The least common multiple of two numbers is 60 and one of the numbers is seven less than the other number what are the numbers? justify your answer
well, x(x+7) is a multiple of 60.
5*12 = 60
guess that works, eh?
To find the two numbers, we can set up equations using the given information.
Let's assume the two numbers are x and y, with x being the larger number.
According to the problem, one of the numbers is seven less than the other. We can write this as an equation:
x = y + 7 ---(Equation 1)
The least common multiple (LCM) of two numbers is the smallest number that both numbers divide evenly into. In this case, the LCM is 60.
To find the LCM of two numbers, we can use the following equation:
LCM(x, y) = (x * y) / GCD(x, y) ---(Equation 2)
where GCD(x, y) represents the Greatest Common Divisor of x and y.
Since we know the LCM (60), we can substitute the values into Equation 2:
60 = (x * y) / GCD(x, y)
To simplify the equation, we need to find the GCD of x and y. However, we don't have enough information to determine the GCD directly.
But, we can still make progress.
Let's consider the relationship between the LCM and the numbers themselves. The LCM is the product of two numbers divided by their GCD.
Since the GCD should divide both numbers evenly, it means that the GCD should also divide their difference (x - y) evenly.
Therefore, the GCD of (x - y) and y should be the same as the GCD of x and y.
Hence, we can write:
GCD(x, y) = GCD(x - y, y)
Now, let's find the GCD using Euclid's algorithm:
Step 1: Set r1 = x and r2 = y
Step 2: Divide r1 by r2 to get q1 and r3 (r1 = q1 * r2 + r3)
Step 3: Replace r1 with r2 and r2 with r3 (r1 = r2 and r2 = r3)
Step 4: Repeat steps 2 and 3 until r3 = 0
Step 5: The GCD will be the last non-zero remainder (r2)
Using Equation 1, we can substitute x = (y + 7) into Equation 2:
60 = ((y + 7) * y) / GCD((y + 7), y)
Now, we can solve for the GCD by applying Euclid's algorithm to GCD((y + 7), y):
Step 1: Set r1 = (y + 7) and r2 = y
Step 2: Divide r1 by r2 to get q1 and r3 (r1 = q1 * r2 + r3)
Step 3: Replace r1 with r2 and r2 with r3 (r1 = r2 and r2 = r3)
Step 4: Repeat steps 2 and 3 until r3 = 0
Step 5: The GCD will be the last non-zero remainder (r2)
Once we find the GCD, we can substitute the values back into Equation 2 to find the LCM.
Using this approach, we can calculate the two numbers involved. Let's solve the equations step-by-step.