Here, some tricks to solve HCF and LCM of related problems....
Concept :
* Highest Common Factor (HCF)
If two or more number are broken into their prime factors (as explained in 2.3), then the product of the maximum common prime factors in the given numbers is the H.C.F. of the numbers. In other words, the HCF of two or more numbers is the greatest number (divisor) that divides all the given numbers exactly.
So, HCF is also called the Greatest Common Divisor (GCD).
Ex : Find the HCF of 72, 60, 96. Here, we first find the prime factors of each given number ?
Sol: Here 72 = 2 × 2 × 2 × 3 × 3
60 = 2 × 2 × 3 × 5
96 = 2 × 2 × 2 × 2 × 2 × 3
and so HCF = product of maximum common prime factors
= 2 × 2 × 3 = 12
Note : The common factors in the given numbers have been encircled.
* LCM (Lowest Common Multiple)
The LCM of two or more than two numbers is the product of the highest powers of all the prime factors that occur in these numbers.
Ex : Find the LCM of 36, 48, 64 and 72 ?
2| 36, 48, 64, 72
2| 18, 24, 32, 36
2| 9, 12, 16, 18
2| 9, 6, 8, 9
3| 9, 3, 4, 9
3| 3, 1, 4, 3
| 1, 1, 4, 1
∴ LCM = 2 × 2 × 2 × 2 × 3 × 3 × 4 = 576.
Some shortcuts :
Trick-1:
Product of Two Numbers
∴ HCF of numbers × LCM of numbers = Product of numbers
Ex: If the product of two numbers is 120 and LCM of two numbers is 40.find the HCF ?
Sol: 120 = 40 * HCF
HCF = 3
Trick-2:
Find the GREATEST NUMBER that will exactly divide x, y and z. Required number = HCF of x, y & z
Ex: the greatest number that exactly divide 120,90 and 40 ?
Sol: = HCF of 120,90 and 40
∴ 10 is greatest number divides exactly 120,90 and 40
Trick-3:
Find the GREATEST NUMBER that will divide x, y and z leaving remainders a, b and c respectively. Required nimber = HCF of (x - a), (y - b) and (z -c).
Ex: the greatest number that exactly divide 125,94 and 43 leaves remainders 5,4 and 3 respectively ?
Sol:
= HCF of (125-5),(94-4) and (43-3)
∴ 10 is greatest number divides exactly 120,90 and 40
Trick-4:
Find the LEAST NUMBER which is exactly divisible by x, y and z. Required number = LCM of x, y and z (least dividend)
Ex: Least number which when divided by 35,45,55 ?
Sol: = LCM of ( 35,45,55)
= 3465
Trick-5:
Find the LEAST NUMBER which when dividend by x, y and z leaves the remainders a, b and c respectively.
Then, it is always observed that (x - a) = (y -b) = (z - c) = k (say)
∴ Required number = (LCM of x, y and z) - (K)
Ex: Least number which when divided by 35,45,55 and leaves remainder 18,28,38 ?
Sol: Here the difference between every divisor and remainder is same i.e. 17.
Therefore, required number
= LCM of (35,45,55)-17
= (3465-17)= 3448.
Trick-6:
Find the LEAST NUMBER which when dividend by x, y and z leaves the same remainder 'r' each case. Required number = (LCM of x, y and z) + r.
Ex: Least number which when divided by 5,6,7,8 and leaves remainder 3 ?
Sol: = LCM ( 5,6,7,8 ) + 3
= 843
Trick-7:
Find the GREATEST NUMBER that will divided x, y and z leaving the same remainder in each case. Required number = HCF of (x - y), (y -z) and (z -x)
Trick-7:
Find the GREATEST NUMBER that will divided x, y and z leaving the same remainder R in each case. Required number = HCF of (x - R), (y -R) and (z -R)
Trick-8:
Find the n-digit GREATEST NUMBER which when divided by x, y and z.
leaves no remainder(i.e exactly divisible)
Remainder 'R' =[ n-digit greatest number/ LCM of (x,y and z)
∴ Required number = (n-digit greatest number) - R
Ex: Greater number of 4 digits which is divisible by each one of 12,18,21 and 28 is?
Sol: LCM of 12,18,21,28 = 252 Therefore, required number must be divisible by 252.
Greatest four digit number = 9999 On dividing 9999 by 252 remainder = 171
Therefore, 9999-171 = 9828.
Trick-9:
Find the n-digit GREATEST NUMBER which when divided by x, y and z.
leaves remainder K in each case
Remainder 'R' =[ n-digit greatest number/ LCM of (x,y and z)
∴ Required number = [ (n-digit greatest number) - R ] + K
Ex: Greater number of 4 digits which is divisible by each one of 12,18,21 and 28.leaves 3 as remainder in each case ?
