HCF and LCM is another very important topic from the number system. The concept is not just restricted to the number system but is also helpful in solving some questions from arithmetic which is a very important topic for competitive exams.
In this article, we will take the topic from basics and will try to understand the various types of problems which require HCF and LCM concepts. Many questions on HCF and LCM from competitive exams are solved by applying direct formulas and tricks.
The H.C.F. of two or more than two numbers is the greatest number that divides each of them exactly.
Finding the GCD / H.C.F.
HCF of 4 and 6 via diagram is shown below:
Example: Find the GCD of 150, 210, 375.
Sol:
 Step 1: Writing down the standard form of numbers
⇨ 150 = 5 X 5 X 3 X 2
⇨ 210 = 5 X 2 X 7 X 3
⇨ 375 = 5 X 5 X 5 X 3
 Step 2: Writing Prime factors common to all the three numbers is 5 X 3.
 Step 3: Hence, the HCF will be 5 X 3 = 15.
Express each number as the product of primes and take the product of the least powers of common factors to get the H.C.F.
Steps to solve:
Example: Evaluate the HCF of 60 and 75.
Sol: Write each number as a product of its prime factors.
2^{2 }x 3 x 5 = 60
3 x 5^{2 }= 75
The product of all common prime factors is the HCF.
The common prime factors in this example are 3 & 5.
The lowest power of 3 is 3 and 5 is 5.
So, HCF = 3 x 5 = 15
Example: Find out the HCF of 36 and 48.
Sol:
Step I:Here we need to divide 48 by 36. ie. Dividend = 48 and Divisor = 36
[Divide the larger number by the smaller one].
Step II: Divide the 2 numbers
Step III: When 12 becomes divisor, remainder becomes 0. Therefore, highest common factor = 12.
[The last divisor is the required highest common factor (H.C.F) of the given numbers].
Prime factorization is the process of breaking down a composite number into its prime factors, which are the prime numbers that multiply together to give the original number.
Steps to solve:
Example: Using prime factorization method, find HCF of 18 and 90
Ans: Prime factorization of 18 is given below:
Prime factorization of 90 is given below:
There are 6 common factors of 18 and 90, that are 1, 2, 3, 6, 9, and 18. Therefore, the greatest common factor of 18 and 90 is 18.
Suppose you were required to find the HCF of 39,78 and 195:
Example: The sides of a hexagonal field are 216, 423, 1215, 1422, 2169 and 2223 meters . Find the greatest length of tape that would be able to exactly measure each of these sides without having to use fractions/parts of the tape?
Sol: Before solving this example, remind yourself with the above short trick you read above.
⇨ In this question, we are required to identify the HCF of the numbers 216, 423,1215, 1422, 2169 and 2223.
⇨ In order to do that, we first find the smallest difference between any two of these numbers. It can be seen that the difference between 22232169 = 54.Thus, the required HCF would be a factor of the number 54. The factors of 54 are: 1 × 54, 2 × 27, 3 X 18, 6 X 9
⇨ One of these 8 numbers has to be the HCF of the 6 numbers. 54 cannot be the HCF because the numbers 423 and 2223 being odd numbers would not be divisible by any even number. Thus, we do not need to check any even numbers in the list.
⇨ 27 does not divide 423 and hence cannot be the HCF. 18 can be skipped as it is even.
⇨ Checking for 9: 9 divides 216,423,1215,1422 and 2169.
⇨ Hence, it would become the HCF. (Note: we do not need to check 2223 once we know that 2169 is divisible by 9)
LCM stands for Lowest or Least Common Multiple. The LCM of two or more numbers is the smallest positive integer that is divisible by all the given numbers.
Finding the LCM two numbers E and R
LCM of 4 and 6 via diagram is shown below:
Example : Find the LCM of 150, 210, 375.
