Least Common Multiple Questions,Least Common Factor Questions:-

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What Is HCF:- Highest Common Factor(HCF) of two or more numbers is the greatest number which divides each of them exactly.

• How to Calculate HCF:-

Step 1: Express each number as a product of prime factors.

Step 2: HCF is the product of all common prime factors using the least power of each common prime factor.

Example 1: Find out HCF of 60 and 75

Step 1 : Express each number as a product of prime factors.

60 = 2² × 3 × 5

75 = 3 × 5²

Step 2: HCF is the product of all common prime factors using the least power of each common prime factor.

Here, common prime factors are 3 and 5

The least power of 3 here = 3

The least power of 5 here = 5

Hence, HCF = 3 × 5 = 15

Example 2: Find out HCF of 36, 24 and 12

Step 1: Express each number as a product of prime factors.

36 = 2² × 3²

24 = 2³ × 3

12 = 2² × 3

SO HCF is 2³ × 3=12

• How to find out HCF - by dividing the numbers

Step 1: Write the given numbers in a horizontal line separated by commas.

Step 2: Divide the given numbers by the smallest prime number (write in the left side) which can exactly divide all the given numbers.

Step 3: Write the quotients in a line below the first.

Step 4: Repeat the process until we reach a stage where no common prime factor exists for all the numbers.

Step 5: We can see that the factors mentioned in the left side clearly divides all the numbers exactly and they are common prime factors. Their product is the HCF

Example 1: Find out HCF of 60 and 75

Hence HCF = 3 × 5 = 15.

• How to find out HCF using division method

To find out HCF of two given numbers using division method,
Step 1: Divide the larger number by the smaller number.
Step 2: Divisor of step 1 is divided by its remainder.
Step 3: Divisor of step 2 is divided by its remainder. Continue this process till we get zero as remainder.
Step 4: Divisor of the last step is the HCF.

To find out HCF of three given numbers using division method,
Step 1: Find out HCF of any two numbers.
Step 2: Find out the HCF of the third number and the HCF obtained in step 1.
Step 3: HCF obtained in step 2 will be the HCF of the three numbers.

In a similar way as explained for three numbers, we can find out HCF of more than three numbers also using division method.

Example 3: Find out HCF of 3556 and 3224

Hence HCF of 3556 and 3224 = 4

What Is Least Common Multiple (LCM):-

Least Common Multiple (LCM) of two or more numbers is the smallest number that is a multiple of all the numbers.

Example: LCM of 3 and 4 = 12 because 12 is the smallest number which is a multiple of both 3 and 4 (In other words, 12 is the smallest number which is divisible by both 3 and 4)

• How to find out LCM using prime factorization method

Step 1 : Express each number as a product of prime factors.

Step 2 : LCM = The product of highest powers of all prime factors.

Example 1: Find out LCM of 8 and 14

Step 1 : Express each number as a product of prime factors. (Reference: Prime Factorization)

8 = 2³

14 = 2 × 7

Step 2 : LCM = The product of highest powers of all prime factors.

Here the prime factors are 2 and 7

The highest power of 2 here = 2³

The highest power of 7 here = 7

Hence LCM = 2³ × 7 = 56

Example : Find out LCM of 18, 24, 9, 36 and 90

Step 1 : Express each number as a product of prime factors.

18 = 2 × 3²

24 = 2³ × 3

9 = 3²

36 = 2³ × 3²

90 = 2 × 5 × 3²

Step 2 : LCM = The product of highest powers of all prime factors.

Here the prime factors are 2, 3 and 5

The highest power of 2 here = 2³

The highest power of 3 here = 3²

The highest power of 5 here = 5

Hence LCM = 2³ × 3² × 5 = 360

How to find out LCM using division Method (shortcut method)

Step 1: Write the given numbers in a horizontal line separated by commas.

Step 2: Divide the given numbers by the smallest prime number which can exactly divide at least two of the given numbers.

Step 3: Write the quotients and undivided numbers in a line below the first.

