HCF and LCM with solved examples

HCF and LCM

HCF:
H.C.F is the highest common factor if two or more numbers is the greatest number that divides each one of them exactly. HCF is also called GCM i.e. greatest common measures.

Methods of calculation of HCF

1. Prime factorization method

Step 1. Break the given numbers into prime factor
Step 2. Find the product of all the prime factors common to all the numbers.
Step 3. The product of common prime factors with the least powers gives HCF.

Ex. Find the HCF of 24,30 and 42.
Sol.Factors of 24 = 2*2*2*3= 2³*3¹

Factors of 30 = 2*3*5

Factors of 42= 2*3*7

The product of common prime factors with the least powers = 2*3=6

2. Division Method

Step 1. Divide the larger number by the smaller ones.
Step 2. Divide the divisor by the remainder.
Step 3. Repeat step 2 until the remainder becomes zero. The last divisor is the required HCF.

Ex. find the HCF of 26 and 455.

Sol. 

  THE required HCF is 13.

To find the HCF of  more than two numbers:

Rule: find the HCF of any two of the numbers and then find the HCF of this HCF and the third number and so on. The last HCF will be the required HCF.

Ex. Find the HCF of 1365,1560 and 1755.

Sol.

The required HCF is 195.

Another method:

The work of finding the HCF may sometimes be simplified By the following devices:

i. Any obvious factor which is common to both numbers may be removed before the rule is applied. Care should however be taken to multiply this factor into the HCF of the quotients.

ii. If one of the numbers has a prime factor not contained in the other, it may be rejected.

iii. At any stage of the work, any factor of the divisor not contained in the dividend may be rejected. This is because any factor which divides only one of the two cannot be a portion of the required HCF. 

Ex. Find the HCF of 42237 and 75582.

Sol. 42237 = 9* 4693

75582 = 2*9*4199

We may reject 2 which is not common. But 9 is a common factor. We, therefore, set it aside and find the HCF of 4199 and 4693.

494 is divisible by 2 but 41299 is not. we, therefore, divide 494 by 2 and proceed with 247 and 4199.

The HCF is 247.

LCM:

LCM is the least common multiple of two or more numbers is the least or lowest number which is exactly divided by them.

Methods of Calculation of LCM.

1. Prime factorization method

Step 1. Resolve the given numbers in the prime factors.
Step 2. Find the product of the highest powers of all the factors that occur in the given number.
Step 3. This product in terms of the highest powers of all factors is the required LCM.

Ex. Find the LCM of 8,12 and 15.

Sol. Factors of 8 = 2*2*2 = 2³

Factor of 12 = 2*2*3 = 2² * 3

Factor of 15 = 3*5 

Required LCM = 2³ * 3¹ * 5¹ = 8*3*5 = 120

2. Division method.

Step 1. Write down the given numbers in a line, separating them by commas.
Step 2. Divide by any one of the prime numbers which are exactly divided at least any two of the given numbers.
Step 3. Set down the quotients and the divided numbers in a line below 1st.
Step 4. Repeat the process until you get a line of numbers that are prime to one another.
Step 5. The product of all the divisors and the numbers in the last line will be the required LCM.

Ex. Find the LCM of 15,24,32 and 45.

Sol.

2	15,24,32,45
2	15,12,16,45
2	15,6,8,45
3	5,1,4,15
5	1,1,4,3

HCF and LCM of fractions:

HCF of fraction = HCF of Numerators / LCM of Denominator
LCM of fraction = LCM of Numerators / HCF of Denominator

Point to Remember:

i.Product of two numbers = product of their HCF and LCM.

ii. The Greatest number divides the numbers x,y, and z leaving the remaining a,b, and c, respectively. HCF of x-a, x-b, x-c.

iii. The greatest number that will divide x, y, and z leaving the same remainder in each case is given by  HCF of x-y, y-z, z-c.

iv. When the HCF of each pair of n given numbers is a and their LCM is b then the product of these numbers is given by (a)^n-1 * b or (HCF)^n-1 * LCM.

v.On dividing a number by a, b, and c If we get a-k, b-k, and c-k respectively then that number will be n * LCM of [a,b,c]-k

vi. On dividing a number by a, b and c if we get k as remainder always, then that number will be n * 

LCM of [a,b,c]+k.

vii. If a number after adding k is exactly divisible by a, b, and c then that number will be n* LCM [a,b,c]– k

Ex.1.On dividing a number by 5,6,7 we get 3,4,5 as the remainder. Find the number.

Sol. Now the number will be n * LCM of [a,b,c]-k

5-2= 3

6-2=4

7-2=5

k = 2

the number will be  n * LCM of [5,6,7]-2

210-2 = 208 Ans.

Here we take n =1

Ex.2 On dividing a number by 5,6,7 if we get 2 as remainder always, find that number.

Sol. That number will be n * LCM of [a,b,c]+k.

 n *  LCM of [5,6,7]+2.

210+2 = 212 Ans,

Ex.3 Find a number which after adding 7 is divisible by 10,11 and 12.

Sol.That number will be n* LCM [a,b,c]– k

n* LCM [10,11,12]– 7

If n=1 then

660-7 = 653 Ans.

Ex 4. What is the greatest number that will divide 2400 and 1810 and leave remainders 6 and 4 respectively?

Sol. Since on dividing 2400 a remainder 6 is left, 

The required number must divide (2400-6) or 2394.

similarly, it must divide 1810-7 or 1806 exactly.

Hence the greatest required number should be the HCF of 2394 and 1806 i.e 42 Ans.

Ex.5 Find the greatest number which will divide 410,751 and 1030 to leave the remainder 7 in each case.

Sol. the required greatest number = HCF of 410-7, 751-7, and 1030-7

31 Ans.

Ex.6 The numbers 11284 and 7655, when divided by a certain number of three digits, leave the same remainder. Find that number of three-digit.

Sol. The required number must be a factor of 11284-7655 = 3629

Now 3629 = 19*191

191 is the required number

Ex 7. LCM of two numbers is 2079 and their  HCF is 27. If one of the numbers is 189. the other number is.

Sol. HCF * LCM = n1 *n2

 

2079 * 27 = 189 * x

x = (2079*27)/189

x=297 Ans.

Ex.8.The product of two numbers is 2160 and their HCF is 12. The number of such possible pairs is.

Sol. HCF = 12, let the pair of numbers is 12x and 12y

12x * 12y = 2160

xy = 15

15 = 3 * 5, 1* 15

the pairs will be (36,60) and (12,80)