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                                       LCM and HCF Concept

Factor:- If a number ‘a’ divides another number ‘b’ exactly then ‘a’ is called a factor of b.
Multiple:- If a number ‘b’ is exactly divisible by another number ‘a’ then ‘b’ is called multiple of a.
Ex: Consider two numbers 6 & 24
6 is a factor of 24
& is a multiple of 6.

HCF or GCD:-  The HCF of two or more numbers is the greatest number that divides eah of them exactly.

HCF Highest common factorGCD Greatest common divisorSame

Ex: Find the HCF of 24, 363, 60.
The number which divides these numbers exactly is 12.
∴  HCF ⟶ 12      Answer
There are three methods to find HCF of given set of numbers:
(i). Factorization Method
(ii).
Division Method
(iii).
Difference Method
● Rules of divisibility are very helpful in finding the HCF or finding the answer using given options in competitive exams. So learn them properly.

(i). Factorization Method:- Express each of the given number as the product of prime factors. Product of least powers of common prime factors gives HCF.
Ex: Find HCF of 24, 64.

24126312222643216842222212224 = 2×364 = 2

∴ HCF = 2³ = 8  Answer

Ex: Find HCF of 96, 232

964824122222232116582912222996 = 2×3232 = 2×2963331

∴ HCF = 2³ = 8          Answer

Ex: Find HCF of 390, 1001

39019565132351323211658291222291001 = 7×11×131390 = 2×3×5×13

∴ HCF = 13          Answer

Ex: Find HCF of 151200, 65824, 19008

15200756003780018900222265824329121645682284114222215200 = 2×11×179450151200 = 2×3×5×747251575525175357123335572111117205718717119008950447522376118859429799331112222223331119008 = 2×3×11

∴ HCF = 2⁵ = 32        Answer

(ii ) Division Method:- Suppose we have to find the HCF of two given numbers. Divide the larger number by the smaller one. Now divide the divisor by the remainder. Repeat this recursive process of dividing the prededing by the remainder last obtained till zero is obtained as the remainder. The last divisor is required HCF.
Finding the HCF of more than two numbers: Suppose we are to find the HCF of three numbers then HCF of [(HCF of any two numbers) and (the third number)] gives the HCF of three given numbers.
Similarly, the HCF of more than three numbers may be obtained.

Ex: Find the HCF of 24, 64

24)64(248)162416(18)1616(20HCF∴ HCF = 8 Answer

Ex: Find the HCF of 96, 132

96)132(196)369672(224)3624(124HCF∴ HCF = 12 Answer12)24(20

Ex: Find HCF of 390, 1001

390)1001(1780)221390221(1169)221169(1HCF∴ HCF = 13 Answer52)156(1316913)1331(0

Ex: Find the HCF of 1512000, 65824, 19008

65824)15120001448128(63872)1(165824638721952)(3263872624641408)1952(11408)54414081088(2)320544320(1224)320224(196)224192(1)3232(1320HCF of 65824, 1512000
32)1900819008(5940now HCF of 32, 19008

∴ HCF of these numbers is 32     Answer

(iii) Difference Method:- From the given set of numbers take those two number whose difference is least. Let this difference be d.
The HCF is either ‘d’ or a number less than ‘d’ which is a multiple of ‘d’, which divides given numbers exactly.
Ex: Find HCF of 24, 64

24 644020108521No need to check these Since we already found HCFdivides 24 & 64 exactly

Ex: Find HCF of 96, 232

96 23213668348divides 96 & 232 exactly2×2×2×17HCF Answer

Ex: Find HCF of 390, 1001

390 10016114713divides 390 & 1001 exactly13×47HCF Answer

Ex: Find HCF of 1512000, 65824, 19008

65824 19008468162×2×2×2×2×7×11×19

By observation, Given numbers are multiple of 8 so HCF will also be multiple of 8, move further & we find HCF = 8×2×2 = 32               Answer

