FACTS  AND  FORMULAE  FOR  HCF  AND  LCM  QUESTIONS

 

 

I.Factors and Multiples : If a number 'a' divides another number 'b' exactly, we say that 'a' is a factor of 'b'. In this case, b is called a multiple of a.

 

II.Highest Common Factor (H.C.F) or Greatest Common Measure (G.C.M) or Greatest Common Divisor (G.C.D) : The H.C.F. of two or more than two numbers is the greatest number that divides each of them exactly. There are two methods of finding the H.C.F. of a given set of numbers :

1. Factorization Method : Express each one of the given numbers as the product of prime factors.The product of least powers of common prime factors gives H.C.F.

2. Division Method : Suppose we have to find the H.C.F. of two given numbers. Divide the larger number by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is the required H.C.F.

Finding the H.C.F. of more than two numbers : Suppose we have to find the H.C.F. of three numbers. Then, H.C.F. of [(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given numbers. Similarly, the H.C.F. of more than three numbers may be obtained. 

 

III.Least Common Multiple (L.C.M) : The least number which is exactly divisible by each one of the given numbers is called their L.C.M.

1. Factorization Method of Finding L.C.M: Resolve each one of the given numbers into a product of prime factors. Then, L.C.M. is the product of highest powers of all the factors.

2. Common Division Method (Short-cut Method) of Finding L.C.M : Arrange the given numbers in a row in any order. Divide by a number which divides exactly at least two of the given numbers and carry forward the numbers which are not divisible. Repeat the above process till no two of the numbers are divisible by the same number except 1. The product of the divisors and the undivided numbers is the required L.C.M. of the given numbers.

 

IV. Product of two numbers = Product of their H.C.F and L.C.M

 

V. Co-primes : Two numbers are said to be co-primes if their H.C.F. is 1.

 

VI. H.C.F and L.C.M of Fractions :

1. H.C.F = H.C.F. of Numerators / L.C.M of Numerators

2. L.C.M = L.C.M of Numerators / H.C.F of Denominators

 

VII. H.C.F and L.C.M of Decimal Fractions : In given numbers, make the same number of decimal places by annexing zeros in some numbers, if necessary. Considering these numbers without 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.

 

VIII. Comparison of Fractions : Find the L.C.M. of the denominators of the given fractions. Convert each of the fractions into an equivalent fraction with L.C.M. as the denominator, by multiplying both the numerator and denominator by the same number. The resultant fraction with the greatest numerator is the greatest.

Q:

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:

A) 74 B) 94
C) 184 D) 364
 
Answer & Explanation Answer: D) 364

Explanation:

L.C.M. of 6, 9, 15 and 18 is 90.

 

Let required number be 90k + 4, which is multiple of 7.

 

Least value of k for which (90k + 4) is divisible by 7 is k = 4.

 

=>Required number = (90 x 4) + 4   = 364.

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88 51015
Q:

The G.C.D of 1.08, 0.36 and 0.9 is

A) 0.03 B) 0.9
C) 0.18 D) 0.108
 
Answer & Explanation Answer: C) 0.18

Explanation:

Given numbers are 1.08 , 0.36 and 0.90

 H.C.F of 108, 36 and 90 is 18                  [  G.C.D is nothing but H.C.F]

 Therefore, H.C.F of given numbers = 0.18                 

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138 49289
Q:

L.C.M of two prime numbers x and y (x>y) is 161. The value of 3y-x is :

A) -2 B) -1
C) 1 D) 2
 
Answer & Explanation Answer: A) -2

Explanation:

 H. C. F of two prime numbers is 1.  Product of numbers = 1 x 161 = 161.

 Let the numbers be a and b . Then , ab= 161.

 Now, co-primes with  product 161 are (1, 161) and (7, 23).

 Since x and y are prime numbers and x >y , we have x=23 and y=7.

 Therefore,  3y-x = (3 x 7)-23 = -2

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118 47954
Q:

The sum of two numbers is 528 and their H.C.F is 33. The number of pairs of numbers satisfying the above condition is

A) 4 B) 6
C) 8 D) 12
 
Answer & Explanation Answer: A) 4

Explanation:

Let the required numbers be 33a and 33b. 

 

Then 33a +33b= 528   =>   a+b = 16.

 

Now, co-primes with sum 16 are (1,15) , (3,13) , (5,11) and (7,9).

 

Therefore, Required numbers are  ( 33 x 1, 33 x 15), (33 x 3, 33 x 13), (33 x 5, 33 x 11), (33 x 7, 33 x 9)

 

The number of such pairs is 4

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128 47710
Q:

The product of two numbers is 4107. If the H.C.F of these numbers is 37, then the greater number is 

A) 101 B) 107
C) 111 D) 185
 
Answer & Explanation Answer: C) 111

Explanation:

 Let the numbers be 37a and 37b. Then , 37a x 37b =4107  => ab = 3.

 Now co-primes with product 3 are (1,3)

 So, the required numbers are (37 x 1, 37 x 3) i.e, (37,111).

 Therefore, Greater number = 111.

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64 46952
Q:

Product of two co-prime numbers is 117. Their L.C.M  should be

A) 1 B) 117
C) Equal to their H.C.F D) cannot be calculated
 
Answer & Explanation Answer: B) 117

Explanation:

H.C.F of co-prime numbers is 1. So, L.C.M = 117/1 =117

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122 44240
Q:

The H.C.F and L.C.M of two numbers are 11 and 385 respectively. If one number lies between 75 and 125 , then that number is

A) 77 B) 88
C) 99 D) 110
 
Answer & Explanation Answer: A) 77

Explanation:

Product of numbers = 11 x 385 = 4235

 

Let the numbers be 11a and 11b . Then , 11a x 11b = 4235  =>  ab = 35

 

Now, co-primes with product  35 are (1,35) and (5,7)

 

So, the numbers are ( 11 x 1, 11 x 35)  and (11 x 5, 11 x 7)

 

Since one number lies 75 and 125, the suitable pair is  (55,77)

 

Hence , required number = 77

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71 37508
Q:

Three numbers are in the ratio  1 : 2 : 3  and their H.C.F is 12. The numbers are

A) 4, 8, 12 B) 5, 10, 15
C) 10, 20, 30 D) 12, 24, 36
 
Answer & Explanation Answer: D) 12, 24, 36

Explanation:

Let the required numbers be x, 2x, 3x.   Then, their H.C.F =x. so, x= 12

Therefore, The numbers are 12, 24, 36

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73 36477