# HCF and LCM Questions

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 product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:

 A) 1 B) 2 C) 3 D) 4

Explanation:

Let the numbers 13a and 13b.

Then, 13a x 13b = 2028

=>ab = 12.

Now, the co-primes with product 12 are (1, 12) and (3, 4).

[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]

So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).

Clearly, there are 2 such pairs.

52 26814
Q:

The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:

 A) 123 B) 127 C) 235 D) 305

Explanation:

Required number = H.C.F. of (1657 - 6) and (2037 - 5)

= H.C.F. of 1651 and 2032 = 127.

101 24883
Q:

Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:

 A) 40 B) 80 C) 120 D) 200

Explanation:

Let the numbers be 3x, 4x and 5x.

Then, their L.C.M. = 60x.

So, 60x = 2400 or x = 40.

The numbers are (3 x 40), (4 x 40) and (5 x 40).

Hence, required H.C.F. = 40.

46 22154
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

Explanation:

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

51 21549
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

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

60 21201
Q:

A rectangular courtyard 3.78 meters long 5.25 meters wide is to be paved exactly with square  tiles, all of the same size. what is the largest size of the tile which could be used for the purpose?

 A) 14 cms B) 21 cms C) 42 cms D) None of these

Explanation:

3.78 meters =378 cm = 2 × 3 × 3 × 3 × 7

5.25 meters=525 cm = 5 × 5 × 3 × 7

Hence common factors are 3 and 7

Hence LCM = 3 × 7 = 21

Hence largest size of square tiles that can be paved exactly with square tiles is 21 cm.

75 20277
Q:

The L.C.M of two numbers is 495 and their H.C.F is 5. If the sum of the numbers is 100, then their difference is

 A) 10 B) 46 C) 70 D) 90

Explanation:

Let the numbers be x and (100-x).

Then,$x\left(100-x\right)=5*495$

=>  ${x}^{2}-100x+2475=0$

=>  (x-55) (x-45) = 0

=>  x = 55 or x = 45

The numbers are 45 and 55

Required difference = (55-45) = 10

39 19060
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

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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