Ex: Find the HCF of 184, 230, 276

184 230462×23232divides all three exactlyHCF Answer

Ex: Find the HCF of 216, 423, 1215, 1422, 2169, 2223
Take two numbers which have least difference

2169 2223542×3×3×3279HCF Answer

Since sum of digits of each number is divisible by 9 so each number will also be divisible by 9.
Ex: Find HCF of 960 and 1020 

960 1020602×2×3×530HCF AnswerNo need to further calculate

\;\;\;● The HCF of two or more than two numbers is smaller than or equal to the smallest number of given numbers.
● The greatest number which divides a, b and c to leave the remainder R is HCF of (a –

R), (b – R) and (c – R)
The greatest number which divides x, y and z to leave remainders a, b and c respectively is HCF of (x – a), (y – b) and (z – c)
● The HCF of two or more prime numbers will always be 1.
● HCF of co-prime numbers is 1.
● HCF of fraction = \(\frac{{HCF\;of\;Numerator}}{{HCF\;of\;Denominator}}))
Ex: HCF of \(\frac{4}{9}\) and \(\frac{{12}}{{15}}\)
= \(\frac{{HCF\;of(4,12)}}{{LCM\;of(9,15)}}\) = \(\frac{4}{{45}}\)
Ex: Find the HCF of \(\frac{{12}}{{25}}\), \(\frac{{9}}{{10}}\), \(\frac{{18}}{{35}}\), \(\frac{{21}}{{40}}\)

12 = 2×2×39 = 3×318 = 2×3×321 = 3×725 = 5×510 = 2×535 = 5×740 = 2×2×2×5

LCM:- The least number which is exactly divisible by each one of the given numbers is called their LCM.
LCM ⟶ Least common multiple
Ex: Find the LCM of 25, 10, 35, 40

25 5 × 510 2 × 535 5 × 740 2×2×2×5×1400LCM

Ex: LCM of 16, 20

16 2 × 2 ×2 × 220 2 × 2× 5×80LCM

Ex: LCM of 10, 18 and 20

10 2 × 518 2 × 3 × 320 2 × 2 × 5 ×180LCM

● LCM by division method:-
⟶ First, write the numbers and separate them with commas
⟶ Now divide the numbers by the smallest prime number
⟶ If any number is not divisible then write down that number and proceed further.
⟶ Keep on dividing the row of numbers by prime numbers unless we get the results as 1 in the complete row.
⟶ Now LCM of the numbers will be equal to the product of all the prime numbers we obtained in the division method.
Ex: LCM of 14, 20, 30, 21 

2233510 , 18 , 205 9 105 9 55 3 55 1 51 1 1

∴ LCM = 2 × 2 × 3 × 3 × 5 = 180      Answer

Ex: LCM of 14, 20, 30, 21

2235714 , 20 , 30 217 10 15 217 5 15 217 5 5 77 1 1 71 1 1 1

∴ LCM = 2 × 2 × 3 × 5 × 7 = 420     Answer

Factor Method:14 = 2 × 720 = 2 × 530 = 2 × 3 × 521 = 3 × 7LCM = 2 × 3 × 5 × 7 = 140 Answermax power of eachprime number

● The LCM is always greater than or equal to the larger of th given numbers.
● If two numbers are co-prime, their LCM is equal to their product.
● The LCM of two or more numbers is divisible by each of the given numbers.
● If LCM is divisible by a number A then it will also be divisible by all the factors of number A.
● LCM of two distinct prime numbers is their product.
● The HCF of (Sum of Numbers and LCM of numbers) will also be the HCF of these numbers

Ex: The sum and LCM of two numbers are 204 & 864. Find the numbers

HCF of (204, 864) = HCF of numbers

204)864(481648)204(4192)124848(40HCF

∴ HCF of numbers will be 12
Let numbers are 12a & 12b where a & b are co-prime.
12a + 12b = 204
a + b = 17 ………………..(i)
& LCM = 12ab = 864 ⟹ ab = 72 ……………(ii)
a = 8     b = 9
∴ Numbers are : 
12 × 8 = 96
12 × 9 = 108          Answer

