Division Method to Find HCF
By Division method we start with the two numbers and proceed as shown below, till the remainder becomes zero.
Example: Find HCF of 12 and 48
Solution:
12) 48 ( 4
48
0
HCF = 12.
Example: Find HCF of 10 and 25.
Solution:
10)25(2
20
5)10(2
10
0
If there are more than two numbers, we will repeat the whole process with the HCF obtained from two numbers as the divisor and so on. The last divisor will then be the required HCF of the number.
Example: Find HCF of 10,25 and 30.
Solution: We can find the HCF of 10 and 25 i.e. 5. Now we have to find the HCF of 5 and 30 which is 5. so, the HCF of 10,25 and 30 is 5.
Note: If we have to find the greatest number that will exactly divide p,q and r then required number = HCF of p,q and r.
Example: Find the greatest number that will exactly divide 65, 52 and 78.
Solution: Required number = HCF of 65, 52 and 78 = 13.
Note: If we have to find the greatest number that will divide p, q and r leaving remainders a, b and c respectively, then the required number = HCF of (p - a), (q - b) and (r - c).
Example: Find the greatest number that will divide 65, 52 and 78 leaving remainders 5, 2 and 8 respectively.
Solution: Required number = HCF of (65 - 5), (52 - 2) and (78 - 8) = HCF of 60, 50 and 70 = 10.
Note: If we have to find the greatest number that will divide p, q and r leaving the same remainder in each case, then required number is,
= HCF of the absolute values of (p -q), (q - r) and (r - p)
Example: Find the greatest number that will divide 65, 81 and 145 leaving the same remainder in each case.
Solution: Required number is
= HCF of (81 - 65), (145 - 81) and (145 - 65)
= HCF of 16, 64 and 80
= 16
Example: How many numbers below 90 and other than unity exist, such that the HCF of that number and 90 is unity?
Solution: we can write 90 as 32 x 2 x 5
Number of Multiplies of 2 = 45
Number of Multiplies of 3 = 90
Number of Multiplies of 5 = 18
Number of Multiplies of 2 & 5 = 9
Number of Multiplies of 2 & 3 = 15
Number of Multiplies of 3 & 5 = 6
Number of Multiplies of 2 ,3 & 5 =3
Total = (24 + 6 + 12) + (12 + 3 + 6 + 3) = 42 + 24 = 66
24 numbers (including unity) compared with 90 have only '1' as common factor. Hence required result is 24 - 1 = 23.
Here, when the remainder is not zero, divide the previous divisor with that remainder and proceed in the same way until you get the remainder as zero. The last divisor is the required HCF.
Example: Find HCF of 10 and 25.
Solution:
10)25(2
20
5)10(2
10
0
If there are more than two numbers, we will repeat the whole process with the HCF obtained from two numbers as the divisor and so on. The last divisor will then be the required HCF of the number.
Example: Find HCF of 10,25 and 30.
Solution: We can find the HCF of 10 and 25 i.e. 5. Now we have to find the HCF of 5 and 30 which is 5. so, the HCF of 10,25 and 30 is 5.
Note: If we have to find the greatest number that will exactly divide p,q and r then required number = HCF of p,q and r.
Example: Find the greatest number that will exactly divide 65, 52 and 78.
Solution: Required number = HCF of 65, 52 and 78 = 13.
Note: If we have to find the greatest number that will divide p, q and r leaving remainders a, b and c respectively, then the required number = HCF of (p - a), (q - b) and (r - c).
Example: Find the greatest number that will divide 65, 52 and 78 leaving remainders 5, 2 and 8 respectively.
Solution: Required number = HCF of (65 - 5), (52 - 2) and (78 - 8) = HCF of 60, 50 and 70 = 10.
Note: If we have to find the greatest number that will divide p, q and r leaving the same remainder in each case, then required number is,
= HCF of the absolute values of (p -q), (q - r) and (r - p)
Example: Find the greatest number that will divide 65, 81 and 145 leaving the same remainder in each case.
Solution: Required number is
= HCF of (81 - 65), (145 - 81) and (145 - 65)
= HCF of 16, 64 and 80
= 16
Example: How many numbers below 90 and other than unity exist, such that the HCF of that number and 90 is unity?
Solution: we can write 90 as 32 x 2 x 5
Number of Multiplies of 2 = 45
Number of Multiplies of 3 = 90
Number of Multiplies of 5 = 18
Number of Multiplies of 2 & 5 = 9
Number of Multiplies of 2 & 3 = 15
Number of Multiplies of 3 & 5 = 6
Number of Multiplies of 2 ,3 & 5 =3
Total = (24 + 6 + 12) + (12 + 3 + 6 + 3) = 42 + 24 = 66
24 numbers (including unity) compared with 90 have only '1' as common factor. Hence required result is 24 - 1 = 23.
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