The numbers that are circled as well as crossed are the common multiples of 6 and 7. So, the common multiples of 6 and 7 are 42, and 84. We can find the common multiples of two or more numbers by listing the multiples of each number and then finding their common multiples.
What is the multiple of 7 and 6?
To sum up, the lcm of 7 and 6 is 42. In common notation: lcm (7,6) = 42.
What are the first six multiple of 7?
The first six multiples of 7 are 7, 14, 21, 28, 35, and 42.
What are the common multiples of 5 6 and 7?
LCM of 5, 6, and 7 is the smallest number among all common multiples of 5, 6, and 7. The first few multiples of 5, 6, and 7 are (5, 10, 15, 20, 25 . . .), (6, 12, 18, 24, 30 . . .), and (7, 14, 21, 28, 35 . . .)
How many multiple does 7 have?
Multiples of 7 are the numbers obtained by multiplying 7 with other numbers. The first five multiples of 7 are 7, 14, 21, 28, and 35.
What number is a multiple of 5 6 and 7?
The LCM of 5,6,7 5 , 6 , 7 is 2⋅3⋅5⋅7=210 2 ⋅ 3 ⋅ 5 ⋅ 7 = 210 .
What are the first three multiples of 6 and 8?
The first three common multiples of 6 and 8 are = 24, 48, 72.
What are the common multiples of 6 and 7?
We will mark the multiples of 6 by a circle and multiples of 7 by a cross. The numbers that are circled as well as crossed are the common multiples of 6 and 7. So, the common multiples of 6 and 7 are 42, and 84. We can find the common multiples of two or more numbers by listing the multiples of each number and then finding their common multiples.
How to create a list of multiples of 6?
To create a list of multiples of 6, we multiply 6 by 1, 6 by 2, and so on like this: 6 x 1 = 6 6 x 2 = 12 6 x 3 = 18 6 x 4 = 24 Similarly, to create a list of multiples of 8, we multiply 8 by 1, 8 by 2, and so on like this: 8 x 1 = 8 8 x 2 = 16 8 x 3 = 24 8 x 4 = 32
Which is the least common multiple of 6 and 8?
Since 24 is the first number they have in common, 24 is the least common multiple of 6 and 8. Note: Our lists are shortened because they go on forever, but can you see the pattern when you look at the answer in green above?
