lcm and hcf tricks, problems and formulas _ bank exams today

8/12/2014 LCM and HCF tricks, problems and formulas | Bank Exams Today

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LCM and HCFLCM and HCFtricks, problemstricks, problems

and formulasand formulas

By: By: Ramandeep SinghRamandeep Singh on on 7/21/20137/21/2013

08:39:00 PM08:39:00 PM



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and I can put a face to a name!and I can put a face to a name!

Ramandeep Singh

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POPULAR POSTSPOPULAR POSTS

LL CM i.e. least commonCM i.e. least commonmultiple is a number which ismultiple is a number which ismultiple of two or more than twomultiple of two or more than two

numbers. For example: Thenumbers. For example: The

common multiples of 3 and 4 arecommon multiples of 3 and 4 are

12,24 and so on. Therefore,12,24 and so on. Therefore,

l.c.m.is smallest positive numberl.c.m.is smallest positive number

that is multiple of both. Here,that is multiple of both. Here,

l.c.m. is 12.. HCF i.e. highestl.c.m. is 12.. HCF i.e. highest

common factor are those integralcommon factor are those integral

values of number that can dividevalues of number that can divide

that number. LCM and HCFthat number. LCM and HCF

problems are very important partproblems are very important part

of all competitive exams. of all competitive exams.

SSOMEOME IMPORTANTIMPORTANT LL..CC..MM..ANDAND HH..CC..FF. . TRICKSTRICKS::

1) Product of two numbers = Their1) Product of two numbers = Their

h.c.f. * Their l.c.m.h.c.f. * Their l.c.m.

2) h.c.f. of given numbers always2) h.c.f. of given numbers always

divides their l.c.m.divides their l.c.m.

3) h.c.f. of given fractions = 3) h.c.f. of given fractions = h.c.f. h.c.f.

of numerator of numerator

l.c.m. of l.c.m. of

denominatordenominator

4) l.c.m. of given fractions = 4) l.c.m. of given fractions = l.c.m. l.c.m.

Quantitative Aptitude ShortcutsQuantitative Aptitude Shortcuts11

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4) l.c.m. of given fractions = 4) l.c.m. of given fractions = l.c.m. l.c.m.

of numerator of numerator

h.c.f. of h.c.f. of

denominatordenominator

5) If d is the h.c.f. of two positive5) If d is the h.c.f. of two positive

integer a and b, then there existinteger a and b, then there exist

unique integer m and n, such thatunique integer m and n, such that

d = am + bn d = am + bn

6) If p is prime and a,b are any6) If p is prime and a,b are any

integer then integer then P P ,This implies ,This implies P P or or PP

ab a b ab a b

7) h.c.f. of a given number always7) h.c.f. of a given number always

divides its l.c.m.divides its l.c.m.

MMOSTOST IMPORTANTIMPORTANT POINTSPOINTSABOUTABOUT LL..CC..MM. . ANDAND HH..CC..FF..

PROBLEMSPROBLEMS : :

1) Largest number which divides1) Largest number which divides

x,y,z to leave same remainder =x,y,z to leave same remainder =

h.c.f. of y-x, z-y, z-x.h.c.f. of y-x, z-y, z-x.


x,y,z to leave remainder R (i.e.x,y,z to leave remainder R (i.e.

same) = h.c.f of x-R, y-R, z-R.same) = h.c.f of x-R, y-R, z-R.


x,y,z to leave same remainder a,b,cx,y,z to leave same remainder a,b,c

= h.c.f. of x-a, y-b, z-c. = h.c.f. of x-a, y-b, z-c.



= h.c.f. of x-a, y-b, z-c. = h.c.f. of x-a, y-b, z-c.

4) Least number which when4) Least number which when

divided by x,y,z and leaves adivided by x,y,z and leaves a

remainder R in each case = ( l.c.m.remainder R in each case = ( l.c.m.

of x,y,z) + Rof x,y,z) + R

HCF HCF ANDAND LCM LCMQUESTIONSQUESTIONS::

Problem 1Problem 1: Least number which: Least number which

when divided by 35,45,55 andwhen divided by 35,45,55 and

leaves remainder 18,28,38; is?leaves remainder 18,28,38; is?

SolutionSolution: i) In this case we will: i) In this case we will

evaluate l.c.m.evaluate l.c.m.

ii) Here the dierence ii) Here the dierence

between every divisor andbetween every divisor and

remainder is same i.e. 17.remainder is same i.e. 17.

Therefore, required Therefore, required

number = l.c.m. of (35,45,55)-17 =number = l.c.m. of (35,45,55)-17 =

(3465-17)= 3448.(3465-17)= 3448.

Problem 2Problem 2: Least number which: Least number which

when divided by 5,6,7,8 and leaveswhen divided by 5,6,7,8 and leaves

remainder 3, but when divided byremainder 3, but when divided by

9, leaves no remainder?9, leaves no remainder?

