The sum of 'n' consecutive integers is '0'. Which of the following statements is / are true? I. There will be more odd integers than even integers in the above set of integers. II. The value of 'n' will be odd. III. The first and the last integer of the set of integers will be even. A. I and II only B. I and III only C. II and III only D. II only E. I, II and III
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I get II only. I'll explain why.
You can draw a number line from, say, -4 to +4. For the first case, (I), if there is one more odd number than even in a set, the sum will be odd. You can see this by looking at a series going from even (e) to odd (o): o + e + o + e + o The numbers on the number line move from odd to even, so you have this sort of set up. Now, you can pair up the odds with the evens to get (o + o) + (e+e) + o odd + odd is always even, and even + even is always even, so you have (e) + (e + e) + o = e + o = o since an even number + odd number is odd. Since the sum is 0 and zero is even, you cannot have more odd numbers in the set than even. So (I) is gone.
As for (III), it's not necessarily the case that the first and last integer in the set need to be even, since you could have, for example, -3 + -2 + -1 + 0 + 1 + 2 + 3 Here you have consecutive integers that sum to 0 but don't start and end on even numbers, so III is out.
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You can see that the value of n will be odd since, looking at our number line, you pair off a number an its opposite, like -4 and 4 -3 and 3 -2 and 2 -1 and 1 when adding, BUT you also have 0 in your set, so the number of elements in your set will be odd, so II is in.