Sol: LCM of 12,18,21,28 = 252 Therefore, required number must be divisible by 252.
Greatest four digit number = 9999 On dividing 9999 by 252, remainder = 171
Therefore, 9999-171 + 3= 9831.
Trick-10:
Find the n-digit SMALLEST NUMBER which when divided by x, y and z.leaves no remainder(i.e exactly divisible)
Remainder 'R' =[ n-digit smallest number/ LCM of (x,y and z)
∴ Required number = (n-digit smallest number) + [L - R ]
Ex: Smallest number of 4 digits which is divisible by each one of 12,18,21 and 28 is?
Sol: LCM of 12,18,21,28 = 252 Therefore, required number must be divisible by 252.
Smallest four digit number = 1000 On dividing 1000 by 252 remainder = 244
Therefore, 1000 + (252-244) = 1008.
Trick-11:
Find the n-digit SMALLEST NUMBER which when divided by x, y and z.
leaves remainder K in each case
Remainder 'R' =[ n-digit greatest number/ LCM of (x,y and z)
∴ Required number = [ (n-digit greatest number) +(L - R) ] + K
Ex: Smallest number of 4 digits which is divisible by each one of 12,18,21 and 28.leaves 3 as remainder in each case ?
Sol: LCM of 12,18,21,28 = 252 Therefore, required number must be divisible by 252.
Smallest four digit number = 1000 On dividing 1000 by 252, remainder = 171
Therefore,1000+(252-244)+3= 1011.
Trick-12:
Find the HCF of fractions
= HCF of numerators / LCF of denominators
Trick-13:
Find the LCM of fractions
= LCM of numerators / HCM of denominators
Trick-14:
Find the HCF of decimal numbers ?
Step-1: Find the HCF of the given numbers without decimal.
Step-2: Put the decimal point (in the HCF of step 1) from right to left according to the MAXIMUM decimal places among the given numbers.
Trick-15:
Find the LCM of decimal numbers ?
Step-1: Find the LCM of the given numbers without decimal.
Step-2: Put the decimal point (in the LCM of step 1) from right to left at the place equal to the MINIMUM decimal places among the given numbers
Concept :
* Highest Common Factor (HCF)
If two or more number are broken into their prime factors (as explained in 2.3), then the product of the maximum common prime factors in the given numbers is the H.C.F. of the numbers. In other words, the HCF of two or more numbers is the greatest number (divisor) that divides all the given numbers exactly.
So, HCF is also called the Greatest Common Divisor (GCD).
Ex : Find the HCF of 72, 60, 96. Here, we first find the prime factors of each given number ?
Sol: Here 72 = 2 × 2 × 2 × 3 × 3
60 = 2 × 2 × 3 × 5
96 = 2 × 2 × 2 × 2 × 2 × 3
and so HCF = product of maximum common prime factors
= 2 × 2 × 3 = 12
Note : The common factors in the given numbers have been encircled.
* LCM (Lowest Common Multiple)
The LCM of two or more than two numbers is the product of the highest powers of all the prime factors that occur in these numbers.
Ex : Find the LCM of 36, 48, 64 and 72 ?
2| 36, 48, 64, 72
2| 18, 24, 32, 36
2| 9, 12, 16, 18
2| 9, 6, 8, 9
3| 9, 3, 4, 9
3| 3, 1, 4, 3
| 1, 1, 4, 1
∴ LCM = 2 × 2 × 2 × 2 × 3 × 3 × 4 = 576.
Some shortcuts :
Trick-1:
Product of Two Numbers
∴ HCF of numbers × LCM of numbers = Product of numbers
Ex: If the product of two numbers is 120 and LCM of two numbers is 40.find the HCF ?
Sol: 120 = 40 * HCF
HCF = 3
Trick-2:
Find the GREATEST NUMBER that will exactly divide x, y and z. Required number = HCF of x, y & z
Ex: the greatest number that exactly divide 120,90 and 40 ?
Sol: = HCF of 120,90 and 40
∴ 10 is greatest number divides exactly 120,90 and 40
Trick-3:
Find the GREATEST NUMBER that will divide x, y and z leaving remainders a, b and c respectively. Required nimber = HCF of (x - a), (y - b) and (z -c).
Ex: the greatest number that exactly divide 125,94 and 43 leaves remainders 5,4 and 3 respectively ?
Sol:
= HCF of (125-5),(94-4) and (43-3)
∴ 10 is greatest number divides exactly 120,90 and 40
Trick-4:
Find the LEAST NUMBER which is exactly divisible by x, y and z. Required number = LCM of x, y and z (least dividend)
Ex: Least number which when divided by 35,45,55 ?
