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 smallest number which when diminished by 7,  is divisible by  12, 16, 18, 21 and 28 is

A) 1008 B) 1015
C) 1022 D) 1032
 
Answer & Explanation Answer: B) 1015

Explanation:

Required Number = (L.C.M  of 12, 16, 18,21,28)+7

= 1008 + 7

= 1015

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101 29629
Q:

The H.C.F of two numbers is 11 and their L.C.M is 7700. If one of the numbers is 275 , then the other is

A) 279 B) 283
C) 308 D) 318
 
Answer & Explanation Answer: C) 308

Explanation:

Other number = 11×7700275= 308

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54 23830
Q:

The H.C.F and L.C.M of two numbers are  84 and 21 respectively.  If the ratio of the two numbers is 1 : 4 , then the larger of the two numbers is 

A) 12 B) 48
C) 84 D) 108
 
Answer & Explanation Answer: C) 84

Explanation:

Let the numbers be x and 4x. Then,  x×4x=84×21  x2=84×214 x=21 

Hence Larger Number = 4x = 84

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26 23253
Q:

The difference of two numbers is 14. Their LCM and HCF are 441 and 7. Find the two numbers ?

A) 63 and 49 B) 64 and 48
C) 62 and 46 D) 64 and 49
 
Answer & Explanation Answer: A) 63 and 49

Explanation:

Since their HCFs are 7, numbers are divisible by 7 and are of the form 7x and 7y

Difference = 14
=> 7x - 7y = 14
=> x - y = 2

product of numbers = product of their hcf and lcm
=> 7x * 7y = 441 * 7
=> x * y = 63

Now, we have
x * y = 63 , x - y = 2
=> x = 9 , y = 7

The numbers are 7x and 7y
=> 63 and 49

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Exam Prep: AIEEE , Bank Exams , CAT , GATE
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25 22499
Q:

Which of the following has the 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 2 x 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 and 101

 

Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176

 

Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.

 

Hence, 176 has the most number of divisors.

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35 22362
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) 180 B) 360
C) 540 D) 1260
 
Answer & Explanation Answer: A) 180

Explanation:

4×27×3125=22×33×55

8×9×25×7=23×32×52×7

16×81×5×11×49=24×34×5×72×11

Therefore, H.C.F = 22×32×5 = 180

 

 

\inline \fn_jvn \therefore   \inline \fn_jvn H.C.F =2^{2}\times 3^{2}\times 5=180

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74 22259
Q:

If HCF of two numbers is 8,which of the following can never be their LCM?

A) 32 B) 48
C) 60 D) 152
 
Answer & Explanation Answer: C) 60

Explanation:

60 is not a multiple of 8

 

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51 21080
Q:

The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:

A) 1677 B) 1683
C) 2523 D) 3363
 
Answer & Explanation Answer: B) 1683

Explanation:

L.C.M. of 5, 6, 7, 8 = 840.

 

 Required number is of the form 840k + 3

 

Least value of k for which (840k + 3) is divisible by 9 is k = 2.

 

 Required number = (840 x 2 + 3) = 1683.

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25 19801