Number theory -1

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guerrero: Master | Next Rank: 500 Posts; Posts: 227; Joined: Sun Apr 08, 2012 4:53 amFollowed by: 3 members; Thanked: 12 times

by guerrero » Sun Mar 31, 2013 12:05 am

Five consecutive integers are labeled n1, n2, n3, n4, and n5. Which of these numbers should be removed, in order for the sum of the other four to become 4/5th of the original sum?
(A)n1
(B)n2
(C)n3
(D)n4
(E)n5

OA
C

Last edited by guerrero on Sun Mar 31, 2013 12:36 am, edited 1 time in total.

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Source: Beat The GMAT — Problem Solving | Sun Mar 31, 2013 12:05 am

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Anju@Gurome: GMAT Instructor; Posts: 511; Joined: Wed Aug 11, 2010 9:47 amFollowed by: 86 members; Location: Delhi, India; Thanked: 344 times

by Anju@Gurome » Sun Mar 31, 2013 12:13 am

guerrero wrote:Five consecutive integers are labeled n1, n2, n3, n4, and n5. Which of these numbers should be removed, in order for the sum of the other four to become 4/5 of the original sum?

A very good problem to illustrate how careful selection of numbers can reduce calculation and save a lot of time.

Let us assume the integers are -2, -1, 0, 1, and 2.
Why did I picked these numbers? Because it makes the sum zero.
Now, it is just matter of second to identify the number which can be removed so that the sum remains zero as 4/5 of zero = zero.

The number must be zero because if we remove any other number the sum will be either positive or negative.

The correct answer is C.

Last edited by Anju@Gurome on Sun Mar 31, 2013 5:34 am, edited 1 time in total.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sun Mar 31, 2013 3:28 am

guerrero wrote:Five consecutive integers are labeled n1, n2, n3, n4, and n5. Which of these numbers should be removed, in order for the sum of the other four to become 4/5th of the original sum?
(A)n1
(B)n2
(C)n3
(D)n4
(E)n5

OA
C

Let the five integers be 1, 2, 3, 4, and 5.
Original sum = number*median = 5*3 = 15.
New sum = 4/5 of the original sum = (4/5)15 = 12.
Integer that must be removed = original sum - new sum = 15-12 = 3.

The correct answer is C.

Algebraic solution:
Let the five integers be n, n+1, n+2, n+3, and n+4.
Original sum = number*median = 5(n+2) = 5n+10.
New sum = 4/5 of the original sum = (4/5)(5n+10)= 4n+8
Integer that must be removed = original sum - new sum = (5n+10) - (4n+8) = n+2, which is the value of the third integer.

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hemant_rajput: Master | Next Rank: 500 Posts; Posts: 447; Joined: Sun Apr 22, 2012 7:13 amFollowed by: 13 members; Thanked: 46 times; GMAT Score:700

by hemant_rajput » Sun Mar 31, 2013 11:33 am

all no.s are consecutive numbers so they'll be of form n, n+1, n+2, n+3 and n+4

sum of 5 consecutive no.s will be 5n + 10 or 5(n + 2).

4/5 of sum of 5 consecutive no.s is 4/5 *5(n+2) = 4(n+2) or 4n + 8

subtract this sum from the original sum
(5n +10) - (4n + 8) = n+2

n+2 is the no. that you have to remove from the original list to make the sum 4/5 of the original sum.

Hope this helps,

Kudos,
Hemant

I'm no expert, just trying to work on my skills. If I've made any mistakes please bear with me.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Sun Mar 31, 2013 12:01 pm

guerrero wrote:Five consecutive integers are labeled n1, n2, n3, n4, and n5. Which of these numbers should be removed, in order for the sum of the other four to become 4/5th of the original sum?
(A)n1
(B)n2
(C)n3
(D)n4
(E)n5

One more approach.

When numbers are equally spaces (as they are with consecutive numbers), the mean and median are equal.
So, the mean of the 5 numbers is n3 (the middle value).
So, we have 5 numbers with a mean of n3.
Removing the mean (n3) will result in 4 numbers with a mean of n3 as well.
So, the sum of these 4 numbers must be 4/5 of the sum of the original 5 numbers.

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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