Sol: = LCM of ( 35,45,55)
= 3465
Trick-5:
Find the LEAST NUMBER which when dividend by x, y and z leaves the remainders a, b and c respectively.
Then, it is always observed that (x - a) = (y -b) = (z - c) = k (say)
∴ Required number = (LCM of x, y and z) - (K)
Ex: Least number which when divided by 35,45,55 and leaves remainder 18,28,38 ?
Sol: Here the difference between every divisor and remainder is same i.e. 17.
Therefore, required number
= LCM of (35,45,55)-17
= (3465-17)= 3448.
Trick-6:
Find the LEAST NUMBER which when dividend by x, y and z leaves the same remainder 'r' each case. Required number = (LCM of x, y and z) + r.
Ex: Least number which when divided by 5,6,7,8 and leaves remainder 3 ?
Sol: = LCM ( 5,6,7,8 ) + 3
= 843
Trick-7:
Find the GREATEST NUMBER that will divided x, y and z leaving the same remainder in each case. Required number = HCF of (x - y), (y -z) and (z -x)
Trick-7:
Find the GREATEST NUMBER that will divided x, y and z leaving the same remainder R in each case. Required number = HCF of (x - R), (y -R) and (z -R)
Trick-8:
Find the n-digit GREATEST NUMBER which when divided by x, y and z.
leaves no remainder(i.e exactly divisible)
Remainder 'R' =[ n-digit greatest number/ LCM of (x,y and z)
∴ Required number = (n-digit greatest number) - R
Ex: Greater number of 4 digits which is divisible by each one of 12,18,21 and 28 is?
Sol: LCM of 12,18,21,28 = 252 Therefore, required number must be divisible by 252.
Greatest four digit number = 9999 On dividing 9999 by 252 remainder = 171
Therefore, 9999-171 = 9828.
Trick-9:
Find the n-digit GREATEST NUMBER which when divided by x, y and z.
leaves remainder K in each case
Remainder 'R' =[ n-digit greatest number/ LCM of (x,y and z)
∴ Required number = [ (n-digit greatest number) - R ] + K
Ex: Greater number of 4 digits which is divisible by each one of 12,18,21 and 28.leaves 3 as remainder in each case ?
Sol: LCM of 12,18,21,28 = 252 Therefore, required number must be divisible by 252.
Greatest four digit number = 9999 On dividing 9999 by 252, remainder = 171
Therefore, 9999-171 + 3= 9831.
Trick-10:
Find the n-digit SMALLEST NUMBER which when divided by x, y and z.leaves no remainder(i.e exactly divisible)
Remainder 'R' =[ n-digit smallest number/ LCM of (x,y and z)
∴ Required number = (n-digit smallest number) + [L - R ]
Ex: Smallest number of 4 digits which is divisible by each one of 12,18,21 and 28 is?
Sol: LCM of 12,18,21,28 = 252 Therefore, required number must be divisible by 252.
Smallest four digit number = 1000 On dividing 1000 by 252 remainder = 244
Therefore, 1000 + (252-244) = 1008.
Trick-11:
Find the n-digit SMALLEST NUMBER which when divided by x, y and z.
leaves remainder K in each case
Remainder 'R' =[ n-digit greatest number/ LCM of (x,y and z)
∴ Required number = [ (n-digit greatest number) +(L - R) ] + K
Ex: Smallest number of 4 digits which is divisible by each one of 12,18,21 and 28.leaves 3 as remainder in each case ?
Sol: LCM of 12,18,21,28 = 252 Therefore, required number must be divisible by 252.
Smallest four digit number = 1000 On dividing 1000 by 252, remainder = 171
Therefore,1000+(252-244)+3= 1011.
Trick-12:
Find the HCF of fractions
= HCF of numerators / LCF of denominators
Trick-13:
Find the LCM of fractions
= LCM of numerators / HCM of denominators
Trick-14:
Find the HCF of decimal numbers ?
Step-1: Find the HCF of the given numbers without decimal.
Step-2: Put the decimal point (in the HCF of step 1) from right to left according to the MAXIMUM decimal places among the given numbers.
Trick-15:
Find the LCM of decimal numbers ?
Step-1: Find the LCM of the given numbers without decimal.
Step-2: Put the decimal point (in the LCM of step 1) from right to left at the place equal to the MINIMUM decimal places among the given numbers
I remember when I first time did that question on my own so that was a remarkable day of my life as my father, teacher, all praised me a lot because I was the first in the class who learned HCF and LCM. After that I also took the custom essays writing services to make unique contents on my own and I did succeed in that process as well.
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