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The largest number in a series of consecutive even integers

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goodyear2013
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The largest number in a series of consecutive even integers [#permalink] New post   Updated on: 07 Feb 2014, 05:00
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73% (01:41) correct 27% (01:35) wrong based on 411 sessions
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The largest number in a series of consecutive even integers is w. If the number of integers is n, what is the smallest number in terms of w and n?

A. w– 2n
B. w–n + 1
C. w– 2(n– 1)
D. n– 6 + w
E. w – n/2

OE
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(A): w– 2n = 10 – 2(2) = 10 – 4 = 6 ≠ 8
(B): w–n + 1 =10 – 2 +1 = 9 ≠ 8
(C): w– 2(n– 1) = 10 – 2(2 – 1) = 10 –2 = 8 = 8
(D): n– 6 + w = 2 – 6 + 10 = 6 ≠ 8
(E): w – n/2 = 10 – 2/2 = 10 – 1 = 9 ≠ 8

Only (C) matches target number.
Conceptual approach is a bit trickier:
In a set of n consecutive integers in descending order, series would be {w, w-1, w-2, w-3…} - that is, each subsequent integer is w minus integer's position in sequence below w, or w - (n - 1).
For example, set {4, 3, 2, 1} may be written as {4, (4 -1), (4 - 2), (4 - 3)}.

Since this is a series of EVEN numbers, next smaller number in series is w - 2, and next is w - 4 - in other words, each subsequent integer subtracts 2 times number of its position below w, or w - 2(n -1).
For example, set {8, 6, 4} may be written as {8, (8 - 2(2-1), (8 - 2(3-1)}.


Hi, I want to request the solution for this question, please.
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C

Originally posted by goodyear2013 on 06 Feb 2014, 10:46.
Last edited by Bunuel on 07 Feb 2014, 05:00, edited 1 time in total.
Edited the question.
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KarishmaB
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The largest number in a series of consecutive even integers [#permalink] New post   Updated on: 17 Oct 2022, 00:29
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goodyear2013 wrote:
The largest number in a series of consecutive even integers is w. If the number of integers is n, what is the smallest number in terms of w and n?

w– 2n
w–n + 1
w– 2(n– 1)
n– 6 + w
w – n/2

OE
Show Spoiler
(A): w– 2n = 10 – 2(2) = 10 – 4 = 6 ≠ 8
(B): w–n + 1 =10 – 2 +1 = 9 ≠ 8
(C): w– 2(n– 1) = 10 – 2(2 – 1) = 10 –2 = 8 = 8
(D): n– 6 + w = 2 – 6 + 10 = 6 ≠ 8
(E): w – n/2 = 10 – 2/2 = 10 – 1 = 9 ≠ 8

Only (C) matches target number.
Conceptual approach is a bit trickier:
In a set of n consecutive integers in descending order, series would be {w, w-1, w-2, w-3…} - that is, each subsequent integer is w minus integer's position in sequence below w, or w - (n - 1).
For example, set {4, 3, 2, 1} may be written as {4, (4 -1), (4 - 2), (4 - 3)}.

Since this is a series of EVEN numbers, next smaller number in series is w - 2, and next is w - 4 - in other words, each subsequent integer subtracts 2 times number of its position below w, or w - 2(n -1).
For example, set {8, 6, 4} may be written as {8, (8 - 2(2-1), (8 - 2(3-1)}.


Hi, I want to request the solution for this question, please.



Another method would be to use the concept of Arithmetic Progressions. Consecutive even integers form an AP where the common difference is 2.
In an AP, Last term = First term + (n-1)*common difference
w = First term + (n-1)*2
First term = w - 2(n-1)

Originally posted by KarishmaB on 06 Feb 2014, 21:42.
Last edited by KarishmaB on 17 Oct 2022, 00:29, edited 1 time in total.
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tatachalikyan
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Re: The largest number in a series of consecutive even [#permalink] New post   06 Feb 2014, 11:20
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Ciao

I would pick a random set of 3 consecutive even integers e.g. {2, 4, 6} and try all the answers choices.
IMO will be quicker.

Thanx
TC
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goodyear2013
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Re: The largest number in a series of consecutive even [#permalink] New post   06 Feb 2014, 17:44
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{2, 4, 6} --> i.e. W = 6, n = 3 to get smallest one; 2
6 - 6 = 0
6 - 3 + 1 = 4
6 - 2(2) = 2 --> Answer
3 - 6 + 6 = 3
6 - 3/2 = 4.5

Cool. Thanks!
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Re: The largest number in a series of consecutive even [#permalink] New post   06 Feb 2014, 23:03
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[quote="goodyear2013"]The largest number in a series of consecutive even integers is w. If the number of integers is n, what is the smallest number in terms of w and n?

w– 2n
w–n + 1
w– 2(n– 1)
n– 6 + w
w – n/2

Let us say the consecutive even integers are 0, 2, 4 where w = 4, n = 3

By plugging in these values our answer should be 0

(A) 4 - 6
(B) 4 - 3 + 1
(C) 4 - 2(2)
(D) 3 - 6 + 4
(E) 4 - 3/2

Only option C satisfies that
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Re: The largest number in a series of consecutive even integers [#permalink] New post   21 Jul 2014, 13:31
Since the integers are even numbers, let's assume that the first term is 2k. As the series is consecutive even integers, its 2nd term will be 2k+2, 3rd term will be 2K+4, and so on thus giving the nth term as 2k + 2(n-1).

Now,

2k+2(n-1) = w

or, 2k = w-2(n-1)
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Re: The largest number in a series of consecutive even integers [#permalink] New post   06 Feb 2022, 13:14
Don't know if my method is the fastest way to solve it but I'll note it down anyway (TTP way).

Let x = smallest value and thus the value we are solving for

(W-X)/2 + 1 = N
W-X/2 = N-1
W-X = 2N - 2
-X = 2N-2-W
X = -(2N-2-W)
X = -2N+2+W

Now rearrange in order of answer choices (seems like W is the first element in choices), so:
X = W-2N+2
X=W-2(N-1)
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Re: The largest number in a series of consecutive even integers [#permalink] New post   17 May 2023, 22:52
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Re: The largest number in a series of consecutive even integers [#permalink]
17 May 2023, 22:52
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