Sol:
 Step 1: Writing down the standard form of numbers:
⇨ 150 = 5 × 5 × 3 × 2 = 5^{2 }× 3 × 2
⇨ 210 = 5 × 2 × 7 × 3
⇨ 375 = 5 × 5 × 5 × 3 = 5^{3 }× 3
 Step 2: Write down all the prime factors: that appear at least once in any of the numbers: 5, 3, 2, 7.
 Step 3: Raise each of the prime factors to their highest available power (considering each to the numbers).
The LCM = 2 × 3 × 5 × 5 × 5 × 7 = 5250.
Steps involved:
Example: Find the least common multiple (LCM) of 60 and 90 using prime factorization.
Sol: Let us find the LCM of 60 and 90 using the prime factorization method.
 Step 1: The prime factorization of 60 and 90 are: 60 = 2 × 2 × 3 × 5 and 90 = 2 × 3 × 3 × 5
 Step 2: If we write these prime factors in their exponent form it will be expressed as, 60 = 2^{2} × 3^{1} × 5^{1} and 90 = 2^{1} × 3^{2} × 5^{1}
 Step 3: Now, we will find the product of only those factors that have the highest powers among these. This will be, 2^{2} × 3^{2} × 5^{1} = 4 × 9 × 5 = 180
=>LCM(60,90) = 180
Steps involved:
Example : Find the L.C.M. of 72, 240, 196.
Sol: (i) Using Prime Factorisation method:
⇨ 72 = 2 × 2 × 2 × 3 × 3 = 2^{3 }× 3^{2}
⇨ 240 = 2 × 2 × 2 × 2 × 3 × 5 = 2^{4 }× 3 × 5
⇨ 196 = 2 × 2 × 7 × 7 = 2^{2 }× 7^{2}L.C.M. of the given numbers = Product of all the prime factors of each of the given number with greatest index of common prime factors
= 2^{4 }× 3^{2 }× 5 × 7^{2} = 16 × 9 × 5 × 49 = 35280.(ii) Using the Division method:
2  72, 240, 196
2  36, 120, 98
2  18, 60 , 49
3  9 , 30 , 493  3 , 10 , 49
7  1 , 10 , 49
7  1 , 10 , 1
10  1 , 10 , 1
 1 , 1 , 1
L.C.M. of the given numbers:
= Product of divisors and the remaining numbers
= 2 × 2 × 2 × 3 × 3 × 10 × 49
= 35280
Coprimes: Two numbers are said to be coprimes if their H.C.F. is 1.
Let us assume a and b are the two numbers, then the formula that expresses the relationship between their LCM and HCF is given as:
Product of Two numbers = (HCF of the two numbers) x (LCM of the two numbers) GCD (P, Q) × LCM (P, Q) = P × Q 
Note: This rule is applicable only for two numbers.
In given numbers, make the same number of decimal places by annexing zeros in some numbers, if necessary. Considering these numbers without a decimal point, find H.C.F. or L.C.M. as the case may be. Now, in the result, mark off as many decimal places as are there in each of the given numbers.
Example: Find the HCF and LCM of 3, 2.7, 0.09
Sol:
 Step1: Write all the numbers with same number of digits after decimal point.
3.00, 2.70, 0.09 Step2: Now count the number of digits after decimal point (value is 2 for above problem) and calculate 10 power of the obtained value. Let the number be n = 10^{2} = 100.
Step3: Now remove the decimal point and find the LCM and HCF of the numbers.
LCM(300, 270, 9) and HCF(300, 270, 9). 300 = 22 x 31 x 52
270 = 21 x 33 x 51
9 = 20 x 32 x 50
LCM(300, 270, 9) = 22 x 33 x 52 = 2700
HCF(300, 270, 9) = 20 x 31 x 50 = 3
For finding the LCM and HCF, we should write the number in the power of prime numbers as written above. We should ensure that all the numbers should be written as power of prime numbers of same number. Example : 9 can be written as 3^{2} but the other two numbers also contain 2 and 5 as primes. So we can write other two numbers as powers of 0. So 9 can be written as 2^{0} x 3^{2} x 5^{0} and it won’t change the value of the number. Since we have our numbers in form of power of primes, Now the LCM is the number formed as the product of primes with its power is maximum value of the power of the same prime in given numbers.
