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If z1, z2, z3,..., zn is a series of consecutive positive [#permalink]
 02 Mar 2010, 09:23
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Please solve (with explanation): If z1, z2, z3,..., zn is a series of consecutive positive integers, is the sum of all the integers in this series odd? 1. [(z1+z2+z3+...zn)/n] is an odd integer. 2. n is odd. Source: Kaplan
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Re: DS - Sum of integers [#permalink]
02 Mar 2010, 09:46
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swethar wrote: Please solve (with explanation):
If z1, z2, z3,..., zn is a series of consecutive positive integers, is the sum of all the integers in this series odd?
1. [(z1+z2+z3+...zn)/n] is an odd integer.
2. n is odd.
st 1) [(z1+z2+z3+...zn)/n] is the avg arithmetic mean of the series - for consequetive numbers, if the total number is even, then mean is the avg of the middle two numbers(which is not an interger), or if the total number is odd, then mean is the middle number. As its given mean is an odd interger, the middle number is an odd integer and we will have the same number of positive integers to the right of mean as to the left of mean. and the sum of the remaining integers except mean will be even. ( as for every odd number to the right of mean, there would be an odd number to the left of mean). So the sum of all the numbers in the series is odd. Sufficient st 2) n is odd - sum could be even if the middle number(mean/median) is even [3,4,5] or sum could be odd if the middle number(mean/median) is odd [6,7,8] Not sufficient A
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Re: DS - Sum of integers [#permalink]
03 Mar 2010, 14:23
swethar wrote: Please solve (with explanation): If z1, z2, z3,..., zn is a series of consecutive positive integers, is the sum of all the integers in this series odd? 1. [(z1+z2+z3+...zn)/n] is an odd integer. 2. n is odd. Source: KaplanThere is an important property of n consecutive integers: • If n is odd, the sum of consecutive integers is always divisible by n. Given \{9,10,11\}, we have n=3 consecutive integers. The sum of 9+10+11=30, therefore, is divisible by 3. • If n is even, the sum of consecutive integers is never divisible by n. Given \{9,10,11,12\}, we have n=4 consecutive integers. The sum of 9+10+11+12=42, therefore, is not divisible by 4. (1) \frac{z_1+z_2+z_3+...z_n}{n}=odd, as the result of division the sum over the number of terms n is an integer, then n must be odd --> z_1+z_2+z_3+...z_n=odd*n=odd*odd=odd. Sufficient. (2) n is odd. Sum can be odd as well as even. Not sufficient. Answer: A. Hope it helps.
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Re: DS - Sum of integers
[#permalink]
03 Mar 2010, 14:23
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