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:

H.C.F of 4 x 27 x 3125, 8 x 9 x 25 x 7 and 16 x 81 x 5 x 11 x 49 is :

A) 360 B) 180
C) 90 D) 120
 
Answer & Explanation Answer: B) 180

Explanation:

4 x 27 x 3125 = 22 × 33 × 55 ;

 

8 x 9 x 25 x 7 = 23 × 32 × 52 × 7

 

16 x 81 x 5 x 11 x 49 = 24 × 34 × 5 × 11 × 72

 

H.C.F = 22 × 32 × 5 = 180.

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22 9152
Q:

A drink vendor has 368 liters of Maaza, 80 liters of Pepsi and 144 liters of Sprite. He wants to pack them in cans, so that each can contains the same number of liters of a drink, and doesn't want to mix any two drinks in a can. What is the least number of cans required ?

A) 47 B) 46
C) 37 D) 35
 
Answer & Explanation Answer: C) 37

Explanation:

The number of liters in each can = HCF of 80, 144 and 368 = 16 liters.
Number of cans of Maaza = 368/16 = 23
Number of cans of Pepsi = 80/16 = 5
Number of cans of Sprite = 144/16 = 9
The total number of cans required = 23 + 5 + 9 = 37 cans.

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31 9028
Q:

HCF of 1, 2, 3?

A) 3 B) 2
C) 1 D) 0
 
Answer & Explanation Answer: C) 1

Explanation:

Given numbers are 1, 2 and 3

Factors of 1 = 1

Factors of 2 = 1, 2

Factors of 3 = 1, 2 and 3

Common factors = 1

 

Hence, the HCF of 1, 2 and 3 = 1.

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31 8939
Q:

If N is the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. What is the sum of the digits of N ?

A) 6 B) 8
C) 4 D) 9
 
Answer & Explanation Answer: C) 4

Explanation:

N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)

= H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

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17 7845
Q:

Which of the following has most number of divisors?

A) 99 B) 101
C) 176 D) 182
 
Answer & Explanation Answer: C) 176

Explanation:

99  = 1 x 3 x 3 x 11

101= 1 x 101

176= 1 x 2x 2 x 2 x 2 x 11

182= 1 x 2 x 7 x 13

So, divisors of 99 are 1, 3, 9, 11, 33, 99

      divisors of 101 are 1,101

      divisors of 176 are 1, 2, 4, 8, 16, 22, 44, 88, 176

      divisors of 182 are 1, 2, 7, 13, 14, 26, 91, 182

Hence , 176 hasthe most number of divisors. 

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8 7361
Q:

A student can divide her books into groups of 5, 9 and 13. what is the smallest possible number of the books ?

A) 487 B) 585
C) 635 D) 705
 
Answer & Explanation Answer: B) 585

Explanation:

The smallest possible number of books = L.C.M of 5, 9 and 13.
Therefore, L.C.M of 5,9 and 13 is = 5 x 9 x 13 = 585.

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5 7120
Q:

H.C.F of 3240, 3600 and a third number is 36 and their L.C.M is 24 x 35 x 52 x 72 . What is the third number?

A) 47628 B) 49874
C) 24157 D) 42146
 
Answer & Explanation Answer: A) 47628

Explanation:

hcf_of_three_numbers_3240,_3600_and_third_number_is_361542439249.jpg image

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17 6991
Q:

If the 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 equal to ?

A) 13/125 B) 14/57
C) 11/120 D) 16/41
 
Answer & Explanation Answer: C) 11/120

Explanation:

Let the numbers be a and b.
We know that product of two numbers = Product of their HCF and LCM
Then, a + b = 55 and ab = 5 x 120 = 600.
=> The required sum = (1/a) + (1/b) = (a+b)/ab
=55/600 = 11/120

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23 6590