 LCM = 2 power of max(2, 1, 0) x 3 power of max(1, 3, 2) x 5 power of max(2, 1, 0) = 22 x 33 x 52 = 2700 .
HCF calculation is similar but with only one change. Instead of taking max in power we take min in power. HCF = 2 power of min(2, 1, 0) x 3 power of min(1, 3, 2) x 5 power of min(2, 1, 0) = 20 x 31 x 50 = 3
 Step4: Now divide the obtained answer with our number n in step 2. The value we obtain is our required answer.
LCM(3, 2.7, 9) = 2700/100 = 27
HCF(3, 2.7, 9) = 3/100 = 0.03
Example: Find the H.C.F. and the L.C.M. of 1.20 and 22.5
Sol: Converting each of the following decimals into like decimals we get; 1.20 and 22.50
Now, expressing each of the numbers without the decimals as the product of primes we get
120 = 2 × 2 × 2 × 3 × 5 = 2^{3} × 3 × 5
2250 = 2 × 3 × 3 × 5 × 5 × 5 = 2 × 3^{2} × 5^{3}
Now, H.C.F. of 120 and 2250 = 2 × 3 × 5 = 30
Therefore, the H.C.F. of 1.20 and 22.5 = 0.30 (taking 2 decimal places)
L.C.M. of 120 and 2250 = 2^{3} × 3^{2} × 5^{3} = 9000
Therefore, L.C.M. of 1.20 and 22.5 = 90.00 (taking 2 decimal places)
Example: Find the H.C.F. and the L.C.M. of 0.48, 0.72 and 0.108
Sol: Converting each of the following decimals into like decimals we get;
0.480, 0.720 and 0.108
Now, expressing each of the numbers without the decimals as the product of primes we get
480 = 2 × 2 × 2 × 2 × 2 × 3 × 5 = 2^{5} × 3 × 5
720 = 2 × 2 × 2 × 2 × 3 × 3 × 5 = 2^{4} × 3^{2} × 5
108 = 2 × 2 × 3 × 3 × 3 = 2^{2} × 3^{3}
Now, H.C.F. of 480, 720 and 108 = 2^{2} × 3 = 12
Therefore, the H.C.F. of 0.48, 0.72 and 0.108 = 0.012 (taking 3 decimal places)
L.C.M. of 480, 720 and 108 = 2^{5} × 3^{3} × 5 = 4320
Therefore, L.C.M. of 0.48, 0.72, 0.108 = 4.32 (taking 3 decimal places)
To calculate the Highest Common Factor (HCF) and Lowest Common Multiple (LCM) of fractions, find the HCF and LCM of their numerators and denominators separately.
(a) HCF: To calculate the Highest Common Factor (HCF) of fractions, find the HCF of their numerators and denominators individually.
With fractions 3/6 and 5/15, the HCF of the numerators (3 and 5) is 1, and the HCF of the denominators (6 and 15) is 3. Thus, the HCF of the fractions is 1/3.
(b) LCM: To calculate the Lowest Common Multiple (LCM) of fractions, determine the LCM of their numerators and denominators separately.
The LCM of the numerators (3 and 5) is 15, and the LCM of the denominators (6 and 15) is 30. Consequently, the LCM of the fractions is 15/30, which simplifies to 1/2.
The product of LCM and HCF of any two given natural numbers is equivalent to the product of the given numbers. LCM × HCF = Product of the Numbers.
Suppose A and B are two numbers, then.
LCM (A & B) × HCF (A & B) = A × B
For Example: If 3 and 8 are two numbers.
LCM (3,8) = 24
HCF (3,8) = 1
LCM (3,8) x HCF (3,8) = 24 x 1 = 24
Also, 3 x 8 = 24
Hence, proved.
Note: This property is applicable for only two numbers.
HCF of coprime numbers is 1.
