Well, see, that's going to be a problem, since there are an infinite number of them,
and we're a little busy just now.
But you can easily generate the list for yourself. Here's how:
-- Start at ' 1 ', and count.
-- Throw out the first number, then write down the second number.
-- Throw out a number, then write down the next number.
-- Throw out a number, then write down the next number.
-- Throw out a number, then write down the next number.
-- Throw out a number, then write down the next number.
Keep going like that. The list you write down will be a list of numbers that have ' 2 '
as a factor. You can just keep right on going for as long as you want, and make
that list as long as you want it to be.
You can also stop whenever you want to, and then you'll only have another
infinite number left to go.
256 is not prime. 256 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2
There are 32C3 = 32*31*30/(3*2*1) = 4960 combinations. I do not have the inclination to list them all.
Yes. Not counting one as a factor, you could have numbers like 6 and 10 which only share one common factor, 2.
the factor of (x^2-2) could be looked at if we write (x^2-2) as (x^2-(sqrt(2)^2 now we have the difference of squares so we can factor is... The key point here is where can your factors come from? Most often we factor over the Integers and (X^2-2) cannot be factored over the integers If we allow our field, or set of numbers over which we factor to be the real numbers, then the method I suggested works. If we allow our field to be the complex numbers then we cal always factor an expression. This is an important theorem in algebra.
Factorize the LCM 20 = 2 x 2 x 5 Factorize the HCF 300 = 2 x 2x3x5x5 The HCF is the common factor for both numbers whereas the LCM contains all the unique factors of both numbers. So, 2 x 2x5 or 20 is a common factor for both numbers. The factors remaining are 3 and 5. So, 3 is a factor of one number and 5 is the factor of the second number. The first number = 20x 3 = 60 The second number = 20x 5 = 100 So 60 and 100 are two numbers Answer = 60 and 100
It would be impossible to list all the even numbers.
All numbers with 12 as a factor have 2 as a factor, but not all have 10 as a factor.
2 is a divisor of all even numbers, and a divisor is a factor of all numbers which can be divided by it.
Not AllTo find out if all numbers that have 5 as a factor have 2 as a factor, you can list the multiples of each. 5's multiples are 5, 10, 15, 20...2's multiples are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20...5 and 15 are multiples of 5, but not 2, so all numbers that have 5 as a factor DO NOT have 2 as a factor.It is not really necessary to write that all out because the fact that 5 is an odd number and 2 is only a factor of even numbers indicates that all numbers that have 5 as a factor would not have 2 as a factor. Only the multiples of 5 that are even have 2 as a factor also.
All even numbers greater than 0 have 2 as a factor.
Yes. All even numbers have 2 as a factor.
Yes, every number on the product board can be obtained by multiplying two numbers from the factor list. This is because each number on the product board is the product of two numbers from the factor list.
Possiblilty is almost 1/2 cause every two numbers is a number with a factor of 2.From 2: 2,3,4,5,6,7,8,9,10...
Actually they do because only even numbers have 2 as a factor.
It is, because 2 is a common factor for all even numbers.
No.
Yes, all even numbers have 2 as a factor. I'm not sure how to explain that statement.