LCM and HCF
Basic Formulas of LCM and HCF with Definition
LCM stands for Least Common Factor
LCM or least common factor of two numbers 4, 6 is denoted as LCM(4, 6). And the LCM is the
smallest positive integer that is divisible by both 4 and 6, which is 12.
HCF stands for Highest Common Factor
Greatest Common Divisor or gcd of two or more positive integers is defined as the largest positive
integer that divides the numbers without leaving the remainder.
HCF in case of only two numbers:
Suppose the question asks to calculate the HCF of any two numbers. Then Divide the larger
number by the smaller one.
Now divide the divisor by the remainder.
Keep dividing the preceding divisor with the remainder until a Zero remainder is obtained.
The last divisor obtained will be the HCF of those two given number.
HCF in case of more than two numbers:
The HCF of these two numbers and the third number will be the HCF of all three numbers
and so forth.
Choose any two numbers and find their HCF by applying the above-mentioned method.
The rule for Solving HCF and LCM Questions and Answers:
Calculate the multiples or the factors of the larger number until one of them is also a
multiple of the smaller number.
Then multiply all these factors, and the resultant will be the LCM of the given numbers.
LCM and HCF Formulas
HCF and LCM FormulaProduct of Two numbers = (HCF of the two numbers) x (LCM of the two
numbers)
How to find HCFH.C.F. of Two numbers = Product of Two numbers/L.C.M of two numbers
How to find LCML.C.M of two numbers = Product of Two numbers/H.C.F. of Two numbers
HCF by Prime Factorization Method
Take an example of finding the highest common factor of 100, 125 and 180.
Now let us write the prime factors of 100, 125 and 180.
100 = 2 × 2 × 5 × 5
125 = 5 × 5 × 5
180 = 3 × 3 × 2 × 2 × 5
The common factors of 100, 125 and 180 are 5
Therefore, HCF (100, 125, 180) = 5
HCF by Division Method
Steps to find the HCF of any given numbers:
1. Larger number/ Smaller Number
2. The divisor of the above step / Remainder
3. The divisor of step 2 / remainder. Keep doing this step till R = 0(Zero).
4. The last step’s divisor will be HCF.
Example:
Let’s take two number 120 and 180
120) 180 (1
120
---------
60) 120 (2
120
---------
000
LCM by Prime Factorization Method
A technique to find the Least Common Multiple (LCM) of a set of numbers by breaking down each
number into its prime factors and then multiplying the highest powers of each prime factor.
Lets take two numbers i.e., 25 and 35, now to calculate the LCM:
List the prime factors of each number first.
25 = 5 × 5
35 = 7 × 5
Then multiply each factor the most number of times it occurs in any number.
If the same multiple occurs more than once in both the given numbers, then multiply the factor by
the most number of times it occurs.
The occurrence of Numbers in the above example:
5: two times
7: one time
LCM = 7 × 5 × 5 = 175
LCM by Division Method
Let us see with the same example, which we used to find the LCM using prime factorization.
Solve LCM of (25,35) by division method.
5 | 25, 35
----------
5 | 5, 7
---------
7 | 1, 7
---------
| 1, 1
Therefore, LCM of 25 and 35 = 5 x 5 × 7 = 175
Questions and Answers of HCF and LCM
Question:
Calculate the highest number that will divide 43, 91 and 183 and leaves the same remainder in
each case
Options
A. 4
B. 7
C. 9
D. 13
Solution:
Here the trick is :
1. Find the Differences between numbers
2. Get the HCF ( that differences)
We have here 43, 91 and 183
So differences are
183 – 91 = 92,
183 – 43 = 140,
91 – 43 = 48.
Now, HCF (48, 92 and 140)
48 = 2 × 2 × 2 × 2 × 3
92 = 2 × 2 × 23
140 = 2 × 2 × 5 × 7
HCF = 2 × 2 = 4
And 4 is the required number.
Correct Answer : A
Question:
Which of the following is greatest number of four digits which is divisible by 15, 25, 40 and 75 is:
Options
A. 9700
B. 9600
C. 9800
D. 9650
Solution: Greatest number of 4-digits is 9999.
Now , find the L.C.M. of 15, 25, 40 and 75 i.e. 600.
On dividing 9999 by 600, the remainder is 399.
Hence, Required number (9999 – 399) = 9600.
Alternatively,
99996006009999 = 16.66500
Ignore the decimal points, required number would be 16 * 600 = 9600
Correct Answer : B
Question:
The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12
m 95 cm is:
Options
A. 25 cm
B. 15 cm
C. 35 cm
D. 55 cm
Solution: Required length = H.C.F. of 700 cm, 385 cm and 1295 cm = 35 cm.