Step 4: Repeat the process until we reach a stage where no prime factor is common to any two numbers in the row.

Step 5: LCM = The product of all the divisors and the numbers in the last line.

Example : Find out LCM of 8 and 14

Hence Least common multiple (L.C.M) of 8 and 14
= 2 × 4 × 7
= 56

Example : Find out LCM of 18, 24, 9, 36 and 90

Hence Least common multiple (L.C.M) of 18, 24, 9, 36 and 90
= 2 × 2 × 3 × 3 × 2 × 5
= 360

Ex. Find the H.C.F. and L.C.M. of 0.63, 1.05 and 2.1.
Sol. Making the same number of decimal places, the given numbers are 0.63, 1.05 and 2.10. Without decimal places, these numbers are 63, 105 and 210.
Now, H.C.F. of 63, 105 and 210 is 21.
H.C.F. of 0.63, 1.05 and 2.1 is 0.21.
L.C.M. of 63, 105 and 210 is 630.
L.C.M. of 0.63, 1.05 and 2.1 is 6.30.

Ex. 1. Find the H.C.F. of 2³ X 3² X 5 X 7⁴, 2² X 3⁵ X 5² X 7³,2³ X 5³ X 7²

Sol. The prime numbers common to given numbers are 2,5 and 7.

H.C.F. = 2² x 5 x7² = 980.

Ex. 2. Find the H.C.F. of 108, 288 and 360.

Sol. 108 = 2² x 3³, 288 = 2⁵ x 3² and 360 = 2³ x 5 x 3².

H.C.F. = 2² x 3² = 36.

Ex. Two numbers are in the ratio of 15:11. If their H.C.F. is 13, find the numbers.
Sol. Let the required numbers be 15.x and llx.
Then, their H.C.F. is x. So, x = 13.
The numbers are (15 x 13 and 11 x 13) i.e., 195 and 143.
Ex. 11. TheH.C.F. of two numbers is 11 and their L.C.M. is 693. If one of the
numbers is 77,find the other.
Sol. Other number = 11 X 693 = 99
77

Ex. Find the greatest possible length which can be used to measure exactly the lengths 4 m 95 cm, 9 m and 16 m 65 cm.
Sol. Required length = H.C.F. of 495 cm, 900 cm and 1665 cm.
495 = 32 x 5 x 11, 900 = 22 x 32 x 52, 1665 = 32 x 5 x 37.
H.C.F. = 32 x 5 = 45.
Hence, required length = 45 cm.

Ex. Find the largest number of four digits exactly divisible by 12,15,18 and 27.
Sol. The Largest number of four digits is 9999.
Required number must be divisible by L.C.M. of 12,15,18,27 i.e. 540.
On dividing 9999 by 540,we get 279 as remainder .
Required number = (9999-279) = 9720.

Ex.Find the smallest number of five digits exactly divisible by 16,24,36 and 54.
Sol. Smallest number of five digits is 10000.
Required number must be divisible by L.C.M. of 16,24,36,54 i.e 432,
On dividing 10000 by 432,we get 64 as remainder.
Required number = 10000 +( 432 – 64 ) = 10368.

Ex. Find the least number which when divided by 20,25,35 and 40 leaves remainders 14,19,29 and 34 respectively.
Sol. Here,(20-14) = 6,(25 – 19)=6,(35-29)=6 and (40-34)=6.
Required number = (L.C.M. of 20,25,35,40) – 6 =1394.

Ex.Find the least number which when divided by 5,6,7, and 8 leaves a remainder 3, but when divided by 9 leaves no remainder .
Sol. 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

Ex.The traffic lights at three different road crossings change after every 48 sec., 72 sec and 108 sec.respectively .If they all change simultaneously at 8:20:00 hours,then at what time they again change simultaneously .
Sol. Interval of change = (L.C.M of 48,72,108)sec.=432sec.
So, the lights will agin change simultaneously after every 432 seconds i.e,7
min.12sec
Hence , next simultaneous change will take place at 8:27:12 hrs.

LCM HCM Quiz:-