HCF and LCM problems play a major role
quantitative aptitude test. There is no competitive exam without the
questions based on H.C.F and L.C.M . We have already learned this topic in our
lower
classes.Even though we have been already taught this topic in our lower
classes, we need to learn some more short cuts which are being used to
solve HCF and LCM problems .

The only thing we have to do is, we need to apply the appropriate short cut and solve the problems in a limited time. This limited time will be one minute or less than one minute in most of the competitive exams

On this web page of HCF and LCM problems, first we are going to see HCF

**Highest Common Factor (H.C.F) or Greatest Common Divisor (G.C.D):**

The H.C.F of two or more numbers is the greatest number that divides each of them exactly.

**Methods to find H.C.F: **

There are two methods to find H.C.F of given set of numbers

**1. Factorization Method:**

Express each one of the given numbers as the product of prime factors. The product of least powers of common prime factors gives H.C.F

**Example :**

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

Let us see, how H.C.F can be found for the given three numbers.

First let us write the given numbers as the product of prime factors.

108 = 2²X3³

288 = 2⁵X3²

360 = 2³X5X3²

When we look in to the prime factors of the given numbers, we find 2 and 3 in common of all the three numbers. The least power of 2 is 2 and 3 is also 2.

Now, to find the H.C.F, we just have to multiply 2² and 3².

**Hence, the H.C.F = 36**

**2. Division Method:**

To find the H.C.F of two given numbers using division method, please follow the following steps.

**Step 1 :** Divide the larger number by the smaller one. You will get some remainder

**Step 2 :****Now, divide the divisor (smaller one in the above) by the remainder of step 1.When you do so, again, you will get some remainder. **

**Step 3 :** Again you have to divide the divisor (remainder of step1) by the remainder of step 2.

We have to continue the same process, until we get the remainder zero.

It has been clearly shown in the following HCF and LCM problems.

**Example :**

**Find the H.C.F of 1134 and 1215. **

In the above example, the larger number is 1215 and the smaller number is 1134. As we explained above,we do the following steps.

**Step 1 :**

We divide the lager number 1215 by the smaller number 1134. When we do so, we get the remainder 81.

**Step 2 :**

Now we divide the divisor (1215) by the remainder of step 1 (that is 81).

**Step 3 :**

On continuing this process, we get remainder zero when we divide the divisor of step1 (that is 1215) by the remainder of step2 (that is 81).

**Here the H.C.F is 81**. Because we get the remainder zero when we divide by 81.

From the above example, it is very clear that H.C.F is nothing but the divisor for which we get the remainder is zero.

**How to find H.C.F for more than two numbers: **

If we want to find H.C.F of three numbers, first find H.C.F of any two numbers. Then, H.C.F of [H.C.F of two numbers and the third number] gives H.C.F of three numbers. In the same manner, H.C.F of more than three numbers may be obtained..

On this webpage of HCF and LCM problems, next, we are going to see L.C.M

**Least Common Multiple (L.C.M):**

The least number which is exactly divisible by each one of the given numbers is called their L.C.M.

**1. Factorization Method:**

Resolve each one of the given numbers in to a product of prime factors. Then, L.C.M is the product of highest powers of all the factors.

**Example:**

**Find the L.C.M of 108, 288 and 360.**

Let us see, how L.C.M can be found for the given three numbers.

First let us write the given numbers as the product of prime factors.

108 = 2²X3³

288 = 2⁵X3²

360 = 2³X5X3²

The prime factors we find above are 2,3 and 5. The highest power of 2 is 5, 3 is 3 and 5 is 1

To find L.C.M, we just have to multiply 2⁵, 3³ and 5

**Hence, the L.C.M = 4320**

**2. Common Division Method (Short-cut Method) :**

To find the L.C.M of two given numbers using division method, please follow the following steps.

**Step 1 : **Arrange the given numbers in a row in any order.

** Step 2 : **Divide by a number which divides exactly at least two of the given numbers and write the remaining numbers as it is.

** Step 3 : **Repeat the same process till no two of the numbers are divisible by the same number except 1.

**Step 4 : **The product of the divisors and undivided numbers is the required L.C.M of the given numbers.

It has been clearly shown in the following example.

**Example:**

**Find the L.C.M of 16, 24, 36 and 54 **

In the above calculation, the divisors are 2,2,2,3,3 and undivided numbers are 2,1,13. To find the L.C.M, we just have to multiply the divisors and undivided numbers.

Product = 2X2X2X3X3X2X1X1X3 = 432

**Hence, the L.C.M is 432. **

**Co.Primes :**

If the H.C.F of two numbers is 1, they are called as co-primes.

**Example:**

3 and 4 co-primes.Because, the H.C.F of 3 and 4 is 1 That is, there is no common factor between them except 1.

**H.C.F and L.C.M of Fractions :**

**Click hereto get HCF and LCM word problems**

HTML Comment Box is loading comments...