# LCM AND HCF

Factor: one number is said to be a factor when it divides the other number exactly. Thus 3 and 4 are factors of 12

Multiple: one number is said to be a multiple of other number when it is exactly divisible by the other.

Common factor: A common factor of two or more numbers is a number that divides each of them exactly. Thus 4 is a common factor of 12,16,24,72

Common multiple: A common multiple of two or more numbers is a number which is exactly divisible by each of them. Thus 18 is a common multiple of 2,3,6 and 9.

Highest Common Factor (HCF): HCF of two or more given numbers is the greatest number that divides each of them exactly. Thus 5 is the HCF of 25 and 35.

HCF is also called Highest Common Divisor or Greatest Common Divisor.

Methods to find HCF

Method of prime factors: Break the given numbers into prime factors and then find the product of the prime factors common to all the numbers. This product will be the required HCF.

Example: HCF of  20, 35 and 45 is

20  =  2 × 2 × 5

35  =  5 × 7

45  =  3 × 3 × 5

The factor common to all the numbers is 5, hence 5 is the HCF.

Method of Division: Divide the greater number by the smaller number, if there is remainder then divide the divisor by the remainder, if again there is remainder, then again divide the divisor by the next remainder and so on until no remainder is left. The last remainder is the required HCF. Methods to find LCM

Method of prime factors: Divide the given numbers into their prime factors and then find the product of the highest powers of all the factors that occur in the given numbers, and this product will be the required LCM.

Example: LCM of 4, 8 and 24 is

4   =  2 × 2                =  22

8   =  2 × 2 × 2         =  23

24 =  2 × 2 × 2 × 3  =  23 × 3

The prime factors that occur here are 2 and 3. the highest powers of these prime factors are 23 and 31 respectively.

Therefore the required LCM  is 23 × 31 = 24.

Regular method: Write all the given numbers in a line and divide them by a number which will exactly divide at least any two of the numbers. write down the quotients and the undivided numbers in a line below the first. Repeat the process until we get a line of numbers which are prime

to each other. The product of all the divisors and the numbers in the last line will be the required LCM.

Example:
LCM of 8, 12 and 32 is Therefore required LCM is  2 × 2 × 2 × 3 × 7 × 4 = 672

LCM of decimals: First find LCM of the given numbers without decimals and then put the decimal in the result after the number of digits which is equal to the minimum digits after the decimal in the given numbers from right to left.

Example:

LCM of 2.4, 0.012  and 0.32 is

LCM of 24, 12 and 32 is 96

In the given numbers the minimum digits from right to left is in 2.4 i.e., 1

Therefore the required LCM  is 9.6. 1. Factors and Multiples:

If number a divided another number b exactly, we say that a is a factor of b.

In this case, b is called a multiple of a.

2. 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:

I. Factorization Method: Express the 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.

Example:

HCF (18, 42), we find the prime factors of 18 = 2 × 3 × 3 and 42 = 7 × 2 × 3 and notice the "common" of the two expressions is 2 × 3; So HCF (18, 42) = 6.

II. Division Method: Suppose we have to find the H.C.F. of two given numbers, divide the larger 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 required H.C.F.

Example: 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 number.

Similarly, the H.C.F. of more than three numbers may be obtained.

3. 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.

There are two methods of finding the L.C.M. of a given set of numbers:

I. Factorization Method: 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.

Example: To find the value of LCM (9, 48, and 21). First, find the factor of each number and express it as a product of prime number powers.

Like 9 = 32, 48 = 24 × 3, 21 = 3 × 7

Then, write all the factors with their highest power like 32, 24, and 7. And multiply them to get their LCM.

Hence, LCM (9, 21, and 48) is 32 × 24 × 7 = 1008.

II. Division Method (short-cut): Arrange the given numbers in a row in any order. Divide by a number which divided 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.

Example: To find the value of LCM (72, 240 and 196). L.C.M. of the given numbers

= product of divisors and the remaining numbers

= 2×2×2×3×3×10×49

= 72×10×49 = 35280.