Therefore, the LCM of given coprime numbers is equal to the product of the numbers.
LCM of Coprime Numbers = Product Of The Numbers
For Example: Let us take two coprime numbers, such as 21 and 22.
LCM of 21 and 22 = 462
Product of 21 and 22 = 462
LCM (21, 22) = 21 x 22
H.C.F. and L.C.M. of Fractions
LCM of fractions = LCM of Numerators / HCF of Denominators.
HCF of fractions = HCF of Numerators / LCM of Denominators.
For Example: Let us take two fractions 4/9 and 6/21. 4 and 6 are the numerators & 9 and 12 are the denominators
LCM (4, 6) = 12
HCF (4, 6) = 2
LCM (9, 21) = 63
HCF (9, 21) = 3
Now as per the formula, we can write:
LCM (4/9, 6/21) = 12/3 = 4. Then HCF (4/9, 6/21) = 2/63
HCF of any two or more numbers is never greater than any of the given numbers.
For Example: HCF of 4 and 8 is 4. Here, one number is 4 itself and another number 8 is greater than HCF (4, 8), i.e.,4.
LCM of any two or more numbers is never smaller than any of the given numbers.
For Example: LCM of 4 and 8 is 8 which is not smaller to any of them.
There are generally four types of questions on finding remainders that appear in the exam; these require the concepts of HCF and LCM.
Let us understand these formulas and their working with the help of examples, one on each type.
The greatest number that will divide p, q and r leaving the same remainder in each case, then required number = HCF of the absolute values of (pq), (qr), and (rp).
Let us understand this with the help of following example
Example: Find the greatest number that will divide 65, 81, and 145 leaving the same remainder in each case.
Sol: Required number = HCF of (8165), (14581), and (14565)
= HCF of 16, 64, and 80 = 16.
Direct Formula: HCF = a^{HCF(m,n)} 1
Question: Find the HCF of 2^{120}1 and 2^{50}1
Sol: HCF of (120, 50) = 10.
On applying the direct formula, we get the required HCF = 2^{10}1.
Example 1: How many pairs of integers (x, y) exist such that the product of x, y and HCF (x, y) = 1080?
a. 8
b. 7
c. 9
d. 12
Sol: Option 3
We need to find ordered pairs (x, y) such that xy * HCF(x, y) = 1080.
Let x = ha and y = hb where h = HCF(x, y) => HCF(a, b) = 1.
So h^{3}(ab) = 1080 = (2^{3})(3^{3})(5).
We need to write 1080 as a product of a perfect cube and another number.
Four cases:
1. h = 1, ab = 1080 and b are coprime. We gave 4 pairs of 8 ordered pairs (1, 1080), (8, 135), (27, 40) and (5, 216). (Essentially we are finding coprime a,b such that a*b = 1080).
2. h = 2, We need to find a number of ways of writing (3^{3}) * (5) as a product of two coprime numbers. This can be done in two ways  1 and (3^{3}) * (5) , (3^{3}) and (5)
number of pairs = 2, number of ordered pairs = 4
3. h = 3, number of pairs = 2, number of ordered pairs = 4
4. h = 6, number of pairs = 1, number of ordered pairs = 2
Hence total pairs of (x, y) = 9, total number of ordered pairs = 18.
The pairs are (1, 1080), (8, 135), (27, 40), (5, 216), (2, 270), (10, 54), (3, 120), (24, 15) and (6, 30).Hence the answer is "9"
Example 2: Find the smallest number that leaves a remainder of 4 on division by 5, 5 on division by 6, 6 on division by 7, 7 on division by 8 and 8 on division by 9?
a. 2519
b. 5039
c. 1079
d. 979
Sol: Option 'a' is correct
Explanation: When a number is divided by 8, a remainder of 7 can be thought of as a remainder of 1. This idea is very useful in a bunch of questions. So, N = 5a  1 or N + 1 = 5a
N = 6b  1 or N + 1 = 6b
N = 7c  1 or N + 1 = 7c
N = 8d  1 or N + 1 = 8d
N = 9e  1 or N + 1 = 9e
N + 1 can be expressed as a multiple of (5, 6, 7, 8, 9)
N + 1 = 5a*6b*7c*8d*9e
Or N = (5a*6b*7c*8d*9e)  1
Smallest value of N will be when we find the smallest common multiple of (5, 6, 7, 8, 9)
or LCM of (5, 6, 7, 8, 9)
N = LCM (5, 6, 7, 8, 9)  1 = 2520  1 = 2519.
