GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 22 Oct 2019, 05:37 GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  Let a, b, c, d, and e be positive integers with a + b + c + d + e = 20

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 58428
Let a, b, c, d, and e be positive integers with a + b + c + d + e = 20  [#permalink]

Show Tags

1
13 00:00

Difficulty:   95% (hard)

Question Stats: 14% (02:28) correct 86% (02:31) wrong based on 77 sessions

HideShow timer Statistics

Let a, b, c, d, and e be positive integers with a + b + c + d + e = 2010 and let M be the largest of the sum a + b, b + c, c + d and d + e. What is the smallest possible value of M?

A. 670
B. 671
C. 802
D. 803
E. 804

_________________
GMAT Tutor G
Joined: 24 Jun 2008
Posts: 1802
Re: Let a, b, c, d, and e be positive integers with a + b + c + d + e = 20  [#permalink]

Show Tags

2
1
To make the largest sum as small as possible, we'll want to make all of the sums equal (or as close to equal as we're allowed to make them), because if, say, d+e were the largest sum, and if it were larger than b+c, we'd be able to reduce d or e, and increase b or c by a corresponding amount, and thereby make d+e, the largest sum, smaller. So we wouldn't have found the smallest possible largest sum in this case.

And if all of the sums are equal, so a+b = b+c, b+c = c+d and so on, we find that a=c=e and b=d. If we permit zero, we'd find this arrangement makes M smallest:

670, 0, 670, 0, 670

but since zero is not allowed (the letters must be positive integers), the answer will be 671, which we can get with an arrangement like this:

669, 2, 668, 2, 669
_________________
GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com
Intern  B
Joined: 24 Jun 2019
Posts: 4
Let a, b, c, d, and e be positive integers with a + b + c + d + e = 20  [#permalink]

Show Tags

IanStewart wrote:
To make the largest sum as small as possible, we'll want to make all of the sums equal (or as close to equal as we're allowed to make them), because if, say, d+e were the largest sum, and if it were larger than b+c, we'd be able to reduce d or e, and increase b or c by a corresponding amount, and thereby make d+e, the largest sum, smaller. So we wouldn't have found the smallest possible largest sum in this case.

And if all of the sums are equal, so a+b = b+c, b+c = c+d and so on, we find that a=c=e and b=d. If we permit zero, we'd find this arrangement makes M smallest:

670, 0, 670, 0, 670

but since zero is not allowed (the letters must be positive integers), the answer will be 671, which we can get with an arrangement like this:

669, 2, 668, 2, 669

I have a doubt. Why can't it be 669,1,669,1,669?
And why the answer can't be 670?
Director  D
Joined: 19 Oct 2018
Posts: 989
Location: India
Re: Let a, b, c, d, and e be positive integers with a + b + c + d + e = 20  [#permalink]

Show Tags

669,1,669,1,669
Their sum is not equal to 2010

Krish728 wrote:
IanStewart wrote:
To make the largest sum as small as possible, we'll want to make all of the sums equal (or as close to equal as we're allowed to make them), because if, say, d+e were the largest sum, and if it were larger than b+c, we'd be able to reduce d or e, and increase b or c by a corresponding amount, and thereby make d+e, the largest sum, smaller. So we wouldn't have found the smallest possible largest sum in this case.

And if all of the sums are equal, so a+b = b+c, b+c = c+d and so on, we find that a=c=e and b=d. If we permit zero, we'd find this arrangement makes M smallest:

670, 0, 670, 0, 670

but since zero is not allowed (the letters must be positive integers), the answer will be 671, which we can get with an arrangement like this:

669, 2, 668, 2, 669

I have a doubt. Why can't it be 669,1,669,1,669?
And why the answer can't be 670? Re: Let a, b, c, d, and e be positive integers with a + b + c + d + e = 20   [#permalink] 11 Oct 2019, 13:33
Display posts from previous: Sort by

Let a, b, c, d, and e be positive integers with a + b + c + d + e = 20

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  