**HCF and LCM Formula**

To find HCF and LCM with shortcut tricks is very necessary to know while you are preparing for any competitive or entrance exam because it safes your time and the chances of accuracy in short time arises. So, let’s start with the topic of **HCF and LCM Formula** in which we will discuss about shortcuts of HCF and LCM with examples.

Sometimes, it taken two long to find HCF and LCM to solve the questions of calculative part in the exam but if you follow some guidelines of HCF and LCM Shortcut Tricks presented on this page then you will definitely solve HCF and LCM in less time. Here in this article, you will find the basic ideas of how to solved question on HCF and LCM with short tricks.

Get the detailed information about HCF and LCM Formula and its shortcut tricks with examples by continue reading this page that is prepared by the expert team members of www.recruitmentinboxx.com for the ease and comfort of readers.

**HCF and LCM Meaning**

LCM is also known as *Least Common Multiple*. Least Common Multiple is a number which is a multiple of two or more than two numbers. LCM is smallest positive number that is multiple of both. HCF is also known as *Highest Common Factor.* Highest Common Factors are those integral values of number that can divide that number.

LCM can be find in two ways that is Factorization Method which expresses each one of the given numbers as the product of prime factors. The product of highest powers of all the factors gives LCM. Second way is Division Method or Decimals Method where we have to convert decimal into integers by making the number of digits after decimal point is common.

Highest Common Factor can be find through Prime Factorization Method which expresses each one of the given numbers as the product of prime factors. The product of least powers of common prime factor gives HCF. Also, you can find Highest Common Factor by Long Division Method.

Know Here: __How to Find HCF And LCM by Using Division Method, Prime Factorization__

**HCF and LCM Formula**

**HCF and LCM Tricks**

- Product of two numbers = Their HCF* Their LCM
- HCF of given numbers always divides their LCM
- HCF of given fractions = HCF of numerator/ LCM of denominator
- LCM of given fractions = LCM of numerator/ HCF of denominator
- If d is the HCF of two positive integer a and b, then there exist unique integer m and n, such that d = am + bn
- If p is prime and a, b are any integer then P/ ab ,This implies P/ a or P/b
- HCF of a given number always divides its LCM

**Points for HCF and LCM Problems**

- Largest number which divides x,y,z to leave same remainder = HCF of y-x, z-y, z-x.
- Largest number which divides x,y,z to leave remainder R (i.e. same) = HCF of x-R, y-R, z-R.
- Largest number which divides x,y,z to leave same remainder a,b,c = HCF of x-a, y-b, z-c.
- Least number which when divided by x,y,z and leaves a remainder R in each case = ( LCM of x,y,z) + R

Get Here Basic/ Vedic Formulas Topic Wise: __Maths Formulas PDF__

**Highest Common Factor**

__Prime Factorization Method__

Example: Find the HCF of (36,48)

⇒ 36=2² х 3²

⇒ 48=2⁴ х 3

⇒ So 2² х 3 is the common least factor and the answer is 12

__Long division Method__

**Least Common Multiple**

__Factorization Method__

__Division Method__

Also Read Out: __How to Prepare Maths For Competitive Exams__

**Shortcut Tricks To Find HCF and LCM With Examples**

**Problem 1:** Least number which when divided by 35,45,55 and leaves remainder 18,28,38; is?

Solution: In this case we will evaluate LCM.

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.

**Problem 2:** Least number which when divided by 5,6,7,8 and leaves remainder 3, but when divided by 9, leaves no remainder?

Solution: LCM of 5,6,7,8 = 840

required number = 840 k + 3

Least value of k for which (840 k + 3) is divided by 9 is 2

Therefore, required number = 840*2 + 3 = 1683

**Problem 3:** Greater number of 4 digits which is divisible by each one of 12,18,21 and 28 is?

Solution: LCM of 12,18,21,28 = 254

therefore, required number must be divisible by 254.

Greatest four digit number = 9999

on dividing 9999 by 252, remainder = 171

Therefore, 9999-171 = 9828

**Problem 4:** The sum of two numbers is 216 and their HCF is 27. How many pairs of such numbers are there?

Solution: In this two numbers x and y and HCF is 27

27x+27y=216

27(x+y) =216

x+y=216/27

x+y=8

Now we check prime multiplication and it is

7+1=8 ⇾1st pair co-prime

5+3=8⇾2nd pair

4+4=8⇾ not right pair

Read Out: __Aptitude Questions And Answers__

**Problem 5:** Find the HCF of (42,49,56) that leaves different remainders(0,1,2)

Solution:

HCF of [(42-0) (49-1) (56-2)] HCF of [42, 48, 54] Now we take HCF of these numbers and we get the answer as 6.

**Problem 6:** Find the HCF of (201,333,339) that leaves same remainder 3 in each the case

Solution:

HCF of [(201-3)(333-3)(339-3)] HCF of[(198)(330)(396)] Now we take HCF of these numbers and we get the answer as 66

**Problem 7:** Find the greatest number that can divide (62,132,237) that leaves same remainder in each case

Solution:

HCF of [(132-62)(237-132)(237-62)] HCF of [70,105,175] Now we take HCF of these numbers and we get the answer as 35.

**Problem 8:** Six bells commence tolling together and toll at intervals of 2,4,6,8,10 and 12 seconds respectively. In 30 minutes how many times do they toll together?

Solution:

Take LCM of 2,4,6,8,10 and 12 is 120

So the bell will toll together every 120sec = 2min

In 30 minutes they will toll together (30/2)+1

And the answer is 16 times.

Choose Best institutes For Study: __Best Institutes for Mathematics__

**Problem 9:** Traffic lights at three different road crossing change after every 48 sec, 72sec, and 108sec respectively. If they all change simultaneously at 8:20:00 hours then they will again change simultaneously at what time?

Solution:

Take LCM of 48,72,108 is 432 sec

Now 432 sec =7 min 12 sec

Now add the time (8hr 20min 0sec)+(7min 12 sec)

So, signal will again change simultaneously at 8:27:12.

**Problem 10:** Find the LCM of (8,9,12,15) that leaves same remainder 1 in each the case.

Solution:

⇒LCM of [8,9,12,15]
⇒2³*3²*5

⇒360

⇒à[360]+1

⇒Now we get the answer as 361

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