# LCM and HCF Tricks

Candidates who are going to appear in any competitive exam can check LCM and HCF Shortcut Tricks and Formulas from here. As you all know that LCM and HCF is important topic and asked in many of exams, so who are engaged in preparing for government jobs can download PDF LCM and HCF Tricks in Hindi from here.

Lowest Common Multiple (LCM): LCM of two or more numbers is the smallest number that is exactly divisible by each of the given numbers.

Highest Common Factor (HCF): HCF of two or more numbers is the greatest number that divides each of them exactly.

LCM and HCF Important terms:

Factors: Factor is a number which exactly divides other number.

Example: 4 divides 24, so 4 is a factor of 24; while 5 does not divide 24 completely, so 5 is not a factor of 24

Factors of 24 are 1, 24, 2, 12, 3, 8, 4 and 6

Multiple: A number is said to be multiple of another number, when it is exactly divisible by other number.

Example: 14 is exactly divisible by 7, 14 is a multiple of 7

Common multiple: A common multiple of two or more numbers is a number which is exactly divisible by each of them.

Example: 18 is a common multiple of 2, 3, 6 and 9

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## Problems on LCM and HCF Shortcut Tricks with Formulas

 Trick 1

LCM and HCF of Fractions

Formula:  Solution:

LCM of numerators 2, 3 and 6 is 6

HCF of denominators 5, 10 and 25 is 5  समाधान:

4, 10 और 20 अंशो का HCF 2 है

9, 21 और 63 LCM 63 है Trick 2

Formula:

Product of two numbers = LCM of two numbers x HCF of two numbers

Example: The LCM of two numbers is 864 and their HCF is 144. If one number is 288, then find the other number.

Solution:

Product of two numbers = LCM of two numbers x HCF of two numbers

288 x Other Number = 864 x 144 ∴ Other Number =   = 432

उदाहरण: दो संख्याओं का LCM 225 तथा HCF 5 है. यदि उसमे  से एक संख्या 25 है, तो दूसरी संख्या ज्ञात करे |

समाधान:

दो संख्याओं का उत्पाद = दो संख्याओं का एलसीएम x दो संख्याओं का एचसीएफ

25 x दूसरी संख्या = 225 x 5 ∴ दूसरी संख्या =   = 45

Check Here – How to Prepare Maths for Competitive Exams

 Trick 3

Find the greatest number that will divide x, y and z leaving remainders a, b and c respectively.

Required number = HCF of (x – a), (y – b) and (z – c)

Example: Find the greatest number that will divide 148, 246 and 623 leaving remainders 4, 6 and 11 respectively.

Solution: The required greatest number = HCF of (148 – 4), (246 – 6) and (623 – 11)

= HCF of 144, 240 and 612 HCF of 144, 240 and 612 = 2 x 2 x 3 = 12

उदाहरण: वह अधिकतम संख्या क्या जिससे 989 तथा 1327 में भाग देने पर क्रमशः 5 और 7 शेष बचता हो?

समाधान:

आवश्यक सबसे बड़ी संख्या = (989 – 5) तथा (1327 – 7) का एचसीएफ

= 984 तथा 1320 का एचसीएफ 984 और 1320 का एचसीएफ = 24

 Trick 4

Find the least number which when divided by x, y and z leaves the remainders a, b and c, respectively. It is always observed that (x – a) = (y – b) = (z – c) = k (say)

Required number = (LCM of x, y and z) – k

Example: Find the least number which when divided by 36, 48 and 64 leaves the remainders 25, 37 and 53, respectively.

Solution: Since (36 — 25) = (48 – 37) = (64 – 53) = 11

Therefore, the required smallest number = (LCM of 36, 48 and 64) – 11 LCM of 36, 48 and 64 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 = 576

∴ (LCM of 36, 48 and 64) – 11 = 576 – 11 = 565

उदाहरण: कम से कम संख्या ज्ञात कीजिए, जिसको 36, 24 तथा 48 से भाग देने पर क्रमशः 33, 21 तथा 45 शेष बचे?

समाधान:

चूंकि (36 – 33) = (24 – 21) = (48 – 45) = 3

इसलिए, आवश्यक छोटी संख्या = (36, 24 तथा 48 का एलसीएम) – 3 36, 24 तथा 48 का एलसीएम = 2 x 2 x 2 x 2 x 3 x 3 = 144

∴ (36, 24 तथा 48 का एलसीएम) – 3 = 144 – 3 = 141

Also Check – Maths Formulas

 Trick 5

Find the least number which when divided by x, y and z leaves the same remainder r in each case.

Required number = (LCM of x, y and z) + r

Example: Find the least number which when divided by 12, 16 and 18, will leave in each case a remainder 5.

Solution: The required smallest number = (LCM of 12, 16 and 18) + 5 LCM of 12, 16 and 18 = 2 x 2 x 2 x 2 x 3 x 3 = 144

∴ (LCM of 12, 16 and 18) + 5 = 144 + 5 = 149

उदाहरण: कम से कम संख्या ज्ञात कीजिए, जिसको 6, 15 तथा 18 से भाग देने  पर प्रत्येक मामले में 5 शेष बचे?

समाधान:

आवश्यक छोटी संख्या = (6, 15 तथा 18 का LCM) + 5 6, 15 तथा 18 का LCM = 2 x 3 x 3 x 5 = 90

∴ (6, 15 तथा 18 का LCM) + 5 = 90 + 5 = 95

Download PDF – Problems on एलसीएम और एचसीएफ शॉर्टकट ट्रिक्स

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