Loading...
If the sum of \(\mathrm{n}\) consecutive integers is \(0\), which of the following must be true?
Let's start by understanding what we're dealing with. Consecutive integers are integers that follow one after another without any gaps - like 3, 4, 5, 6 or -2, -1, 0, 1.
The problem tells us that the sum of n consecutive integers equals 0. This means if we add up all these consecutive numbers, we get zero.
Now let's understand what each statement is claiming:
Process Skill: TRANSLATE - Converting the problem language into mathematical understanding
Let's create some actual examples to see what patterns emerge when consecutive integers sum to zero.
Example 1: Let's try n = 3 (odd number)
We need 3 consecutive integers that sum to 0.
Try: -1, 0, 1
Sum: \((-1) + 0 + 1 = 0\) ✓
Example 2: Let's try n = 5 (odd number)
We need 5 consecutive integers that sum to 0.
Try: -2, -1, 0, 1, 2
Sum: \((-2) + (-1) + 0 + 1 + 2 = 0\) ✓
Example 3: Can we make n = 2 work? (even number)
We need 2 consecutive integers that sum to 0.
Try any two consecutive integers: \(5, 6 \rightarrow \mathrm{Sum} = 11\)
Try: \(-1, 0 \rightarrow \mathrm{Sum} = -1\)
Try: \(0, 1 \rightarrow \mathrm{Sum} = 1\)
No matter what we try, two consecutive integers never sum to 0!
Example 4: Can we make n = 4 work? (even number)
Try: \(-2, -1, 0, 1 \rightarrow \mathrm{Sum} = -2\)
Try: \(-1, 0, 1, 2 \rightarrow \mathrm{Sum} = 2\)
Again, we can't make 4 consecutive integers sum to 0!
Now let's understand why these patterns work the way they do.
For odd numbers of consecutive integers:
When we have an odd number of consecutive integers, there's always a "middle" integer. If we arrange the integers symmetrically around 0, they balance out perfectly:
For even numbers of consecutive integers:
With an even number of consecutive integers, there's no single "middle" value. The integers can't balance around a central point to sum to zero.
Think about it: if we have 4 consecutive integers starting from any integer k, we get: \(\mathrm{k}, \mathrm{k}+1, \mathrm{k}+2, \mathrm{k}+3\)
Their sum is: \(\mathrm{k} + (\mathrm{k}+1) + (\mathrm{k}+2) + (\mathrm{k}+3) = 4\mathrm{k} + 6\)
For this to equal 0: \(4\mathrm{k} + 6 = 0\), so \(\mathrm{k} = -1.5\)
But k must be an integer, so this is impossible!
Process Skill: INFER - Drawing the non-obvious conclusion about the symmetry requirement
Now let's check each statement using our examples and understanding:
Statement I: "n is an even number"
From our examples, we saw that even values of n (like 2, 4) don't work. We proved that consecutive integers summing to 0 requires odd n.
Statement I is FALSE.
Statement II: "n is an odd number"
All our working examples had odd n (3, 5, 7...) and our structural analysis shows why only odd n works.
Statement II is TRUE.
Statement III: "The average of the n integers is 0"
Let's check: If the sum of n integers is 0, then their average = \(\mathrm{Sum} \div \mathrm{n} = 0 \div \mathrm{n} = 0\)
This is always true regardless of the value of n.
Statement III is TRUE.
Process Skill: CONSIDER ALL CASES - Systematically checking each statement
Since Statement II (n is odd) and Statement III (average is 0) are both true, while Statement I (n is even) is false, the correct answer is E. II and III.
This matches our given correct answer, confirming our solution is complete and accurate.
Faltering Point 1: Misunderstanding what "consecutive integers" means
Students may think consecutive integers must be positive (like 1, 2, 3, 4) and fail to consider that consecutive integers can include negative numbers and zero (like -2, -1, 0, 1, 2). This limits their ability to find examples where the sum equals zero.
Faltering Point 2: Not recognizing the need to test specific cases
Students might try to work with general algebraic expressions immediately instead of testing concrete examples first. Without testing actual numbers, they miss the clear pattern that emerges and struggle to see why only odd values of n work.
Faltering Point 3: Treating the Roman numeral statements as mutually exclusive
Students may think that since statements I and II contradict each other (n can't be both even and odd), they must choose between them rather than recognizing that multiple statements can be evaluated independently, and the answer could include statement III along with either I or II.
Faltering Point 1: Giving up too quickly when testing even values of n
When testing whether even numbers work (like n=2 or n=4), students might try only a few examples with positive integers and conclude incorrectly that it's possible, rather than systematically testing or understanding why it's structurally impossible for any consecutive integers to sum to zero when n is even.
Faltering Point 2: Arithmetic errors when calculating sums
Students may make simple addition mistakes when computing sums like \((-2) + (-1) + 0 + 1 + 2\), especially with negative numbers. These calculation errors can lead them to incorrect conclusions about which values of n actually work.
Faltering Point 1: Incorrectly concluding that Statement III is always true without verification
While Statement III (average equals 0) is indeed always true when the sum is 0, students might assume this without actually checking the relationship: \(\mathrm{average} = \mathrm{sum} \div \mathrm{n} = 0 \div \mathrm{n} = 0\). They could mistakenly think the average might be undefined or non-zero in some cases.
Faltering Point 2: Selecting answer choice D (I and III) due to incomplete analysis
Students who don't thoroughly test both even and odd values of n might incorrectly conclude that both even and odd numbers work, leading them to think both statements I and III are true, when in fact only statements II and III are correct.