Count the evens
An even integer N can be written in the form N = 2 k , where k is an integer. How many even integers are there in the range − 1 1 ≤ x ≤ 1 1 ?
The answer is 11.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
6 solutions
How can 0 be an even integer?
Log in to reply
Every even integer n can 'always' be expressed in the form n = 2 k for some integer k . And, 0 = 2 × 0 . So, 0 is an even integer.
Log in to reply
But then k=n which is not what we stated.
Log in to reply
@Daniel Alfaro – So what if k = n ?
Refer to the definition, for k = 0 , we have 0 = 2 × 0 , hence 0 is an even integer.
Even integers between -11 and 11 are : − 1 0 [ 2 × ( − 5 ) ] − 8 [ 2 × ( − 4 ) ] − 6 [ 2 × ( − 3 ) ] − 4 [ 2 × ( − 2 ) ] − 2 [ 2 × ( − 1 ) ] 0 [ 2 × 0 ] 2 [ 2 × 1 ] 4 [ 2 × 2 ] 6 [ 2 × 3 ] 8 [ 2 × 4 ] 1 0 [ 2 × 5 ]
Oops! Forgot the main answer in a hurry! So, the number of even integers between -11 and 11 is 1 1
Hey, I got an idea after solving this: The number of even numbers between -n and n is: *1. (n - 1) iff n is even * *2. n iff n is odd *
The number of odd numbers between -n and n is: *1. n iff n is even * *2. (n + 1) iff n is odd *
Between number -11 and 11, there are 11 even numbers.
yeah...........!
The numbers are -10,-8,-6,-4,-2,0,2,4,6,8,10
In this problem zero is counted as even integers ,it is true
N must be an even number because the 2 in 2k makes in even. So from -11 to 11, you ask yourself, How many even numbers are there? The answer is -10, -8, -6, -4, -2, 0, 2, 4, 6, 8 ,10. Therefore the answer is 11.
N=2K then N=22......so that 22=2k ......22/2=k......Answer is K=11
include<iostream>
using namespace std; int main() { int j; int n=2*j; int count++; for(int i=-11;i<11;i++) { if(n%2==0 && i>0) count++; } cout<<count; }
Listing integers k gives {0,1,2,3,4,5,6 etc.}
Therefore, 2k gives {0,2,4,6,8,10,12 etc.}
These integers are all even and six of them are less than 11.
However, there are also the negative even integers {-2,-4,-6,-8,-10,-12 etc.} and so there are five more solutions here (0 cannot be counted twice).
This gives 5 + 6 or 1 1 solutions.