HCF & LCM
Important Facts And Formulae
1. FACTORS AND MULTIPLES
If a divides b exactly, then we say that a is a factor of b and that b is a multiple of a. Ex. (i) 2 is a factor of 8—> 8 is a multiple of 2. (ii) 5 is a factor of 15 —> 15 is a multiple of 5. 2. Product of two numbers =(Their H.C.F) (Their L.C.M) 3. Product of n numbers= L.C.M 4. COPRIMES : Two numbers are coprime if their H.C.F is 1. 5. H.C.F AND L.C.M OF FRACTIONS (i) H.C.F = (ii) L.C.M =

ILLUSTRATIVE EXAMPLES
1. The product of two numbers is 1280 and their H.C.F. is 8. The L.C.M of these numbers is: (SSC. 2007) 
Answer: Let the LCM be x. Then,
(H.C.F) (L.C.M) = product of two numbers
Therefore, L.C.M = = 160
2. The L.C.M of two numbers is 12 times their H.C.F. The sum of H.C.F and L.C.M is 403. If one of the numbers is 93, then the other number is: (SSC. 2008) 
Answer: Let H.C.F be x. Then, L.C.M=12x
Therefore, x + 12x = 403
13x = 403 x= 31
Therefore, H.C.F = 31 and L.C.M = (12 31) = 372
Let the other number be y,
Then, 31 372 = 93 y
y = = 124
Therefore, Other number = 124
3. The product of two numbers is 4107 and their H.C.F is 37. The larger number is: (SSC. 2008) 
Answer: Let the number be 37 a and 37 b, where a and b are coprimes.
Then, 37 a 37 b = 4107
ab = .
Let us take a = 1, b= 3. Then,
The numbers are 37 1 and 37 3 i.e. 37 and 111.
Hence, the larger number is 111.
4. The number nearest to 43582 divisible by each of 25, 50, 75 is : (SSC. 2007) 
Answer: We have
L.C.M. of 25, 50, 75 = (25 2 3 ) = 150
On dividing 43582 by 150 we get 82 as remainder.
Therefore, Required number = (43582 – 82) = 43500
5. The H.C.F of two numbers, each having 3digits is 17 and their L.C.M is 714. The sum of the numbers is : (SSC. 2007) 
Answer: Let the two numbers be 17 a and 17 b, where a and b are coprimes. Then,
L.C.M = 17ab
Therefore, 17 ab =714 ab =
The possible pairs of coprimes are (1, 42) , (2, 21) , (3 , 14) , (6, 7).
Possible pairs of numbers are:
(17 1,17 42), (17 2, 17 21), (17 3, 17 14) , (17 6, 17 7)
Out of these the 3digit numbers are 102 and 119 only. Their sum is 221.
6. The H.C.F of two numbers is 15 and their product is 6300. How many such pairs of numbers are there ? (SSC. 2008) 
Answer: Let the number be 15 a and 15 b , where a and b are coprimes.
Therefore, 15 a 15 b = 6300
ab =
Now, the pairs of two numbers which are relatively coprime and whose product is 28 are (1, 28) and (4, 7) only.
Thus, 2 such pairs exist.
7. The H.C.F and L.C.M of two numbers are 21 and 4641 respectively. If one of the numbers lies between 200 and 300, then the two numbers are: ? (M.B.A. 2006) 
Answer: Let the number be 21 a and 21 b, where a and b are coprimes. Then, (21 a 21 b) = (21 4641) ab=221
Two coprimes with product 221 are 13 and 17.
Therefore, The numbers are (21 13, 21 17 ) i.e. 273 and 357
8. The number nearest to 10000, which is exactly divisible by each of the numbers 3,4,5,6,7,8 is: (S.S.C. 2004) 
Answer: L.C.M of 3, 4, 5, 6, 7, 8 = (2 2 3 5 7 2) = 840
On dividing 10000 by 840, we get quotient = 11 and remainder = 760
Therefore, Required number will be [10000 + (840760)]
i.e., Nearest number is 10080
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