Suppose LCM(a,b) = 20160, a=60. Find the smallest possible value of b.
20160 = 60*336 = 2^6 * 3^2 * 5 * 7 = (2^2 * 3 * 5) * (2^4 * 3 * 7)
Now, LCM(60,336) = 1680
You need to include the other 2^2 * 3 to get all the necessary prime powers.
336*12 = 4032
LCM(60,4032) = 20160
Oh, I see you there, trying to trick me with your math question. But don't you worry, I'm always up for a laugh!
Well, if LCM(a, b) = 20160 and a = 60, let me put on my clown shoes and give you a funny solution.
Now, just like in a circus, we need to find the smallest possible value of b that will make this equation work. So, let's do some math tricks together!
20160 divided by 60 equals 336. So, we know that b must be a multiple of 336 in order to meet the requirements.
Now, what's the funniest multiple of 336 you can think of? How about 336 itself? Yes, sit back and laugh because the smallest possible value of b is 336.
So, to sum it all up, when a = 60 and LCM(a,b) = 20160, the smallest possible value of b is 336. And remember, in the circus of math, laughter is always the best solution! 🤡
To find the smallest possible value of b, we need to determine the greatest common divisor (GCD) of a and b.
Given that a = 60, we can calculate the GCD of a and b using the formula:
GCD(a, b) = (a * b) / LCM(a, b)
Substituting the known values:
GCD(60, b) = (60 * b) / 20160
To find the smallest possible value of b, we need to find the largest possible common factor between 60 and b.
Since b cannot be smaller than 60, we can start by dividing 60 by the LCM and multiplying it by 60 to obtain the possible values of b:
b = (GCD(60, b) * LCM(a, b)) / a
b = (60 * 20160) / 60
b = 3360
Therefore, the smallest possible value of b is 3360.
To find the smallest possible value of b, we need to solve for it using the given information.
We are given that the least common multiple (LCM) of a and b is 20160, and a is equal to 60.
The LCM of two numbers is the smallest multiple that is divisible by both numbers. Therefore, we can say that 20160 must be divisible by both a and b.
Since a = 60, we can simplify the problem by dividing 20160 by 60 to find the LCM of b with respect to 60.
20160 / 60 = 336
So, we can now say that the LCM of 60 and b is 336.
To find the smallest possible value of b, we need to find the prime factorization of 336.
Prime factorization of 336:
336 = 2 * 2 * 2 * 2 * 3 * 7
Now, we look for the highest powers of the prime factors in the prime factorization of 336 that are not present in the prime factorization of 60.
In this case, we can see that the prime factors of 336 are 2, 3, and 7. We already have all these factors in the prime factorization of 60, which is 2 * 2 * 3 * 5.
Therefore, the smallest possible value of b is 1, as all the prime factors of 336 are already present in the prime factorization of 60.
So, the smallest possible value of b is 1.