Correct Answer : C
How To Solve LCM and HCF Questions Quickly
As we know that HCF and LCM is the most underrated topic in mathematics. But it is important to
know that the topic consists multiple level of Questions. Let us review thoroughly How To Solve LCM
and HCF Quickly.
HCFThe highest Common Factor is also known as Greatest Common divisor i.e. the largest positive
integer that divides more than one integer is called as the greatest common divisor.
LCMLeast Common Multiple is also known as Smallest Common Multiple i.e. the smallest positive
integer that is divisible by more than one integer.
How to Solve HCF as well as LCM Questions Quickly
HCF – It is important to note that HCF of the given numbers cannot be greater than any one
of them.
LCM – It is important to note that LCM of the given numbers cannot be smaller than any one
of them.
If 1 is the HCF of 2 numbers, then their LCM will be their product.
For two co prime numbers, the HCF is always 1.
The most common methods to solve HCF and LCM easily is
Type 1: Find the greatest or smallest number.
Question 1. The greatest possible length which can be used to measure
exactly the lengths 5 m, 4 m, 12 m 55 cm is
Options
A. 2
B. 10
C. 25
D. 5
Solution: H.C.F. of (500 cm, 400 cm, 1255 cm) = 5 cm
The factors of 400 are: 1, 2, 4, 5, 8, 10, 16, 20, 25, 40, 50, 80, 100, 200, 400
The factors of 500 are: 1, 2, 4, 5, 10, 20, 25, 50, 100, 125, 250, 500
The factors of 1255 are: 1, 5, 251, 1255
Then the highest common factor is 5. (Since, only 5 is common in all three
Correct option: D
Question 2. The H.C.F. and L.C.M. of two numbers are 10 and 560
respectively. If one of the numbers is 70, find the other number?
Options
A. 80
B. 300
C. 308
D. 280
Solution: We know that, Product of two numbers = H.C.F x L.C.M
70 x a = 10 x 560
Therefore, a = 10×56017517510×560 = 560070705600 = 80
So, the other number = 80
Correct option: A
Question 3. Find the greatest number which on dividing 1484 and 2045
leaves remainders 4 and 5 respectively?
Options
A. 20
B. 30
C. 10
D. 40
Solution: Required number = H.C.F. of (1484 – 4) and (2045 – 5)
H.C.F. of 1480 and 2040
1480 = 2 x 2 x 2 × 5 × 37
2040 = 2 x 2 x 2 × 3 × 5 × 17
H.C.F 1480 & 2040 = 2 x 2 x 2 x 5=40
Correct option: D
Type 2: Find HCF
Question 1. Three numbers are in the ratio of 4: 3: 6 and their L.C.M. are
3600. Find their H.C.F:
Options
A. 350
B. 280
C. 250
D. 300
Solution: Let the numbers be 4x, 3x and 6x
Then, their L.C.M. = (4x * 3x *6x) = 12x
So, 12x = 3600 or x = 300
Therefore, numbers are (4 x 300), (3 x 300) and (6 x 300) = 1200, 900, 1800
Let us do the prime factorization to find HCF
1200=24∗3∗521200=24∗3∗52
900=22∗32∗52900=22∗32∗52
1800=23∗32∗521800=23∗32∗52
Hence, required H.C.F.
=22∗3∗52=300=22∗3∗52=300
Correct option: D
Question 2. Find the HCF of 34, 48, 56, and 74
Options
A. 2
B. 4
C. 34
D. 8
Solution: The factors of 34 are: 1, 2, 17, 34
The factors of 48 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
The factors of 56 are: 1, 2, 4, 7, 8, 14, 28, 56
The factors of 74 are: 1, 2, 37, 74
Therefore, the highest common factor is 2.
Correct option:A
Question 3. Find the HCF of 211,417,65.112,174,56.
Options
A. 19359351
B. 29359352
C. 293932
D. 235352
Solution: We know that
HCF = HCFofnumeratorLCMofDenominatorLCMofDenominatorHCFofnumerator
HCF = HCF(2,4,6)LCM(11,17,5)LCM(11,17,5)HCF(2,4,6)
HCF = 29359352
Correct option:B
Type 3: How to Solve when sum of two numbers is given ,
LCM and HCF is given to find the sum of reciprocals.
Question: Sum of two numbers is 55 and the H.C.F. and L.C.M. of these
numbers are 5 and 120 respectively, then the sum of the reciprocals of the
numbers is equals to:
Options
A. 1112012011
B. 1122022011
C. 2112012021
D. 1132032011
Solution : Let the numbers be a and b.
