Short Tricks LCM HCF In Hindi

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.
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.

Q4Find 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 :-Firstly show the given digits into indivisible multiplication.
Select the indivisible multiplication with the biggest base, which is inserted in any multiplication digit.
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.