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Z is the set of the first n positive odd numbers, where n is [#permalink]
14 Jun 2012, 03:08

6

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

65% (hard)

Question Stats:

64% (03:13) correct
36% (02:24) wrong based on 121 sessions

Z is the set of the first n positive odd numbers, where n is a positive integer. Given that n > k, where k is also a positive integer, x is the maximum value of the sum of k distinct members of Z, and y is the minimum value of the sum of k distinct members of Z, what is x + y?

Re: Z is the set of the first n positive odd numbers, where n is [#permalink]
14 Jun 2012, 03:21

6

This post received KUDOS

Expert's post

Smita04 wrote:

Z is the set of the first n positive odd numbers, where n is a positive integer. Given that n > k, where k is also a positive integer, x is the maximum value of the sum of k distinct members of Z, and y is the minimum value of the sum of k distinct members of Z, what is x + y?

Probably the easiest way to solve this question would be to assume some values for n and k.

Say n=3, so Z, the set of the first n positive odd numbers would be: Z={1, 3, 5}; Say k=1, so X, the maximum value of the sum of K distinct members of Z would simply be 5. Similarly, Y, the minimum value of the sum of K distinct members of Z would simply be 1.

X+Y=5+1=6.

Now, substitute n=3 and k=1 in the options provided to see which one yields 6. Only asnwer choice E fits: 2kn=2*3*1=6.

Answer: E.

Note that for plug-in method it might happen that for some particular number(s) more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only.

Re: Z is the set of the first n positive odd numbers, where n is [#permalink]
19 Jan 2013, 09:03

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1

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In the given case, sum of all n odd numbers is n^2.

If the numbers are arranged in increasing order, the max sum is when k is taken from end. So the sum of k odd numbers taken from the end = n^2 - (n-k)^2 (Sum of all numbers) - (Sum of remaining initial numbers) which is equal to 2nk - k^2 -- (1)

Now for minimum sum, take k numbers from the starting and their sum is k^2 -- (2)

Re: Z is the set of the first n positive odd numbers, where n is [#permalink]
23 Jan 2013, 11:36

Let us name Zi the general term of the set Z. z(i) = 2*i - 1 for i = 1 to n. For instance z(1) = 2*1 - 1 = 1 z(2) = 2*2 - 1 = 3

z(n) = 2*n - 1

y is the sum of the first k terms of the set Z. x is the sum of the terms from order (n-k+1) to n.

Let us express the sums x and y. Each of the sums contains k terms. y = Sum (2*i - 1) for i = 1, ... k x = Sum (2*j - 1) for j = (n-k+1), ,n

By developing y, we obtain y = 2*Sum (i) - k for i=1,....k y = 2*(1/2) [k*(k+1)]/2 - k = k^2 + k - k = k^2 (we can also remember that the sum of the first k odd numbers equals k^2).

By developing x, we obtain x = 2*Sum (j) - k for j=n-k+1, n We can notice that x is the difference of - the sum of the first "n" odd numbers - and the sum of the first (n-k) odd numbers. We deduce x = n^2 - (n-k)^2=2*n*k - k^2

Re: Z is the set of the first n positive odd numbers, where n is [#permalink]
09 Oct 2014, 08:08

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