Now , given a+b = 55
a × b = HCF × LCM = 5 × 120
HCF × LCM = 600
Now, as we know that 1a+1b=a+ba×ba1+b1=a×ba+b
1a+1b=55600a1+b1=60055
1112012011
Correct Option : A
Type 4: How to Solve HCF, LCM Problems related to finding
the biggest container to measure quantities
Question : Suppose there are three different containers contain different
quantities of a mixture of milk and water whose measurements are 403
litres, 434 litres and 465 litres What biggest measure must be there to
measure all the different quantities exactly?
Options :
A. 31 litres
B. 21 litres
C. 41 litres
D. 30 litres
Solution : Prime factorization of 403,434 and 465 is
403=13×31
434=2×7×31
465=3×5×31
H.C.F of 403, 434 and 465=31
Correct Option : A
Type 5 : How to Solve HCF, LCM Problems related to Bell
ring.
Question: 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 ?
Options :
A. 8
B. 16
C. 9
D. 10
Solution : L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
Hence, the bells will toll together after every 120 seconds(2 minutes).
Therefore, in 30 minutes ,number of times bells toll together is 302+1230+1 = 16 (We
added 1 because in starting i.e. 0 mins all the bells would ring once together)
Correct Option B
Type 6 : How to Solve HCF, LCM Problems related to Circle
Based Runner Problem.
Question: Two people P and Q start running towards a circular track of
length 400 m in opposite directions with initial speeds of 10 m/s and 40 m/s
respectively. Whenever they meet, P’s speed doubles and Q’s speed
halves. After what time from the start will they meet for the third time?
Options
A. 30 seconds
B. 26 seconds
C. 10 seconds
D. 20 seconds
Solution : Time taken to meet for the 1st time= 40040+1040+10400=8 sec.
Now P’s speed = 20m/s and Q’s speed=20 m/s.
Time taken to meet for the 2nd time= 40020+2020+20400 = 10 sec.
Now P’s speed =40 m/sec and Q’s speed = 10 m/sec.
Time taken to meet for the 3rd time= 40010+4010+40400=8 sec.
Therefore, Total time= (8+10+8) = 26 seconds.
Correct Option B
HCF and LCM tricks, shortcuts, and tips
HCF and LCM shortcuts, tips, and tricks are not easy to find at the time of
examination. So we came up with a dedicated page to help students at the crucial
moment.
HCF and LCM tricksHCF is the greatest common divisor of more than one integer.
Hence, the largest positive integer that divides more than one integer is known as
the Highest common factor. LCM is the least common multiple of two or more
integers. Let us move forward and look up some of the Tips and Tricks Of HCF and
LCM.
Here, are some easy tips and tricks for you to solve HCF and LCM questions
quickly, easily , and efficiently in competitive exams.
HCF and LCM Tips and Tricks and Shortcuts
The H.C.F of two or more numbers is smaller than or equal to the smallest
number of given numbers
The smallest number which is exactly divisible by a, b and c is L.C.M of a, b, c.
The L.C.M of two or more numbers is greater than or equal to the greatest
number of given numbers.
The smallest number which when divided by a, b and c leaves a remainder R
in each case. Required number = (L.C.M of a, b, c) + R
The greatest number which divides a, b and c to leave the remainder R
is H.C.F of (a – R), (b – R) and (c – R)
The greatest number which divide x, y, z to leave remainders a, b, c is H.C.F
of (x – a), (y – b) and (z – c)
The smallest number which when divided by x, y and z leaves remainder of a,
b, c (x – a), (y – b), (z – c) are multiples of M
Required number = (L.C.M of x, y and z) – M
Type 1: Tips and Tricks and Shortcuts to find the greatest or smallest
number
Question 1. Find the greatest 5 digit number divisible by 5, 15, 20, and 25
Options
A. 99900
B. 99000
C. 99990
D. 90990
Solution: LCM of 5, 15, 20, and 25 is 300
The greatest 5 digit number is 99999
9999930030099999 = 333.33
Therefore, the highest 5 digit number divisible by 300 would be 333 * 300 = 99900
Correct option: A
Type 2: Find the numbers, sum of numbers, product of numbers if
Their ratio and H.C.F. are given.
Product of H.C.F. and L.C.M are given
Question 2. The product of two numbers is 3888. If the H.C.F. of these
numbers is 36, then the greater number is:
Options
A. 110
B. 108
C. 36
D. 120
Solution: Let the two numbers be 36x and 36y
Now, 36x * 36y = 3888
xy =388836×3636×363888
xy = 3
Now, co-primes with product 3 are (1, 3).