Hence the answer is "2519"
Example 3: There are three numbers a,b, c such that HCF (a, b) = l, HCF (b, c) = m and HCF (c, a) = n. HCF (l, m) = HCF (l, n) = HCF (n, m) = 1. Find LCM of a, b, c. (The answer can be "This cannot be determined").
Sol: It is vital not to be intimidated by questions that have a lot of variables in them.
 a is a multiple of l and n. Also, HCF (l,n) =1; => a has to be a multiple of ln, similarly, b has to be a multiple of lm and c has to be a multiple of mn.
 We can assume, a = lnx, b = lmy, c = mnz.
Now given that HCF(a, b) = l, that means HCF(nx, my) = 1. This implies HCF(x, y) = 1 and HCF(m, x) = HCF(n, y) = 1. Similarly, it can also be shown that HCF(y, z) = HCF(z, x) = 1 and others also.
 So, in general, it can be written any two of the set {l, m, n, x, y, z} are coprime.
Now LCM(a, b, c) = LCM (lnx, lmy, mnz) = lmnxyz = abc/lmn. Quite obviously, it is a reasonable assumption that a question in CAT will not be as tough as the last one here. However, it is a good question to get an idea of the properties of LCM and HCF.
Hence the answer is "Cannot be determined"
Example 4: Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together?
Solution: In this question, we have to find the least number which is divisible by 2, 4, 6, 8 10 and 12. and that number has to be the LCM of the given six numbers.
LCM (2, 4, 6, 8, 10, 12) = 120.
That means the first time all six Bells toll together is 120 seconds or 2 minutes.
Therefore, in 30 minutes they will toll =30/2=15 times.
Example 5: There are two clocks, one beat 96 times in 5 minute and the other beat 48 times in 7 minutes. If they beat together exactly at 10 am when do they next beat together?
Solution: The time for each beat is 5/96 minutes and 7/48 minutes, or 5/96 minutes and 14/96 minutes.
Hence, they will next beat together at 35/48 minutes past 10 am.
The LCM of numerators = 70
The HCF of denominators = 90
Therefore, the LCM of the fraction =
Example 6: Rakesh and Brijesh alone can do a work in 12 days and 15 days respectively. In how many days they can complete the work if they work together?
Solution: Let us assume that the work = LCM (12, 15) = 60 units.
Units of work Rakesh can do in one day = 60/12=5 units
Similarly, units of work Brijesh can do in one day = 60/15 = 4 units.
Therefore, together in one day, they can complete 5+4=9 units of work.
Hence, the total time taken by then to complete the work = 60/9 = 20/3 days.
Example 7: Sarita and Namita start running simultaneously from the same point on a circular track of length 120 meters at the speed of 10 m/s and 16 m/s respectively. In how much time they will again be together at the starting point?
Solution: Time taken by Sarita to complete one round = 120/10 = 12 sec
Similarly, the time taken by Namita to complete one round = 120/16 = 15/2 sec
Therefore, time after which both will be again at the starting point = LCM (12, 15/2) = 60 sec = 1 minute
215 videos139 docs151 tests

1. What is the definition of Highest Common Factor (H.C.F.)? 
2. What are the different methods to find the HCF of two numbers? 
3. How do you calculate the Least Common Multiple (L.C.M.)? 
4. What is the relationship between HCF and LCM of two numbers? 
5. How do you find the H.C.F. and L.C.M. of fractions? 

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