m and n are 70 and 90
All multiples of 8 are also multiples of 2, but not all multiples of 2 are multiples of 8.
All multiples of 16 are also multiples of 2.
Since 2 is a multiple of 6, all multiples of 6 are also multiples of 2.
No multiples of 2 are factors of 3.
A multiple of 2 would be 2 x N (2 times/ multiplied by N) with N being any integer.The same goes for 6.So 2, 4, 6, 8= 2x1, 2x2, 2x3, 2x4 and are multiples of 2.6, 12, 18, 24= 6x1, 6x2, 6x3, 6x4 and are multiples of 6
int n = 2 // start at 0 if you really want to while n < some arbitrary limit n = n + 2
2n where n is any positive integer. i.e. when n=1 multiple=2 when n=2 multiple=4 when n=3 multiple=6 etc..
Let n be the number. n x 1 n x 2 n x 3 n x 4 n x 5 are the first five multiples. You can check them with a calculator.
All factors of a number n will be factors of any multiple of n. All multiples of a number n will be multiples of any factor of n.
Yes. By definition a multiple of 8 is any number that can be expressed as 8*n, where n is an integer. But 8n=4*(2*n), and 2*n is an integer, when n is an integer. Because 8n equals four times an integer, 8n is a multiple of 4.
Yes, it is. Consider any two elements of this set, both are multiples of n, so they can be written as pn and qn for some integers p and q. Multiplying them together, we obtain pqn^2, which can be factored into (pqn)n. This result is clearly a multiple of n. Since the product of any two multiples of n is also a multiple of n, the set of all multiples of n is closed under multiplication.
A multiple of 2 would be 2 x N (2 times/ multiplied by N) with N being any integer. The same goes for 6. So 2, 4, 6, 8 = 2x1, 2x2, 2x3, 2x4 and are multiples of 2. 6, 12, 18, 24 = 6x1, 6x2, 6x3, 6x4 and are multiples of 6
m and n are 70 and 90
Factors of 36:36 = 22 * 32Thus, n and 36 has highest common factor as 1, if and only if when n is not a multiple of 2 and 3.Total number of multiples of 2 in the given range=500Total number of multiples of 3 in the given range=333Total number of multiples of 6 in the given range=166Total number of multiples of 2 and 3 in the given range=500+333-166=667Thus, the total number of 'n' in the given range which are not multiples of 2 and 3= 999-667 = 332.332 numbers are there such that it is greater than or equal to 2 and less than or equal to 1000 and the highest common factor of the number and 36 is 1.
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.
You find multiples of a number by multiplying that number by successive counting numbers. Let N equal the number. The first multiple is always the original number (N x 1) The rest will be N x 2, N x 3, N x 4 and so on.