0
Add your answer:
Earn +20 pts
What are the consecutive even number?
They are numbers of the form 2n and 2n+2 where n is any integer.
Is an even number times any number always even?
YES Even Number by definition any integer that can be divided exactly by 2. To get an even number, multiply an integer by 2 or in mathematical statement it is 2n. An even number (2x) is multiplied to any number (y) is 2x (y). So, by associative principle of multiplication 2x(y) = 2 (x*y). If (x*y) now represents any integer then 2(x*y)= 2n. And 2n is an even number.
What would be an algebraic representation of an even number?
2n, where n is an integer.
Can the sum of an odd and even number be odd?
Yes, 1+2=3 and since 1 is odd and 2 is even this shows it can happen. In fact it always does! If we let an even number be denoted by 2n for some integer n and and an odd one 2m+1 for some integer m. Then, 2m+1+2n=2(m+n)+1. Call m+n, p since it is also an integer. 2p+1 is always odd.
Does 6 have any odd multiples?
No. All multiples of an even number are even. Here is why: An even number is one that is a multiple of 2. Thus, it can be written in the form: 2n where "n" is any integer. If you multiply this by another integer, call it "m", you get: 2n x m = 2mn Since the product of two integers (m x n) is an integer, it follows that the result has a factor of 2 - i.e., it is even.
If 2n is an even integer what is the next even integer?
2n + 2 = 2(n+1)
Is 2n a even integer?
In general mathematicians use n to signify any integer. On this basis 2n must be an even integer.But, are you sure it was a mathematician who wrote 2n ? There's no law about this.
What is the product of two even numbers?
The product of two even numbers is always an even number.Here is the proof:We define an even number as a number of the form 2n for some integer n.Now let 2n be one even number and 2m be another.The product is (2n)(2m)=2(2mn) and of course 2mn is an integer since the integers are closed under multiplication. Hence, 2(2mn) is an even number.
What are the consecutive even number?
They are numbers of the form 2n and 2n+2 where n is any integer.
Is an even number times any number always even?
YES Even Number by definition any integer that can be divided exactly by 2. To get an even number, multiply an integer by 2 or in mathematical statement it is 2n. An even number (2x) is multiplied to any number (y) is 2x (y). So, by associative principle of multiplication 2x(y) = 2 (x*y). If (x*y) now represents any integer then 2(x*y)= 2n. And 2n is an even number.
When you multiply an odd number and an even number is your answer always odd?
No, it's always even, and here's the proof: All even numbers can be expressed as 2n, where n is any integer. All odd numbers can be expressed as 2p + 1, where p is any integer. Multiply those two together: 2n(2p + 1) = 2(2np + n). Since both 2np and n are integers, that means 2np + n is an integer; and since that integer is being multiplied by 2, it must be even.
Why is the sum of 2 odd numbers always even?
An odd number can be written in the form '2n+1' where 'n' is an integer, and an even number can be written int the form '2n'. We can then write the sum of 2 odd numbers as: (2n+1) + (2m+1) * Combining and factoring out a 2, we arrive at: 2(n + m + 1) Since 'n' and 'm' are both integers, we know that the value contained int the '()' is also an integer. We can therefore rewrite this equation to be: 2n Which represents an even number as given previously. * 'm' is an integer
Is the sum of an even whole number and an odd whole number always odd?
Yes --------------------------------------------- Let n be an integer. Then 2n is an even number Let m be an integer. Then 2m is an even number and 2m + 1 is an odd number. Then: even + odd = (2n) + (2m + 1) = (2n + 2m) + 1 = 2(n + m) + 1 = 2k + 1 (where k = m + n) which is an odd number.
What would be an algebraic representation of an even number?
2n, where n is an integer.
Can the sum of an odd and even number be odd?
Yes, 1+2=3 and since 1 is odd and 2 is even this shows it can happen. In fact it always does! If we let an even number be denoted by 2n for some integer n and and an odd one 2m+1 for some integer m. Then, 2m+1+2n=2(m+n)+1. Call m+n, p since it is also an integer. 2p+1 is always odd.
The sum of two consective even integers are 494 What is the two numbers?
246 and 248.Any even number can be written in the form 2n, for some integer n. The next consecutive integer is 2n + 2. These two numbers added together equal 494:2n + (2n + 2) = 494so4n + 2 = 4944n = 492n = 123so2n = 246 and 2n +2 = 248.
Four consecutive odd integers whose sum is 336?
Any odd integer can be represented as 2n+1, where n is any integer. The integers, starting with one, always proceed as odd, even, odd, even, etc. Therefore, a consecutive odd would be two more than the first. So 2n+1, 2n+3, 2n+5, and 2n+7 summed to 336. Simple algebra gives n=40, and the four numbers as 81, 83, 85, and 87.
