HCF
Common Factor -A number is called to be a factor of another numbers when it divides other numbers exactly.
eg. 4 is a common factor of 8, 12, 16, 20, 24
Highest Common Factor :-HCF of two or more numbers is the greatest number that divides each of them exactly.
Method of Prime Factors - Break the given numbers into prime factors and find the product of common prime factors, product will be HCF
Find the HCF of 42 and 70
Solution- 42=2*3*7, 70=2*5*7 So HCF will be 2*7 =14
Find the HCF of 24, 45, 60
Solution- 24 = 2*2*2*3
45= 3*3*5
60= 2*2*3*5 so HCF of 24, 45 and 60 is 3
Second Method = Wrting in a row and division by a common divisor of all-
Step 1: Write the numbers in a row
Step 2: Divide by a common divisor of all
Step 3: Write the remainders in second row
Step 4: Continue this process till we get all the remainders prme to one another:
Ex. Find the HCF of 12, 18 and 24
Method of division : When two numbers are lage then method of Factor is not convenient, Then we find thr HCF by division method
Step 1: Divide the greater number by smaller number and find the remainder
Step 2: Now repeat this process with the first remainder as divisor and first divisor as dividend and continue this process until we get zero as remainder. Last divisor is HCF
Ex. Find the HCF of 1365, 1560 and 1755
HCF of decimals
Ex. Find the HCF of 16.5, 0.45 and 15
Solution: These numbers can be written as 16.50, 0.45 and 15.00
Now find the HCF of 1650, 45 and 1500 , we get HCF as 15, now convert to euivalent fraction whivh comes as 00.15 This is our required HCF
HCF of Fractions
HCF of two or more fractions means the highest fraction which exactly divides each of the fractions
Step 1: Express all fractions in their lowest terms
Step 2: Find the HCF of all the numerators
Step 3: Find the LCM of all the Denominators
Ex. Find the HCF of 54/9, 3*(9/17), 36/51
Solution: Express all fractions in their lowest terms6/1, 60/17, 12/17
HCF=HCF of Numenerators/ LCM of Denominators= HCF of (6, 60, 12) / LCM of (1, 17, 17)= 6/17
Ex. Find the greatest number which will divide 410, 751 and 1030 so as to leave the remainde 7 in each case ?
Solution: Greatest number will be = HCF of (410-7), (751-7) and (1030-7)= HCF of 403, 744 and 1023
Ex. Find the greatest possible length which can be used to measure exactly the lengths 25 m 20 cm, 198 m, 9m 36 cm.
Solution : Required length = HCF of 2520 cm , 19800 cm, 936 cm
So required length will be 72 cm
Ex. The maximum number of students among them 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is :
Solution : Required number of students= HCF of 1001 and 91
Important Rules :
(1)= Product of two numbers = Product of their HCF and LCM
(2)= Co-prime numbers: Two numbers are said to be co-primes if their HCF is 1
Ex. LCM of two co-prime numbers x and y where x>y is 161. Find out the value of 3y-x ?
Solution : HCF of x and y = 1, since they are co prime numbers
Now, we know Product of two numbers = Product of their HCF and LCM
So, xy = 1 * 161 = 161
So co-primes can be ( 1, 161) or ( 23, 7 )
Since x > y , so x= 23 and y = 7
so, 3y - x = 3 * 7 - 23 = -2
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