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

For Example:

LCM (8, 28) = 56 & HCF (8, 28) = 4

Now, 8 × 28 = 224 and also, 56 × 4 = 224

5. Co-primes:

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

6. H.C.F. and L.C.M. of Fractions: SOME MODELS with EXPLANATIONS

1. The LCM of two numbers is 240 and HCF is 12. If one of the numbers is 48, find the other number?
a) 60         b) 80         c) 120         d) 24
Sol: HCF × LCM = Product of two numbers
⇒ 12 × 240 = 48 × Second number
⇒ Second number = 12 × 240/48 = 60
Ans: a

2. Find the HCF of 513, 1134 and 1215.
a) 81           b) 9           c) 3          d) 27
Sol: 513 = 33 × 19; 1134 = 2 × 34 × 7 and 1215 = 35 × 5
HCF of (513, 1134, 1215) = 33 = 27
Ans: d

3. The greatest number that exactly divides 2436, 1001 and 105 is...
a) 7            b) 21           c) 1         d) None of these
Sol: 2436 = 22 × 3 × 7 × 29; 1001 = 7 × 11 × 13 and 105 = 3 × 5 × 7
Greatest number = HCF of (2436, 1001, 105)  = 7
Ans: a

4. The HCF of two numbers is 8. Which one of the following can never be their LCM?
a) 24         b) 60         c) 48         d) 56
Sol: Since 8 is not the divisor of 60. Therefore, 60 cannot be LCM of two numbers.
Ans: b

5. The product of the two 2-digit numbers and their HCF is 1536 and 16 respectively. Which of the following cannot be one of the numbers?
|a) 16           b) 48          c) 32          d) 102
Sol: Product of two numbers = HCF × LCM
16 × LCM = 1536 ⇒ LCM = 96
Now, let the numbers be 16x and 16y.
16x × 16y = 96 × 16 ⇒ xy = 6
The pairs satisfying the given condition and prime to each other are (1, 6) and (2, 3).
Thus, the numbers are (16, 96) and (32, 48).
Therefore, 102 cannot be one of them.
Ans: d Ans: c

7. The LCM of 96, 144 and N is 576. If their HCF is 48, then which of the following can be one of the value of N?
a) 168            b) 192           c) 144             d) 244
Sol: From options, only 192 and 144 are divisible by 48. LCM of (96, 144 and 192) = 576
Hence, 192 can be value of N.
Ans: b

8. Find the difference between the LCM and Ans: d

9. The LCM of two numbers is 500 and their HCF is 50. If one of the numbers is 100, the other number is...
a) 250            b) 400            c) 500            d) None
Sol: LCM × HCF = N1 × N2 and N1 = 100, Ans: a ​​​​​​​

10. The HCF and LCM of two numbers are 25 and 500 respectively. If the first number is divided by 2, the quotient is 50. The second number is...
a) 50            b) 100            c) 125            d) 250
Sol: First number = 2 × 50 = 100
2nd number × 100 = 25 × 500 (Since product of two numbers = HCF × LCM)
⇒ 2nd number = 125
Ans: c

11. An orchard has 48 apple trees, 60 mango trees and 96 banana trees. These have to be arranged in rows such that each row has the same number of trees of the same type. Find the minimum possible number of such rows that can be formed.
a) 13             b) 14             c) 16              d) 17
Sol: Total number of trees are 48 + 60 + 96 = 204
Given that each row has the same number of trees and all trees in a row are of the same type. Since we need to minimise the number of rows, we need to maximise the number of trees in each row.
Number of trees in each row = HCF of 48, 60 and 96 = 12
∴ Minimum number of rows = 204/12  = 17
Ans: d

12. If product of two numbers is 2160 and their HCF is 6, then what will be the ratio of HCF to LCM?
a) 21 : 60              b) 60 : 21              c) 1 : 60                 d) 60 : 1
Sol: LCM = 2160/6 = 360
Required ratio = 6 : 360 = 1 : 60
Ans: c 13. If the LCM and HCF of two positive even numbers is 84 and 2 respectively, then the minimum sum of two numbers will be..
a) 30              b) 26            c) 14            d) 34
Sol: Let the numbers be 2a and 2b, where a and b are co-prime natural numbers.
Then, 2a × 2b = 2 × 84  ⇒ a × b = 42
Possible pairs of a and b are (42, 1) (14, 3) (7, 6) (21, 2)
Minimum sum of numbers =  2(a + b) = 2(7 + 6) = 26
Ans: b

