For the integers from 1 to 1000, how many multiples of 9 are also multiples of 6?
6 = 2x3
9 = 3x3
so the LCM of 6 and 9 is 18
so how many multiple so 18 are there from 1 to 1000 ?
...18 ....36 ... etc ... 990
term 1 = 18
term 2 = 36
...
term ? = 990
can you determine which term ? is equal to 990 ?
I think term 55 but I am not sure what that proves or if it shows you the answer?
To find how many multiples of 9 are also multiples of 6 among the integers from 1 to 1000, we need to determine the number of integers that are divisible by both 9 and 6.
First, we can find the least common multiple (LCM) of 9 and 6. The LCM is the smallest number that is divisible by both 9 and 6. To find the LCM, we can use the formula:
LCM(a, b) = (a * b) / GCD(a, b)
Here, a is 9 and b is 6. GCD(a, b) is the greatest common divisor of a and b. In this case, GCD(9, 6) is 3.
So, the LCM(9, 6) = (9 * 6) / 3 = 18.
Now, we need to find how many multiples of 18 exist among the integers from 1 to 1000. We can do this by dividing 1000 by 18 and taking the floor value to account for any remainders. This gives us:
1000 / 18 = 55.56 (approx)
Since we cannot have a fraction of a multiple, we take the floor value of 55.56, which is 55. Therefore, there are 55 multiples of 18 among the integers from 1 to 1000.
Each multiple of 18 is also a multiple of both 9 and 6. Hence, there are 55 multiples of 9 that are also multiples of 6 among the integers from 1 to 1000.