Question

Some important tricks in HCF and LCM concept
(Tutorial) and
Some formula and common mistakes students make.
http://tutorial.math.lamar.edu/pdf/Algebra_Cheat_Sheet.pdf

Answer 1

\[\Huge HCF and LCM!\]

Answer 2

Subtopics to be covered down :-
1)LCM and HCF of fractions
2)Important tricks in HCF and LCM
3)Some intelligent problems. on the learnt concept

Answer 3

\[\Huge HCF=\frac{ LCM of numerator }{ HCF of denominator }\]
\[\Huge LCM=\frac{ LCM of numerator }{ HCF of denominator }\]

Answer 4

NOW , SOME TRICKS IN HCF AND LCM CONCEPT

Answer 5

1) Product of the two numbers = Their HCF * Their LCM

Answer 6

2) HCF of a given number always divides their LCM

Answer 7

3) 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

Answer 8

4) Largest number which divides x,y,z to leave the same remainder =
HCF of y-x , z-y , z-x

Answer 9

5) Largest number which divides x,y,z to leave a Remainder "R" =
HCF of x-R ,y-R , z-R

Answer 10

6) Largest numbers which divides x,y,z to leave the same remainder a,b,c =
HCF of x-a , y-b , z-c

Answer 11

7)Least number which when divided by x,y,z leaves a remainder R in each case=
(L.C.M of x,y,z)+R

Answer 12

\[\Huge PROBLEM1\]

Answer 13

Q) Least number when divided by 35 , 45, 55 and leaves remainder 18 , 28 , 38 is

Answer 14

SOLLUTION:- Here we will evaluate the L.C.M

Here the difference between every divisor and remainder is same ie. 17
Therefore the number we want is

LCM of (35,45,55)-17 = 3448

Answer 15

\[\Huge PROBLEM2\]

Answer 16

Q) Least number which when divided by 5,6,7,8 and leaves a remainder 3 , but when divided by 9 , leaves no remainder

Answer 17

Solution:-
L.C.M of 5,6,7,8 = 840
Required number = 840k+3
Least value of k for which (840k+3) is divided by 9 is 2
Therefore the number would be :-
840*2 + 3
= 1683

Answer 18

\[\Huge PROBLEM 3 \]

Answer 19

Q) Greatest four-digit number which is divisible by each one of 12,18,21 and 28 is ?

Answer 20

Sollution:-
Lcm of 12,18,21 and 28 = 252
Therefore required number must be divisible by 252
Greatest four-digit number = 9999
On dividing 9999 by 252 remainder = 171
therefore Answer is 9999-171 = 9828

Answer 21

@Abhisar please add it to your tutorial list

Answer 22

nice work, very useful... @No.name

Answer 23

Thank you

Answer 24

like it

Answer 25

Thank you

Answer 26

well job