What is the parity of the sum of 101 consecutive integers?
The Correct answer and Explanation is:
To determine the parity of the sum of 101 consecutive integers, we first need to define what “consecutive integers” means. Let’s consider a sequence of 101 consecutive integers starting from some integer nnn. The integers can be represented as follows:
n,n+1,n+2,…,n+100.n, n+1, n+2, \ldots, n+100.n,n+1,n+2,…,n+100.
The sum of these integers can be expressed mathematically:S=n+(n+1)+(n+2)+…+(n+100).S = n + (n + 1) + (n + 2) + \ldots + (n + 100).S=n+(n+1)+(n+2)+…+(n+100).
This expression simplifies to:S=101n+(0+1+2+…+100).S = 101n + (0 + 1 + 2 + \ldots + 100).S=101n+(0+1+2+…+100).
The sum of the first mmm integers is given by the formula:Sum=m(m+1)2.\text{Sum} = \frac{m(m + 1)}{2}.Sum=2m(m+1).
In our case, m=100m = 100m=100:0+1+2+…+100=100⋅1012=5050.0 + 1 + 2 + \ldots + 100 = \frac{100 \cdot 101}{2} = 5050.0+1+2+…+100=2100⋅101=5050.
Thus, we can rewrite the sum SSS as:S=101n+5050.S = 101n + 5050.S=101n+5050.
Next, we need to analyze the parity of SSS. The parity of a number refers to whether it is odd or even. A number is even if it is divisible by 2 and odd if it is not.
- Parity of 101n101n101n: The integer 101101101 is odd. Therefore, the parity of 101n101n101n depends on nnn:
- If nnn is even, then 101n101n101n is even.
- If nnn is odd, then 101n101n101n is odd.
- Parity of 505050505050: The number 505050505050 is even.
To find the overall parity of SSS, we consider the two cases based on the parity of nnn:
- If nnn is even:
- S=even+even=evenS = \text{even} + \text{even} = \text{even}S=even+even=even.
- If nnn is odd:
- S=odd+even=oddS = \text{odd} + \text{even} = \text{odd}S=odd+even=odd.
Since nnn can be either even or odd, the parity of the sum SSS varies depending on the starting integer nnn. However, since we are summing an odd number (101) of integers, the sum will always retain the parity of the integers in the sequence.
In conclusion, the parity of the sum of 101 consecutive integers is the same as the parity of the middle integer in the sequence. If nnn is even, the sum is even; if nnn is odd, the sum is odd. Thus, the answer to the question is that the parity of the sum of 101 consecutive integers can be either even or odd, depending on the starting integer nnn.