Therefore, the required numbers are 36 * 1 = 36
36 * 3 = 108
Therefore the greatest number is 108
Correct option: B
Type 3: Tips , Tricks and Shortcuts when sum of two numbers is given ,
LCM and HCF is given to find the sum of reciprocals.
Question: Sum of two numbers is 60 and the H.C.F. and L.C.M. of these
numbers are 5 and 100 respectively, then the sum of the reciprocals of the
numbers is equals to:
Options
A. 325253
B. 1122022011
C. 2112012021
D. 1132032011
Solution : Let the numbers be a and b.
Now , given a+b = 60
a × b = HCF × LCM = 5 × 100
= 500
1a+1b=a+ba×ba1+b1=a×ba+b
1a+1b=60500a1+b1=50060
325253
Correct Option : A
Type 4: How to Solve HCF, LCM Problems related to finding the biggest
container to measure quantities
Question : Suppose there are three different containers contain different
quantities of a mixture of Sugar and rice whose measurements are 403
grams, 434 grams and 465 grams What biggest measure must be there
to measure all the different quantities exactly?
Options :
A. 31 grams
B. 21 grams
C. 41 grams
D. 30 litres
Solution : Prime factorization of 403, 434 and 465 is
403=13×31
434=2×7×31
465=3×5×31
H.C.F of 403, 434 and 465=31
Correct Option : A
Type 5 :Tips , tricks and Shortcuts of HCF, LCM Problems related to Bell
ring.
Question: 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 ?
Options :
A. 8
B. 16
C. 9
D. 10
Solution : L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
Hence, the bells will toll together after every 120 seconds(2 minutes).
Therefore, in 30 minutes ,number of times bells toll together is 302+1230+1 = 16
Correct Option B
Type 6 : Tips , tricks and Shortcuts of HCF, LCM Problems related to Circle
Based Runner Problem.
Question: Two people P and Q start running towards a circular track of
length 400 m in opposite directions with initial speeds of 10 m/s and 40 m/s
respectively. Whenever they meet, P’s speed doubles and Q’s speed
halves. After what time from the start will they meet for the third time?
Options
A. 30 seconds
B. 26 seconds
C. 10 seconds
D. 20 seconds
Solution : Time taken to meet for the 1st time= 40040+1040+10400=8 sec.
Now P’s speed = 20m/s and Q’s speed=20 m/s.
Time taken to meet for the 2nd time= 40020+2020+20400 = 10 sec.
Now P’s speed =40 m/sec and Q’s speed = 10 m/sec.
Time taken to meet for the 3rd time= 40010+4010+40400=8 sec.
Therefore, Total time= (8+10+8) = 26 seconds.
Correct Option B
Tips and Tricks and Shortcuts to find HCF easily
Question. 3 Find HCF of 12 and 16.
Options
(A) 5
(B) 4
(C) 12
(D) 16
Solution Find the difference between 12 and 16. The difference is 4. Now, check
whether the numbers are divisible by the difference. 12 is divisible by 4 and 16 is
divisible by 4.Hence, the HCF is 4.
Correct Option B
Question. 4 Find HCF of 18 and 22.
Options
(A) 2
(B) 4
(C) 18
(D) 36
Solution Find the difference between 18 and 22. The difference is 4. Now, check
whether the numbers are divisible by the difference. Both 18 and 22 are not divisible
by 4. So take the factors of the difference. The factors of 4 are 2*2*1. Now, check
whether the numbers are divisible by the factors. 18 and 22 are divisible by factor 2.
Hence, the HCF is 2.
Note: If there are more than two numbers, take the least difference.
Correct Option(A)
Tips and Tricks and Shortcuts to find LCM easily
Question 5 Find LCM of 2,4,8,16.
Options
(A) 16
(B) 18
(C) 12
(D) 2
Solution Factorize of above number
2 =2
8 = 23
16 = 24
Choose the largest number. In this example, the largest number is 16. Check
whether 16 is divisible by all other remaining numbers. 16 is divisible by 2, 4, 8.
Hence, the LCM is 16.
Correct Option (A)
Question 6 Find the LCM of 2,3,7,21.
Options
(A) 21
(B) 44
(C) 36
(D) 42
Solution Choose the largest number. The largest number is 21. Check whether 21
is divisible by all other remaining numbers. 21 is divisible by 3 and 7 but not by 2. So
multiply 21 and 2. The result is 42. Now, check whether 42 is divisible by 2, 3, 7. Yes,
42 is divisible. Hence, the LCM is 42.
Correct Option (D)
