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

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

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New post 31 Mar 2019, 21:07
1
13
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A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

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

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Re: Let a, b, c, d, and e be positive integers with a + b + c + d + e = 20  [#permalink]

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New post 21 Apr 2019, 21:54
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
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Let a, b, c, d, and e be positive integers with a + b + c + d + e = 20  [#permalink]

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New post 11 Oct 2019, 11:26
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?
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Re: Let a, b, c, d, and e be positive integers with a + b + c + d + e = 20  [#permalink]

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New post 11 Oct 2019, 13:33
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?
GMAT Club Bot
Re: Let a, b, c, d, and e be positive integers with a + b + c + d + e = 20   [#permalink] 11 Oct 2019, 13:33
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