Question

For some integer $ m $ , every even integer if of the form of $ \_\_\_\_\_\_\_ $
A. $ m $
B. $ m + 1 $
C. $ 2m $
D. $ 2m + 1 $

Answer 1

Hint:
Even integers are the integers which are of the form $ 2m $ and here $ m $ is any integer because even integers are those integers that are divisible completely by $ 2 $ or we can say that $ 2 $ is the factor of the even integer.

Complete step by step solution:
Here we are given that if we have the integer $ m $ then we need to find the form in which we can represent any even integer. So first of all we need to know what the integers are. Integers are all the numbers (whether positive or negative or zero) which in their simplest form are not the fractions.
For example: $ - 1,2,99 $ are all integers while $ \dfrac{2}{5},0.8 $ are all not the integers because when they are converted into the simplest form, they are in the fraction form.
Now if we want the even integers then we need to pick out all the integers that are having $ 2 $ as one of its factors or we can say in the simple terms that every even integer is completely divisible by $ 2 $
So we can say that $ 2,4, - 2, - 8,102 $ are all the even integers as they can be represented as
 $
  2 = 2 \times 1 \\
  4 = 2 \times 2 \\
   - 2 = 2 \times ( - 1) \\
   - 8 = 2 \times ( - 4) \\
  102 = 2 \times 51 \\
  $
Now if we see all these together we can notice that all of them have $ 2 $ as one of their factors. So all these are divisible by $ 2 $ and therefore are all the even integers.
So if we have any integer $ m $ then we can say that if we multiply the integer $ m $ by $ 2 $ then we will get the even integer as it will be completely divisible by $ 2 $
Hence we can say even integer $ = 2 \times m = 2m $

Hence C is the correct option.

Note:
Here the question can be complicated if we are given the integer $ m $ and we need to find the odd integer. So it will not be $ m + 1 $ as $ m $ is not said to be even. After every even number there is a consecutive odd number. So we have to find the even number which will be $ 2m $ and hence we can say that $ 2m + 1 $ or $ 2m - 1 $ will be an odd integer.