14. The HCF and LCM of two 2-digit numbers are 16 and 480 respectively. The numbers are...
a) 40, 48            b) 70, 72           c) 64, 80          d) 80, 96
Sol: H.C.F. of the two 2 - digit numbers = 16 Hence, the numbers can be expressed as  16x and 16y, where x and y are prime to each other. Now, First number × second number
= H.C.F. × L.C.M. ⇒ 16x × 16y = 16 × 480 The possible pairs of x and y, satisfying the condition xy = 30 are : (3, 10), (5, 6), (1, 30), (2, 15) Since the numbers are of 2-digits each. Hence, admissible pair is (5, 6)
∴ Numbers are 16 × 5 = 80 and 16 × 6 = 96
Ans: d

Remainders

1. The remainder when 52563744 is divided by 24 is...
a) 1           b) 0           c) 2            d) 3
Sol: 52563744 = 2 × 3 × 4 × 2190156
It is divisible by 24. Hence, remainder = 0
Ans: b

2. In a division problem, the divisor is 10 times the quotient and 5 times the remainder. If the remainder is 46, the dividend is...
a) 5326           b) 5236          c) 5336            d) 5246
Sol: Divisor = 5 × 46 = 230
∴ Quotient = 230/10 = 23
∴ Dividend = 23 × 230 + 46 = 5336
Ans: c

3. Find the smallest value of N (N > 1) such that when N is divided by 6 or 7, it leaves the remainder 1 in each case.
a) 85           b) 87           c) 43          d) 13
Sol: N = 6x + 1 and N = 7y + 1
⇒ 6x + 1 = 7y + 1
⇒ 6x - 7y = 0
By hit and trial x = 7 and y = 6
Therefore, N = 6 × 7 + 1 = 43
Ans: c

4. What is the remainder when 98765432123 is divided by 9....
a) 1           b) 2             c) 5          d) 4
Sol: A number is divisible by 9 if the sum of its digits is divisible by 9.
Sum of digits = 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 + 2 + 3 = 50
When we divide 50 by 9, then remainder = 5
Ans: c

5. What will be the remainder when 35 + 56 is divided by 4?
a) 1          b) 2          c) 3        d) 0
Sol: 35 when divided by 4 leaves a remainder of 3 whereas 56 leaves a remainder of 1 when divided by 4.
So, the sum of these two numbers will be divisible by 4.
Ans: d

6. What is the smallest three-digit number which when divided by 6 leaves a remainder of 5 and when divided by 5 leaves a remainder of 3?
a) 147          b) 117         c) 183          d) 113
Sol: Let N be the number. Then,
N = 6x + 5 and N = 5y + 3
∴ 6x + 5 = 5y + 3
⇒ 5y - 6x = 2
By hit and trial for smallest three digit number.
x = 18 and y = 22
Hence, N = 6 × 18 + 5 = 113
Ans: d

7. The remainder when 231 is divided by 5 is..
a) 3          b) 4           c) 2           d) 1
Sol: Cyclicity of 2 is 4, then last digit of 231 is 2 × 2 × 2 = 8
When we divide 231 by 5,
we get remainder = 3
Ans: a

8. When n is divided by 4, the remainder is 3. What is the remainder when 2n + 1 is divided by 4?
a) 1          b) 2          c) 3           d) 4
Sol: When n is divided by 4, then remainder = 3 ⇒ Remainder = 3
Ans: c

9. What is the remainder when 983 × 653 is divided by 5?
a) 1          b) 3          c) 2          d) 4
Sol: 92 when divided by 5 leaves a remainder of 1. Therefore, 983 will leave a remainder of 4 when divided by 5. 6 divided by 5 leaves a remainder of 1 therefore, 653 will also leave a remainder of 1 when divided by 5. Total remainder will be (4 × 1) = 4
Ans: d

10. A number when divided by 6 leaves the remainder 3. When the square of the same number is divided by 6, the remainder is...
a) 0          b) 1          c) 2            d) 3
Sol: Let N be the number.
Then, N = 6x + 3
∴ N2 = 36x2 + 9 + 36x
⇒ N2 = 6(6x2 + 1 + 6x) + 3
Hence, N2 is divided by 6, remainder = 3
Ans: d

Posted Date : 26-04-2021

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