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Math Expert V
Joined: 02 Sep 2009
Posts: 61549
Let a, b, c, d, and e be positive integers with a + b + c + d + e = 20  [#permalink]

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Question Stats: 19% (02:31) correct 81% (02:35) wrong based on 129 sessions

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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

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GMAT Tutor P
Joined: 24 Jun 2008
Posts: 2012
Re: Let a, b, c, d, and e be positive integers with a + b + c + d + e = 20  [#permalink]

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1
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
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Intern  B
Joined: 24 Jun 2019
Posts: 13
Let a, b, c, d, and e be positive integers with a + b + c + d + e = 20  [#permalink]

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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?
VP  V
Joined: 19 Oct 2018
Posts: 1314
Location: India
Re: Let a, b, c, d, and e be positive integers with a + b + c + d + e = 20  [#permalink]

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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?
Intern  B
Joined: 05 Jul 2017
Posts: 15
Location: India
Schools: IMD '21
Re: Let a, b, c, d, and e be positive integers with a + b + c + d + e = 20  [#permalink]

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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

Why can't it be 2,502,502,502,502 ?

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GMAT Tutor P
Joined: 24 Jun 2008
Posts: 2012
Re: Let a, b, c, d, and e be positive integers with a + b + c + d + e = 20  [#permalink]

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itspC wrote:
Why can't it be 2,502,502,502,502 ?

Because then the largest sum of any pair of numbers would be 502+502 = 1004, and that's not the smallest possible value of that largest sum.
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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 Re: Let a, b, c, d, and e be positive integers with a + b + c + d + e = 20   [#permalink] 28 Jan 2020, 01:19
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