LCM[(13b3)3, 7b2] = LCM[2197b9, 7b2] = 2197*7*b9 = 15379*b9
18a3bc
The question is somewhat ambiguous: the answers are LCM[3T2, 5T] =15T2 or LCM[(3T)2, 5T] =45T2
LCM is 20, which is the multiple of the highest power of prime factors in the given numbers (22 x 5).The LCM is 20.
LCM is 27, which is the multiple of the highest power of prime factors in the given numbers (33).
LCM(5y3, 25y6) = 25y6
The answer is 16x3
80a^3
The LCM is 72. The third LCM is 216.
LCM[(13b3)3, 7b2] = LCM[2197b9, 7b2] = 2197*7*b9 = 15379*b9
The LCM is 42a2b2.
Just write a method or function that calculates the LCM for two numbers at a time. Then calculate the LCM for the first two numbers, get the LCM of the result with the third number, etc.Just write a method or function that calculates the LCM for two numbers at a time. Then calculate the LCM for the first two numbers, get the LCM of the result with the third number, etc.Just write a method or function that calculates the LCM for two numbers at a time. Then calculate the LCM for the first two numbers, get the LCM of the result with the third number, etc.Just write a method or function that calculates the LCM for two numbers at a time. Then calculate the LCM for the first two numbers, get the LCM of the result with the third number, etc.
11200
3
21
Since 20x7z2 is a multiple of 4x5z2, it is automatically the LCM.
Find the LCM of the first two numbers and then find the LCM of that number and the third one. That answer will be the LCM of all three.