HCF & LCM - Atreya Roy
Content & PR team - MBAtious
Author: Atreya Roy is pursuing his BTech From Kalyani Government Engineering College, Bengal.
HCF (Highest Common Factor)
When two or more numbers are broken into their prime factors, then the product of the maximum common prime factors is the HCF of the numbers.
Find the HCF of 72 and 60
72 = 2^3 * 3^2
60 = 2^2 * 3*5
So HCF = product of max common primes = 2^2 *3 = 12
Hence HCF = 12
Use the same approach when we are given 2 or more numbers.
LCM (Least Common Multiple)
LCM of 2 or more numbers is the product of highest powers of all the prime factors that are present in the numbers.
Find the LCM of 72 and 60
72 = 2^3 * 3^2
60 = 2^2 * 3*5
So, LCM = 2^3 * 3^2 *5 = 8*9*5 = 360
Note : For a set of numbers, HCF x LCM = Product of the numbers.
We can also view HCF as the Highest Divisor that exactly divides the numbers and LCM as the Lowest Divided which is exactly divisible by the numbers.
The Product of 2 numbers are 144. If the LCM is 16. What is the HCF ?
Product of 2 numbers = 144
Product = HCF x LCM
144 = 16 x HCF
So, HCF = 9
But Here, LCM is not a multiple of HCF. Hence this is not a possible case.
Hence the Scenario is not possible.
Note : LCM is always a multiple of HCF.
Problem Type :
We may be asked questions of type : Find the least number which when divided by a,b,c leaves remainder p,q,r
If you see these questions, you will always find a-p = b-q = c-r = x (say)
So the required number is always = LCM (a,b,c) – x
Another type of problem that may come is given below :
Find the least number which when divided by a,b,c leaves the remainder “k” in each case.
Solution to these sums are always : LCM (a,b,c ) + k
What is the greatest number which exactly divides 110, 154 and 242?
The required number is the HCF of 110, 154 & 242.
110 = 2 × 5 × 11
154 = 2 × 7 × 11
242 = 2 × 11 × 11
∴ HCF = 2 × 11 = 22
What is the greatest number, which when divides 3 consecutive odd numbers produces a remainder of 1
If x, y, z be 3 consecutive odd numbers, then the required number will be the
HCF of x – 1, y – 1 and z – 1.
Since x-1, y-1 & z-1 are 3 consecutive even integers, their HCF will be 2.
So answer is 2.
What is the highest 3 digit number, which is exactly divisible by 3, 5, 6 and 7?
The least no. which is exactly divisible by 3, 5, 6, & 7 is LCM (3, 5, 6, 7) = 210.
So, all the multiples of 210 will be exactly divisible by 3, 5, 6 and 7.
So, such greatest 3 digit number is 840. (210 × 4).
In a farewell party, some students are giving pose for photograph, If the students stand at 4 students per row, 2 students will be left if they stand 5 per row, 3 will be left and if they stand 6 per row 4 will be left. If the total number of students are greater than 100 and less than 150, how many students are there?
If ‘N’ is the number of students, it is clear from the question that if N is divided by 4, 5, and 6, it produces a remainders of 2, 3, & 4 respectively. Since (4 – 2) = (5 – 3) = (6 – 4) = 2, the least possible value of N is LCM (4, 5, 6) – 2 = 60 – 2, = 58. But, 100 < N < 150. So, the next possible value is 58 + 60 = 118.
There are some students in the class. Mr.X brought 130 chocolates and distributed to the students equally, then he was left with some chocolates. Mr Y brought 170 chocolates and distributed equally to the students. He was also left with the same no of chocolates as MrX was left. Mr Z brought 250 chocolates, did the same thing and left with the same no of chocolates. What is the max possible no of students that were in the class?
The question can be stated as, what is the highest number, which divides 130, 170 and 250 gives the same remainder, i.e. HCF (170-130) , (250-130) , (250-170)
I.e. HCF (40, 80, 120) = 40.
We have learnt about the HCF and LCM of integers. What about fractions.
Lets look at it
If we are said to find out the HCF of a/b, and x/y , a,b,x,y all being integers , then we will follow the following rule :
HCF of fractions = HCF of the Numerators / LCM of Denominators
If we are said to find out the LCM of a/b, and x/y , a,b,x,y all being integers , then we will follow the following rule :
LCM of fractions = LCM of the Numerators / HCF of Denominators
Be it HCF or LCM of the fractions, we will calculate the required part (HCF/LCM) on the Numerator and the other ( LCM if HCF is written and vice – versa) on the denominator
Numerator as usual will be in its original position and so will denominator.