लघुत्तम महत्तम समापवर्त्य: Short Tricks LCM HCF In Hindi | Old Problems; Formulas हिंदी में लघुत्तम समापवर्त्य विधि एवं महत्तम समापवर्त्य विधि

हिंदी में  लघुत्तम समापवर्त्य विधि एवं महत्तम समापवर्त्य विधि |Short Tricks LCM HCF in Hindi  | Methods of Solving LCM and HCF Problems

In this post we are going to share some most important short tricks of LCM and HCF questions. We are also share short tricks in Hindi as well as in English language. These short tricks is very halpful for those candidates who are participating into various state and national level competitive exam.

LCM of Power and Base

लघुत्तम समापवर्त्य विधि 

LCM Trick -1 

If the base of given digit is same and power is not same or different,then LCM will be of the maximum power of the number.

LCM Trick-2

If the power and exponent are not same or different, then its LCM will get by factorization method.

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Some LCM HCF Questions from Previous competitive Exams
Q1	Find the largest number of four digits exactly divisible by 12, 15, 18 and 27. 4 डिजिट की बड़ी संख्या बटेए जो 12, 15 , 18, और 27 से पूर्णा विभाजित हो
Sol:	The largest number of four digits is 9999. Required number must be divisible by LC.M. of 12, 15, 18, 27 i.e.. 540. On dividing 9999 by 540, we got 279 as remainder. Required number = (9999 — 279) = 9720.
Q2	Find the smallest number of five digits exactly divisible by 16 24, 36 and 64. 5 डिजिट की छोटी संख्या बताइये जो 16, 24, 36, अवाम 64 से पूर्णा विभाजित हो।
Sol:	Smallest number of five digits is is 10000. Required number must be divisible by L.C.M. of 16, 24, 36, 54 i.e., 432. On dividing 10000 by 432, we get 64 as remainder. .•. Required number = 10000 + (432 — 64) = 10368.
Q3	Find the least number which when divided by 20, 25, 35 and 40 leaves remainders 14, 19, 29 and 34 respectively. वह नंबर ज्ञात करो जिसे 20, 25 , 35, और 40 से विभाजित से करने पर शेष्फल 14, 19, 29 और 34 हो।
Sol:	Here, (20 — 14) = 6, (25 — 19) = 6, (35 — 29) = 6 and (40 — 34) = 6. Required number =(L.C.M. of 20, 25, 35, 40)— 6 = 1394.
Q4	Find the least number which when divided by 5 8, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder. वह संख्या बातेए जो 5,8,7, और 8 से विभाजित करने पर 3 शेष्फल लेकिन 9 से विभाजित करने पर कोई शेष्फल न आए
Sol:	. L.C,M. of 5, 6, 7, 8 is 840. Required number is of the form 840k + 3. Least alue of k for which (840k + 3) is divisible by 9 is k = 2.   Required number (840 x 2 + 3) = 1683.
Q5	The traffic lights at three different road crossings change after every 48 sec., 72 sec. and 108 sec. respectively. If they all change simultaneously at 8: 20 : 00 hours, then at what time will they again change simultaneously? 3 अलग अलग रोड की ट्रेफिक सिग्नल की लाइट 48sec, 72sec और 108sec मे बदलती है। यदि वे सभी 8:20:00 बजे एक साथ बदले तो अगली बार कब एक साथ बद्लगी ।
Sol:	Interval of change = (LCM. of 48, 72, 108) sec.= 432 sec. So, the lights will again change simultaneously after every 432 seconds i.e., 7 min. 12 sec. Hence, next simultaneous change will take place at 8: 27: 12 hrs.

Prime Factorization Method

Process :-

1.    Firstly show the given digits into indivisible multiplication.
2. Select the indivisible multiplication with the biggest base, which is inserted in any multiplication digit.
3. Now multiply the selected indivisible multiplication and get the LCM.

EXAMPLE- Find the LCM of 18,28,108 & 105.

       Solution:- Here,   18 = 2 x 3 x 3 = 2 x 3^2

                                   28 = 2 x 2 x 7 = 2^2 x 7

                                  108 = 2 x 2 x 3 x 3 x 3 = 2^2 x 3^3

                         and   105 = 3 x 5 x 7

     Requried LCM  = 2^2 x 3^3 x 5 x 7 

                               = 4 x 27 x 5 x 7 

                               = 3780    Ans

 

Note:- Here, The biggest base number of 2 & 3 is 2^2 & 3^3 , and second indivisible multiplications are 5 & 7.

