The least common multiple (LCM) of two numbers is the smallest number that is a multiple of both numbers. In this case, the LCM of ab and bc would be the product of the two numbers divided by their greatest common divisor (GCD), which is b. Therefore, the LCM of ab and bc is abc.
The least common multiple (LCM) of two numbers is the smallest positive integer that is a multiple of both numbers. In this case, the numbers a-b and b-a are simply the same numbers but written in a different order. Therefore, the LCM of a-b and b-a would be the absolute value of the difference between a and b.
If two numbers are expressed as ab and cb this is easier to work out. Assume that a and c have no common prime factors. Thus, the HCF of the two numbers will be b. The LCM is the two numbers multiplied by each other, divided by the HCF. So the LCM will be abc. b is a factor of abc, and so the HCF will always be a factor of their LCM.
The LCM of 9b and 12 b is 36b. LCM is Least Common Multiple.
Shell program to find LCM and hcf of two no.s tput clear echo "Enter first no" read a echo "Enter 2nd no" read b p= 'expr $a \* $b' while [$b -ne 0] do r= 'expr $a % $b' a=$b b=$r done LCM = 'expr $p / $a' echo "LCM = $LCM" echo "Hcf = $a"
lcm(a,b,c,d) = lcm(lcm(a,b,c),d) = lcm(lcm(a,b),lcm(c,d))
The LCM of a^2b and b^2c is a^2b^2c
It depends on the values of B and C but without that knowledge, it is 7*B*C.
To calculate the least common multiple (lcm) of decimals (integers) and fractions you first need to calculate the greatest common divisor (gcd) of two integers: int gcd (int a, int b) { int c; while (a != 0) { c = a; a = b % a; b = c; } return b; } With this function in place, we can calculate the lcm of two integers: int lcm (int a, int b) { return a / gcd (a, b) * b; } And with this function in place we can calculate the lcm of two fractions (a/b and c/d): int lcm_fraction (int a, int b, int c, int d) { return lcm (a, c) / gcd (b, d); }
The LCM of 2, 3 and 5 is 30. The LCM of any three relatively prime numbers a, b and c is a*b*c, and here, 2 is a, 3 is b and 5 is c. Since 2, 3 and 5 are relatively prime, the LCM is 2*3*5=30.
For this you will need a couple of helper algorithms. The first is the GCD (greatest common divisor) which is expressed as follows:procedure GCD (a, b) isinput: natural numbers a and bwhile ab doif a>blet a be a-belselet b be b-aend ifend whilereturn aThe second algorithm is the LCM (least common multiple) of two numbers:procedure LCM (a, b) isinput: natural numbers a and b return (a*b) / GCD (a, b)Now that you can calculate the GCD and LCM of any two natural numbers, you can calculate the LCM of any three natural numbers as follows:procedure LCM3 (a, b, c) isinput: natural numbers a, b and c return LCM (LCM (a, b), c)Note that the LCM of three numbers first calculates the LCM of two of those numbers (a and b) and then calculates the LCM of that result along with the third number (c). That is, if the three numbers were 8, 9 and 21, the LCM of 8 and 9 is 72 and the LCM of 72 and 21 is 504. Thus the LCM of 8, 9 and 21 is 504.
#include<stdio.h> main() { int a,b,i,lcm,gcf; printf("\n Enter two numbers"); scanf("%d%d",&a,&b); for(i=0;i<=a;i++) { if((b%i==0)&&(a%i==0)) { gcf=i; } } lcm=a*b/gcf; printf("\n GCF is %d and LCM is %d",gcf,lcm); }
C. 12b3
Based on gcd: int LCM (int a, int b) { . int d= gcd (a, b); . if (d==0) return 0; . else return a/d*b; }
The least common multiple (LCM) of two numbers is the smallest number that is a multiple of both numbers. In this case, the LCM of ab and bc would be the product of the two numbers divided by their greatest common divisor (GCD), which is b. Therefore, the LCM of ab and bc is abc.
A and D both have LCMs of 72. The LCM of B is 12 and the LCM of C is 36.
The LCM of and b does not equal the LCM of a and b - a.