number system is very important chapter for all competitive exam . everything is included in it like all numbers,some special number,divisibility concept,unit digit,divisibility remainder concept,decimal and number of factor,key to remainder concept,lcm and hcf etc.
LCM AND HCF
Highest Common Factor (H.C.F): - When a largest(greatest) number divides perfectly two or more numbers it is called hcf.
We can find the H.C.F of two or more number by division method and factorization method
Formula: H.C.F of two or more numbers = product of common prime factors of lowest power.
Most important key
#1st number × 2nd number = LCM × HCF
#Product of n number = LCM × HCF^n-1
(IF more than two numbers)
#HCF of x/a , y/b , z/c = (HCF of x,y,z)/LCM of a,b,c
#LCM of x/a , y/b , z/c = (LCM of x,y,z)/HCF of a,b,c
Example: H.C.F (12,288)
12 = 2^2 x 3 and 288 = 2^5 x 3^2
So, H.C.F=2^2 x 3=12
Lowest Common Multiple (L.C.M): -The least number which is divisible by two or more given numbers, that least number is called L.C.M. of the numbers.
Formula: L.C.M. of two or more numbers = product of highest power of each prime factors
EXAMPLE: L.C.M. (24,288)
12= 2^2 x 3 and 288 = 2^5 x 3^2
So, the L.C.M of (12, 288)= 2^5 x 3^2=288
REMEMBER KEY
L.C.M of fractions = L.C.M of numerators/ H.C.F of denominators
H.C.F of fractions = H.C.F of numerators/L.C.M of denominators
Example:-L.C.M of 5/4, 8/3, 7/2 = L.C.M of (5, 8, 7)/ H.C.F of (4, 3, 2)
So, 280/1 = 280
Example:- H.C.F of 5/4, 8/3, 7/2 = H.C.F of (5, 8, 7)/L.C.M of (4, 3, 2)
So, 1/12
IMPORTANT TRICK : -
* find the least number which is completely divisible by m,n,o = lcm of m,n,p
* When a number is divided by m, n or o leaving same remainder 'r' in each case then that number must be k + r where k is lcm of m,n,o. We can say Or lcm of m,n,o + r.
* When a number is divided by m, n or o leaving remainders p, q or r respectively such that the difference between divisor and remainder in each case is same i.e, (m-p) = (n-q)=(o-r) = x (say), then that (least) number must be in the form of (k-x), where k is LCM of m, n and o. Or find the least no. Which is divided by m,n,o leaves remainder a,b,c respectively in this case always m-a = n-b = o-c =k(let) after that lcm of ( m,n,o )-k.
* The largest number which when divide the numbers m, n and o the remainders are same then that largest number is given by H.C.F. of (m - n), (n-o) and (o-p).
* The largest number which when divide the numbers m, n and o give remainders as p, q, r respectively is given by H.C.F. of (m-p), (n-q) and (o-r).
* Greatest n digit number which when divided by three numbers m, n, o leaves no remainder will be Required Number = (N-digit greatest number)-R, R is the remainder obtained on dividing greatest N digit number by L.C.M of m, n, o.
* The n digit largest number which when divided by m, n, o leaves remainder 'a' will be required number = [N-digit largest number - R] + a where, R is the remainder obtained when N-digit largest number is divided by the L.C.M of m,n,o.
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