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Re: DS - Sum of integers [#permalink]
02 Mar 2010, 08:46
1
This post received KUDOS
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
Re: DS - Sum of integers [#permalink]
03 Mar 2010, 13:23
Expert's post
3
This post was BOOKMARKED
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.
There 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.
Re: If z1, z2, z3,..., zn is a series of consecutive positive [#permalink]
03 Oct 2014, 00:53
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Re: If z1, z2, z3,..., zn is a series of consecutive positive [#permalink]
03 Oct 2014, 05:05
swethar wrote:
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.
A.
1) let the numbers be a,a+1,a+2,a+3,...,a+n from FS1 --> (a+a+1+a+2+...+a+n)/n = odd or (n*a + (n(n+1)/2))/n = odd or (2na + n(n+1))/2n = odd or (2na + n(n+1)) = even*n LHS is nothing bu the sum. n may be even or odd, but RHS would always be even and so would be the LHS. sufficient.
Re: If z1, z2, z3,..., zn is a series of consecutive positive [#permalink]
03 Oct 2014, 06:33
Expert's post
thefibonacci wrote:
swethar wrote:
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.
A.
1) let the numbers be a,a+1,a+2,a+3,...,a+n from FS1 --> (a+a+1+a+2+...+a+n)/n = odd or (n*a + (n(n+1)/2))/n = odd or (2na + n(n+1))/2n = odd or (2na + n(n+1)) = even*n LHS is nothing bu the sum. n may be even or odd, but RHS would always be even and so would be the LHS. sufficient.
Re: If z1, z2, z3,..., zn is a series of consecutive positive [#permalink]
03 Oct 2014, 08:31
Bunuel wrote:
thefibonacci wrote:
swethar wrote:
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.
A.
1) let the numbers be a,a+1,a+2,a+3,...,a+n from FS1 --> (a+a+1+a+2+...+a+n)/n = odd or (n*a + (n(n+1)/2))/n = odd or (2na + n(n+1))/2n = odd or (2na + n(n+1)) = even*n LHS is nothing bu the sum. n may be even or odd, but RHS would always be even and so would be the LHS. sufficient.
Re: If z1, z2, z3,..., zn is a series of consecutive positive [#permalink]
03 Oct 2014, 08:55
Expert's post
thefibonacci wrote:
Bunuel wrote:
thefibonacci wrote:
A.
1) let the numbers be a,a+1,a+2,a+3,...,a+n from FS1 --> (a+a+1+a+2+...+a+n)/n = odd or (n*a + (n(n+1)/2))/n = odd or (2na + n(n+1))/2n = odd or (2na + n(n+1)) = even*n LHS is nothing bu the sum. n may be even or odd, but RHS would always be even and so would be the LHS. sufficient.
Thanks Bunuel. But that would not make the sum odd, still. What am I missing here?
Sorry, but don't know what are you trying to prove there? What's your question? Do you get that n is even? Or ...? If you make last term a + n - 1 instead of a + n you'll get that n is odd. _________________
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