The GCF is 1.
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
(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)
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; } }
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 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); }
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