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Division method for LCM

Process:-

1.        Arrange the given digit in a row and divide it by 2,3,5,7,11 etc. from this the least number which is divisible should be taken and that number should be divisible by at least two digits of the given numbers
2.             After that write the remaining quotient and numbers which are not totally divided in next row.
3.  Repeat the first step again and again. It is possible that we have to repeat any process.
4.    By multiplying all the divisor we will get the required LCM.

Example:     Find the LCM of 36 , 60 , 84 & 90 .

LCM of decimal

Process:-

                  To find out the LCM of given decimal digit , find out the LCM of hole relative digit, then drop the decimal from the right side of the digit as it was earlier in the digit.

          

         Example: find out the LCM of  2.4 , 0.36 & 0.045 .

Note:- here in 2.4 there is one number after the decimal because of this in LCM 360 decimal is dropped from the right side before 0.

LCM of Fractions

FORMULA:

LCM of fraction     =                        LCM of numerator / HCF of denominator

Example: Find out the LCM of  1/3 , 2/9 , 5/6 & 4/27

H.C.F. महत्तम समापवर्त्य विधि 

 HCF of power and base

HCF Trick -1 

                         If the base of given digit is same and power is not same or different, then HCF will be of the maximum power of a number.

Example: Find the HCF of 2^8 , 2^10 and 2^15

HCF Trick-2

                         If the power and exponent both are different, then its HCF will get by factorization method

Example 1 : Find out the HCF of 5^2 & 4^3.

Example 2 : Find out the HCF of 2^-2 , 4^-3 & 6^-2.

Prime factorization method for HCF

 Process:-

1.       Firstly show the given digits into indivisible fractions.
2.       The fractions which are present in all digit, the product of that fractions will be required HCF.

     Example: Find HCF of 28 & 32.

Continued Division Method for HCF

Process:-

1.       From the given digits firstly divide bigger number by smaller number.
2.      Then from the reminder divide it by divisor, hence we get next divisor.
3.      Repeat this process till reminder will be zero.
4.      So that the last divisor will be required HCF.

Example:      find the HCF of 493 and 928.

Method for finding HCF of 3 or more digit by division method

Process:-

1.       Firstly find out the HCF of any two numbers by continuous division method.
2.     Then find HCF from that HCF which we get by the first step and remaining of given digit.
3.             Repeat this process with all remaining digits.
4.      At last, the HCF we get will be the required HCF.

Example: Find the HCF of 828 , 1311 & 1955.

HCF of decimals

Process:-

                  To find out the HCF of given decimal digit , find out the HCF of hole relative digit, then drop the decimal from the right side of the digit as it was earlier in the digit.

Example: Find out the HCF of 1.5 , 0.24 & 0.036.

HCF of fraction

FORMULA:-

HCF of the fractions  =  HCF of numerator / LCM of denominator

Example:  Find out the HCF of 16/21 , 8/15 , 2/3 & 24/27

>

Solved questions of LCM & HCF | Download LCM, HCF Privious year questions and solution

• The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:

• 
  The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is:

Explanation:

Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4. So, the numbers 12 and 16.          L.C.M. of 12 and 16 = 48.

24.	Which of the following fraction is the largest ?

Explanation:

L.C.M. of 8, 16, 40 and 80 = 80.

7	=	70	;	13	=	65	;	31	=	62
8		80		16		80		40		80

Since,	70	>	65	>	63	>	62	, so	7	>	13	>	63	>	31
	80		80		80		80		8		16		80		40

So,	7	is the largest.
	8
	The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:
	Explanation: Let the numbers be 2x and 3x.    Then, their L.C.M. = 6x.     So, 6x = 48 or x = 8.  The numbers are 16 and 24.    Hence, required sum = (16 + 24) = 40.

The H.C.F. of 9 , 12 , 18 and 21 is: 10 25 35 40

Explanation: Required H.C.F. = H.C.F. of 9, 12, 18, 21 = 3 L.C.M. of 10, 25, 35, 40 1400

The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:

Explanation: Let the numbers be 37a and 37b.   Then, 37a x 37b = 4107  ab = 3.     Now, co-primes with product 3 are (1, 3). So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).  Greater number = 111.

• Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:

Explanation:

Let the numbers be 3x, 4x and 5x. Then, their L.C.M. = 60x. So, 60x = 2400 or x = 40.

•   The numbers are (3 x 40), (4 x 40) and (5 x 40).  Hence, required H.C.F. = 40.

The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:

Explanation: Clearly, the numbers are (23 x 13) and (23 x 14).   Larger number = (23 x 14) = 322.

• The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

Explanation:

Greatest number of 4-digits is 9999.    L.C.M. of 15, 25, 40 and 75 is 600. On dividing 9999 by 600, the remainder is 399.    Required number (9999 - 399) = 9600.

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