2)

104 =N x c + r………………………..(3)

Subtract equation (1) from (2)

77 – 59 =18 = N x (b -a)……………………..(4)

Subtract equation (2) from (3)

104 – 77=27 = N x (c – b)…………………….(5)

Subtract equation (1) from (3)

104 -59 =45 = N x (c – a)……………………..(6)

From this we can conclude that the N has to be the factor of 18, 27,45

Now we are asked for the largest number so we will take the H.C.F. of 18, 27, 45 and which is 9 .So 9 is the largest number which divides 59, 77, 104 and gives same remainder in each case.

**TYPE – 2**

When the number is divided by various numbers a, b, c and gives different remainders p, q, r.

**Example 2: **Find the largest number N, which divides, 162,292,325 and gives remainders 1, 5 and 3 respectively?

**Solution: **As per given information

162 = N x a + 1………………………..(1)

292 = N x b + 5………………………..(2)

325 = N x c + 3………………………..(3)

Since the three numbers are 162, 292 and 325 and the remainders are 1, 5, 3 so we can say that the number is totally divisible with 161, 287, 322. So finding largest number, which divides 162,292,325, is equivalent to finding H.C.F of 161, 287 and 322.Since HCF (161, 287, 322)=7 so the 7 is the number which when divided by the 162,292,325 gives the remainder 1,5 and 3 respectively.

**Some points to remember **

1. LCM x HCF of two numbers is equal to the product of numbers.

2. If HCF (H) of two numbers is given, then the numbers can be assume as HCF x A and HCF x B (A and B are integers)

3. HCF of a given set of numbers is always a factor of LCM

**LCM of Fractions:**

** **LCM of two or more fractions is given by as a fraction:

(LCM of numerators of all the fractions) / (HCF of denominator of all the fractions)

For example:

**LCM [ ½, 5/12, 7/6] = {LCM of (1,5,7)}/{HCF of (2,12,6)} = 32/2**

*Assignment:*

**1.**What is the LCM of 18 and 36?

A. 18

B. 36

C. 90

D. 72

**Answer:** Option **B**

**Explanation:**

18= 2x3x3=2^{1}x3^{2}

36=2x2x3x3=2^{2}x3^{2}

Primes involved are 2 and 3 and maximum power involved is 2 for both primes. Hence

2^{2 }x 3^{2} is required LCM.

**2.** What is the HCF of 18 and 36?

A.18

B.9

C.12

D.36

**Answer:** Option A

**Explanation:**

18= 2x3x3=2^{1}x3^{2}

36=2x2x3x3=2^{2}x3^{2}

Primes involved are 2 and 3 and minimum power involved is 1 for both primes. Hence 2×3^{2}=18 is required HCF.

**3.** What is the LCM of 18, 180 and 216?

A. 540

B. 360

C. 720

D. 1080

**Answer:** Option D

**Explanation**

18=2x3x3=2^{1}x3^{2}

180=2x2x3x3x5=2^{2}x3^{2}x5^{1}

72=2x2x2x3x3x3=2^{3}x3^{3}

Primes involved are 2, 3 and 5 and maximum powers involved are 3, 3 and 1 respectively. Hence 2^{3}x3^{3}x5^{1} = 1080 is required LCM

**4.** What is the HCF of 12,18 and 30?

A.10

B.5

C.6

D.8

**Answer:** Option C

**Explanation: **

12=2x2x3x3=22×31

18=2x3x3=21×32

30 =2x3x5=21x31x51

Primes involved are 2, 3 and 5 minimum powers involved are 1, 1 and 0 respectively. Hence

21x31x50=6 is required HCF.

5. What is the LCM of 13 and 11 ?

A.11

B.13

C.143

D.1

**Answer**: Option C

**Explanation:**

Since both are prime numbers so LCM is nothing but their product = 13×11 = 143

lcm of 18,180 and 72 is 360

correct