SolutionSolution: l.c.m. of 5,6,7,8 = 840: l.c.m. of 5,6,7,8 = 840

Required number = 840 Required number = 840

k + 3k + 3

Least value of k for which Least value of k for which

(840 k + 3) is divided by 9 is 2(840 k + 3) is divided by 9 is 2

Therefore, required number =Therefore, required number =

840*2 + 3840*2 + 3

= 1683 = 1683



= 1683 = 1683

Problem 3Problem 3: Greater number of 4: Greater number of 4

digits which is divisible by each onedigits which is divisible by each one

of 12,18,21 and 28 is?of 12,18,21 and 28 is?

SolutionSolution: l.c.m. of 12,18,21,28 =: l.c.m. of 12,18,21,28 =

254254

Therefore, required Therefore, required

number must be divisible by 254.number must be divisible by 254.

Greatest four digit number Greatest four digit number

= 9999= 9999

On dividing 9999 by 252, On dividing 9999 by 252,

remainder = 171remainder = 171

Therefore, 9999-171 = Therefore, 9999-171 =

9828.9828.14,383 people like this. Be the f irst of

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Labels: Labels: prepprep, , prep quantprep quant, , quantquant

7 COMMENTS:7 COMMENTS:

MMUSTUST R READEADPPREVIOUSREVIOUSAARTICLESRTICLES :- :-

Ratio and Proportions :Ratio and Proportions :

Tricks, questions andTricks, questions and

FormulaeFormulae

Quantitative AptitudeQuantitative Aptitude

ShortcutsShortcuts

SBI Specialist OcersSBI Specialist Ocers

Syllabus and Study PlanSyllabus and Study Plan

SSC CGL Tier I SyllabusSSC CGL Tier I Syllabus

and Study Planand Study Plan

Permutation andPermutation and

Combination - ConceptCombination - Concept

and Problemsand Problems

Probability - ShortcutsProbability - Shortcuts

and Questions withand Questions with

AnswersAnswers

Logical ReasoningLogical Reasoning

Concepts by Rahul RanjanConcepts by Rahul Ranjan

1. JASSJASS



October 12, 2013 at 6:10 PMOctober 12, 2013 at 6:10 PM

ReplyReply

plz send me plz send me word word list onlist on

[email protected][email protected]

....................


ReplyReply

2. RAMAN2572RAMAN2572

It is a Trial and ErrorIt is a Trial and Error

method.method.

When yo take K = 1, 843When yo take K = 1, 843

is not divisible by 9is not divisible by 9

but when we take K = 2,but when we take K = 2,

1683 is divisible by 9.1683 is divisible by 9.


ReplyReply

3. NAVAJIT BORPATRANAVAJIT BORPATRA

can u plz explain howcan u plz explain how

did u get value of k=2 indid u get value of k=2 in

problem no. 2problem no. 2


ReplyReply

4. RAMAN2572RAMAN2572

Please downloadPlease download

wordlist from herewordlist from here

https://app.box.com/s/fbut5z9csridsp5nn5ehhttps://app.box.com/s/fbut5z9csridsp5nn5eh

February 16, 2014 at 1:30 PMFebruary 16, 2014 at 1:30 PM

5. Anne EmersonAnne Emerson

hey sir , there ishey sir , there is

contradiction betweencontradiction between

MOST IMP pt 4 ndMOST IMP pt 4 nd



RepliesReplies

ReplyReply

EXAMPLE 1.......+R OR -REXAMPLE 1.......+R OR -R

????????

IN POINT 4 YOU AREIN POINT 4 YOU ARE

SAYING +R bt inSAYING +R bt in

EXAMPLE -R ???????EXAMPLE -R ???????

please clear ..../././..please clear ..../././..

THANKEW SIRTHANKEW SIR

February 16, 2014 at 1:35 PMFebruary 16, 2014 at 1:35 PM

ReplyReply

6. Anne EmersonAnne Emerson

YES sir ,, i conform .././.YES sir ,, i conform .././.

In problem 1 you shouldIn problem 1 you should

add +17 .../././....andadd +17 .../././....and

answer should 3482..././.answer should 3482..././.

February 27, 2014 at 12:19 PMFebruary 27, 2014 at 12:19 PM1. AkhileshAkhilesh

you better checkyou better check

before publishingbefore publishing

wrong thingswrong things

like............like............

4) Least number4) Least number

which whenwhich when

divided by x,y,zdivided by x,y,z

and leaves aand leaves a

remainder R inremainder R in

each case = ( l.c.m.each case = ( l.c.m.

of x,y,z)+ Rof x,y,z)+ R



Copyright 2014 Copyright 2014 Bank Exams TodayBank Exams Today. All Rights Reserved. Authored By . All Rights Reserved. Authored By Ramandeep SinghRamandeep Singh

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