800
if the gcd and lcm are given and one of the numbers are also given,multiply the gcd and lcm and divide them by the given number
I think you mean either the GCD or the LCM? Not sure which since they are relatively prime, the LCM will the the product of the three numbers and the GCD is 1
If you have the gcd or the LCM of two numbers, call them a and b, you can use the relationship that gcd(a,b) = (a multiplied by b) divided by LCM (a,b) where LCM or gcd (a,b) means the LCM or a and b. This means the gcd multiplied by the LCM is the same as the product of two numbers. Let's assume you have neither. There are several ways to do this. One way to approach both problems at once is to factor each number into primes. You can use these prime factorizations to find both the LCM and gcd To compute the Greatest common divisor, list the common prime factors and raise each to the least multiplicities that occurs among the several whole numbers. To compute the least common multiple, list all prime factors and raise each to the greatest multiplicities that occurs among the several whole numbers.
If we multiply the gcd and the LCM, we get the numbers.Call the numbers a and b. So 16(LCM)=ab3584=ab let's all the LCM, x 16x=a(3584/a)using the information above.x= 1/16(3584)or x=224 So the LCM is 224 we can just say the (gcd)LCM=ab=3584, so just divide 3584 by 16.
You need at least two numbers to find either of those.
If you have two numbers m and n and their gcd (or gcf), g then their LCM = m*n/g so LCM = 72*252/36 = 2*252 = 504.
GCD = 39 LCM = 1,755
The GCD is: 1The LCM is: 780
The smallest number that is a multiple of two given numbers is called the least common multiple (LCM) of those two numbers. It can be found by taking the product of the two numbers and dividing it by their greatest common divisor (GCD).
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.
The following function will return the GCD or LCM of two arguments (x and y) depending on the value of the fct argument (GCD or LCM). enum FUNC {GCD, LCM}; int gcd_or_lcm(FUNC fct, int x, int y) { int result = 0; switch (fct) { case (GCD): result = gcd (x, y); break; case (LCM): result = lcm (x, y); break; } return result; }
GCD: 1 LCM: 360