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The greatest common divisor (GCD), of two or more non-zero integers, is the largest positive integer that divides the numbers without a remainder
The lowest common multiple (LCM), of two integers and is the smallest positive integer that is a multiple both of and of .
General rule
To find the HCF of 24 of 36, express both using prime factorization:
24 --> 2 * 2 * 2 * 3
36 --> 2 * 2 * 3 * 3
The HCF is thus 2 * 2* 3 = 12 (pick out the common terms)
To find the LCM, again express both using prime factorization:
The LCM is 2 * 2 * 2 * 3 * 3 = 72(pick out the common terms, then multiply in the remaining terms)
Properties of GCD and LCM
For any two consecutive integers the GCM can only be 1.
GCD X LCM = Product of two numbers
LC
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The least common multiplier of A and B is 120, the ratio of A and B is 3:4, what is the largest common divisor?
Soln
Let the nos be 3x and 4x . Hence their GCD will be x
GCD X LCM = Product of two numbers
120 X x = 3x X 4x
x=10
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1. What is the smallest possible common multiple of two integers, which are both greater than 250.
1) 251
2) 252
3) 502
4) 750
5) 884
2. What is the greatest possible common divisor of two different positive integers which are less than 144?
a) 143
b) 142
c) 72
d) 71
e) 12
1. Smallest possible number greater than 250 is 251.
Now we have to look for smallest multiple of 251. Why ? Because then 251 will be a factor of that number .
Now smallest multiple of 251 = 251*2 = 502
So the two numbers are 251 and 502 respectively and their LCM is 502. Hence C.
2. Greatest number less than 144 is 143 = 11*13
So maximum possible value of GCD of 143 and any number less that 143 will be = 13
Now consider the largest number less than 143 which is = 142 = 2*71
Now maximum possible value of GCD of 142 and any number less that 142 will be = 71
The number which is less than 144 as well as GCD of which and of 142 is 71 will be =71
Hence D.
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What is the greatest common divisor of positive integers m and n?
1) m is prime number
2)m and n are consecutive integers
1) m is prime number
then it is divisible by 1 and itself
now let n = multiple of m, therefor GCD is m
let is not a multiple of m then GCD = 1
so we can not determine
2) m and n are consecutive integers
let n = (m+1) GCD will always be 1
Hence B