● Let two numbers a & b then
product of two numbers = HCF × LCM
● Product of ‘n’ numbers = (HCF)ⁿ⁻¹ × LCM
Ex: The LCM of 5 numbers is 240 and their HCF is 3. Find the multiplication of all the numbers.
Multiplication of all numbers = (HCF)⁵⁻¹ × LCM
= (3)⁴ × 240
19440         Answer
● LCM of fraction = \(\frac{{LCM\;of\;Numerator}}{{HCF\;of\;Denominator}}\)
Ex: Find LCM of \(\frac{{12}}{{25}}\), \(\frac{{20}}{{27}}\), \(\frac{{18}}{{25}}\)
\( = \frac{{LCM\;of\;(12,20,18)}}{{HCF\;of\;(25,27,25)}}\)
= \(\frac{{180}}{1}\)
= 180           Answer

(Q). Find LCM of 2.4, 0.36 and 7.2
Sol: 
To calculate the LCM of decimal integers, make the decimal integers into whole numbers multiplying with any friendly numbers i.e. 10, 100, 1000, ……….. etc.
Now find out LCM of these whole numbers.
Finally divide this LCM by that friendly number which we used earlier.

2.4 × 100 240 0.36 × 100 367.2 × 100 72024102×2×2×3= 2×3×52×52×2×3×3= 2×372102×2×2×3×325= 2×3×5

∴ LCM = 2⁴ × 3² × 5 = 720
∴ required LCM = \(\frac{{720}}{{100}}\) = 7.2           Answer

(Q). Find the HCF of 48, 168, 324 and 1400.
Sol:
Method(1):

222224824126312223716884422171223333241628127913322255140070035017535771

∴ LCM = 2⁴ × 3⁴ × 5² × 7 = 226800     Answer

Method(2):

4816832414002428 × 332= 2×3424276238423141002710102525= 2×3×7= 3×2= 2×5×7GCD = 2×3×5×7 = 226800 Answer

Method(3):

2222222222248 168 324 40024 84 162 70012 42 81 3506 21 81 1753 21 81 1751 7 27 1751 7 9 1751 7 3 1751 7 1 1751 7 1 351 7 1 71 1 1 1

LCM = 2⁴ × 3⁴ × 5² × 7 = 226800           Answer

(Q). Find the HCF of 48, 168, 324, 1400
Sol:
Method(1)

222224824126312223716884422171223333241628127913322255140070035017535771

2² is common in all
∴ HCF = 2² = 4           Answer

Method(2):

4816832414002428 × 332= 2×3424276238423141002710102525= 2×3×7= 3×2= 2×5×7HCF = 2 = 4 Answer
Method(3):48)(1681443)24(48482024)32424(138472)1224(242012)1400(116122 0128 0728)(1284)8(8012HCFAnswer
Method (4):48 168 324 1400120156362×2×3×3181294divides allHCF AnswerNote: This method is useful when options are given.

(Q). Find the least number which when divided by 48, 18, 42, 72 leaves the remainder 41, 11, 35, 65 respectively.
Sol: 
In this type of question the difference between the divisor & the corresponding remainder is same in each case
48 – 41 = 17,     18 – 11 = 7,     42 – 35 = 7,     72 – 65 = 7
∴ required number = LCM of (48, 18, 42, 72) – 5

481842722323237418232

∴ LCM = 2⁴ × 3² × 7 = 1008
∴ required LCM = 1008 – 7 = 1001      Answer
● If two numbers have the same remainder when divided by a divisor, then their difference is divisible by that divisor.

(Q). Find the greatest number which is such that when 12288, 19139 and 28200 are divided by it, the remainder are all same.
Sol:
In this type of question, first find out the difference of these numbers & then find HCF of the numbers found out in differences

12288282001913968519061Now find the HCF of (6851, 9061) which will be the required numberTo find HCF of (6851, 9061)6851)(906168512210)(6851663013211)22102210(100Answe HCF