There are many multiples. 24, 48, 72, 96, 120, etc. Pretty much any number that is divisible by 24.
24, 48, 72, and so forth.
No multiples of 24 are factors of 6.1, 2, 3, 4 and 8 are factors of 24 that are not multiples of 6.
A common multiple of 8 and 6 has the form 2 * 2 * 2 * 3 * n = 24n, where n is any whole number. The other factors are necessary to ensure that what you have is indeed a multiple of both 8 (2 * 2 * 2) and 8 (2 * 3). So for the first 3 common multiples you need to replace n with 1, 2 and 3, giving you 24, 48 and 72.
Yes, for example the multiples of 2 are: 2, 4, 6, 8, 10, 12, 14, ......etc. (all even numbers are multiples of 2) or multiples of 3 are 3, 6, 9, 12, 15, 18, 21, ....... etc.
The first three common multiples of 6 and 8 are the first three multiples of the least common multiple of 6 and 8. 6 = 2 x 3 and 8 = 2 x 2 x 2. Therefore, the LCM of 6 and 8 is 2 x 2 x 2 x 3 or 24. The first three multiples of 24 are 24, 48 and 72.
1/8, 2/8, 3/8, 4/8
What are three common multiples of 2 6 8
All multiples of 8 are also multiples of 2, but not all multiples of 2 are multiples of 8.
lcm(2, 7) = 14 → first 3 common multiples are: 14, 28, 42; lcm(2, 8) = 8 → first 3 common multiples are: 8, 16, 24; lcm(7, 8) = 56 → first 3 common multiples are: 56, 112, 168; lcm(2, 7, 8) = 56 → first 3 common multiples are: 56, 112, 168.
3 multiples and they are all 2 2 * 2 * 2 = 8
Assuming you mean that you want the number of multiples of each, then for 1-100: number of multiples of 2 = 50 number of multiples of 3 = 33 number of multiples of 4 = 25 number of multiples of 6 = 16 number of multiples of 8 = 12 number of multiples of 9 = 11 Assuming you mean that you want the numbers that are multiples of 2, 3, 4, 6, 8 or 9, then some numbers may be multiples of more than one (for example 12 is a multiple of 2, 3, 4 and 6) and so a straight addition of the number of multiples of each cannot be done: Consider 2, 4 and 8 Every multiple of 4 or 8 is also a multiple of 2, so all the multiples of 4 and 8 are counted by the multiples of 2. Consider 3 and 9 Every multiple of 9 is also a multiple of 3, so all the multiples of 9 are counted by the multiple of 3 Consider 2, 3 and 6. Every multiple of 6 is an even multiple of 3, so are counted in both the multiples of 2 and 3. So the total number of multiples of 2, 3, 4, 6, 8 or 9 is the number of multiples of 2 plus the number of multiples of 3 minus the number of multiples of 6: For 1 to 100, Number of multiples of 2 = 50 Number of multiples of 3 = 33 Number of multiples of 6 = 16 So number of multiples of 2, 3, 4, 6, 8 or 9 in 1-100 is 50+33-16 = 67. Assuming you mean that they are multiples of all of 2, 3, 4, 6, 8 and 9, then they must be multiples of the lowest common multiple of 2, 3, 4, 6 ,8, 9 2 = 21, 3 = 31, 4 = 22, 6 = 2131, 8 = 23, 9 = 32 LCM = highest power of the primes used = 2332 = 72 Thus all numbers that are multiples of 2, 3, 4, 6, 8 and 9 are multiples of 72, which means between 1 and 100 only 1 number is a multiple of all of them, namely 72
They are the multiples of 3*8 = 24
Yes, multiples of 8 are also multiples of 2 because 8 is a multiple of 2 itself.
4 and 8 are multiples of 2. 6 and 9 are multiples of 3. 40 and 50 are multiples of 10.
No multiples of 24 are factors of 6.1, 2, 3, 4 and 8 are factors of 24 that are not multiples of 6.
multiples of 8 are also multiples of 2 because anything you times by 8 is an even number
2/8 and 3/12 are equivalent multiples of 1/4, along with an infinite number of others.
yes, for 8 is a multiple of 2