40 c;
# include<stdio.h> main() { int a,b,c; print f("enter the values of a,b,c"); scan f("%d%d%d",&a,&b,&c); if((a>b)&&(a>c)) print f("Greatest value is a =%d",a); else if((b>a)&&(b>c)) print f("Greatest value is b=%d",b); else print f("Greatest value is c=%d",c); }
dim a,b,c a=cint(inputbox("enter value for a")) b=cint(inputbox("enter value for b")) c=cint(inputbox("enter value for c")) if((a>b)and(a>c)) then msgbox "greatest number is a="&a else if ((b>a)and(b>c)) then msgbox "greatest number is b="&b else msgbox "greatest number is c="&c end if end if
public class FindLeastAndGreatest { public static void main(String[] args) { // number can't be equal with each other int a = 7; int b = 7; int c = 6; System.out.println(least(a,b,c)); System.out.println(greatest(a,b,c)); } public static int least(int a, int b, int c) { int least = 0; if(a < b && a < c) { least = a; } else if(b < a && b < c) { least = b; } else { least = c;} return least; } public static int greatest(int a, int b, int c) { int greatest = 0; if(a > b && a > c) { greatest = a; } else if(b > a && b > c) { greatest = b; } else { greatest = c;} return greatest; } }
(a/b) divided by (c/d) =(a/b) x (d/c) <-- notice divisor is inverted (turned upside down) now multiply both numerators and both denominators = (ad)/(bc)
If the greatest common factor/divisor of A and B is 1 then they are coprime - they do not share any prime factors. Multiplying both through by C means, obviously, that each number now divides by C. In fact, C is their greatest common divisor, since AC and BC do not have further common factors after C is taken out. Hence the GCF of AC and BC is not merely a factor of C - it is C. (The question makes sense only if A, B and C are integers.)
40 c;
The greatest factors of A, B, and C, respectively, are the absolute values of A, B, and C. The greatest common factor of A, B, and C is 1.
A,b,c
The GCF is 1.
A/B = C A is the dividend B is the divisor C is the quotient
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 greatest common multiple of any set of integers is infinite.
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 answer depends on whether or not a is a factor of c.
the only common factor is 1 b/c 11 is a prime number.
# include<stdio.h> main() { int a,b,c; print f("enter the values of a,b,c"); scan f("%d%d%d",&a,&b,&c); if((a>b)&&(a>c)) print f("Greatest value is a =%d",a); else if((b>a)&&(b>c)) print f("Greatest value is b=%d",b); else print f("Greatest value is c=%d",c); }