The least common multiple of two numbers is 60, and one of the numbers is 7 less than the other number. What are the numbers? Justify your answer.
60 = 10*6 = 5*12
12-5 = 7 remarkable
6, 10
5, 12
To find the two numbers, we can start by setting up a system of equations based on the given information. Let's call the unknown number x, and the number that is 7 less than x as y.
We are given that the least common multiple (LCM) of the two numbers is 60. To find the LCM, we need to find the smallest multiple that is divisible by both x and y.
So, we can write the first equation as:
LCM(x, y) = 60
Now, we also know that one of the numbers (y) is 7 less than the other number (x). We can form our second equation based on this information:
y = x - 7
To solve the system of equations, we can substitute the value of y from the second equation into the first equation. So, our system of equations becomes:
LCM(x, x - 7) = 60
x - 7 = y
Now, let's simplify the first equation:
To find the LCM of two numbers, we need to find their greatest common divisor (GCD). We can find the GCD by prime factorizing the numbers and taking the highest power of each common prime factor. However, since we are given the LCM directly, we can skip this step and use the values directly in our equation.
LCM(x, x - 7) = 60
To find the LCM, we need to consider all the prime factors of 60, which are 2, 3, and 5. Now we can check the possibilities for x and x - 7 to see which one results in an LCM of 60.
Let's start with x = 60 and substitute it into the equation. We get:
LCM(60, 60 - 7) = LCM(60, 53) = 53, which is not equal to 60.
Now let's try x = 30:
LCM(30, 30 - 7) = LCM(30, 23) = 30 * 23 = 690
690 is not equal to 60, so x = 30 is not the solution.
Continuing in this manner, let's try smaller values for x.
If x = 20:
LCM(20, 20 - 7) = LCM(20, 13) = 20 * 13 = 260
260 is not equal to 60, so x = 20 is not the solution.
If x = 10:
LCM(10, 10 - 7) = LCM(10, 3) = 10 * 3 = 30
30 is equal to 60.
Therefore, the number x must be 10. Substituting this value back into the second equation:
y = x - 7 = 10 - 7 = 3
So the two numbers are x = 10 and y = 3.
To justify the answer, we can check if the LCM of these two numbers is indeed 60:
LCM(10, 3) = LCM(10, 10 - 7) = LCM(10, 3) = 30, which is equal to 60.
Hence, the numbers 10 and 3 satisfy the given conditions, and the answer is justified.