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If a, b, and c are distinct positive integers, and a + b + c = 31, wha

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If a, b, and c are distinct positive integers, and a + b + c = 31, wha  [#permalink]

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New post 20 May 2020, 06:11
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If a, b, and c are distinct positive integers, and a + b + c = 31, wha  [#permalink]

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New post Updated on: 24 May 2020, 07:25
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Bunuel wrote:
If a, b, and c are distinct positive integers, and a + b + c = 31, what is the greatest possible value of a*b*c?

A. 1024
B. 1056
C. 1072
D. 1080
E. 1200


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To find out the greatest possible value of the product of a, b and c, these three integers must be as close as possible to each other.
Their values must lie around \(\frac{31}{3} = 10.33\)

Remember that these three integers have to be distinct,
by hit and trial, let a = 10, b = 9 and c = 12 (c cannot be 11 since 11+10+10 = 31 - there will be repitition).

Therefore, product of a, b and c = 10*9*12 = 1080 (D)

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Originally posted by SiffyB on 20 May 2020, 07:18.
Last edited by SiffyB on 24 May 2020, 07:25, edited 1 time in total.
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If a, b, and c are distinct positive integers, and a + b + c = 31, wha  [#permalink]

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New post 20 May 2020, 11:45
I think Answer is D

Multiplication of number near to average of any sum should give highest value.

31/3=~ 10

So 9 , 10 and 12 will be number.

Since integers should be distinct so we can not repeat 10 two times to get 1100

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Re: If a, b, and c are distinct positive integers, and a + b + c = 31, wha  [#permalink]

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New post 24 May 2020, 07:16
Bunuel wrote:
If a, b, and c are distinct positive integers, and a + b + c = 31, what is the greatest possible value of a*b*c?

A. 1024
B. 1056
C. 1072
D. 1080
E. 1200



The greatest product of any number of positive integers when they are given a fixed sum is when the numbers are closest to one another. Since there are 3 numbers and their sum is 31, each number should be around 10. Therefore, the greatest product would be 10 x 10 x 11 = 1100 if the 3 numbers are not necessarily distinct. However, since they have to be distinct, we can decrease one of the 10s to 9 and increase 11 to 12. Therefore, the greatest product is 9 x 10 x 12 = 1080 given that the 3 numbers are distinct and their sum is 31.

Answer: D

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If a, b, and c are distinct positive integers, and a + b + c = 31, wha  [#permalink]

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New post 24 May 2020, 15:16
We try finding three consecutive numbers and sum it to 31.
That let a be the greatest of such positive integer such that
a+a-1+a-2=31
3a=34
a=11.333
Take the ceiling function of a we have 12 and floor function of a-1 and a-2 we have
10 and 9 respectively. Hence the greatest product is
9×10×12=1080. Hence OPTION D.
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If a, b, and c are distinct positive integers, and a + b + c = 31, wha   [#permalink] 24 May 2020, 15:16

If a, b, and c are distinct positive integers, and a + b + c = 31, wha

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