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Bunuel
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If a, b, and c are distinct positive integers, and a + b + c = 31, wha 20 May 2020, 07:11
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D
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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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D
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If a, b, and c are distinct positive integers, and a + b + c = 31, wha Updated on: 24 May 2020, 08:25
Originally posted by SiffyB on 20 May 2020, 08:18.
Last edited by SiffyB on 24 May 2020, 08:25, edited 1 time in total.
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Bunuel
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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If a, b, and c are distinct positive integers, and a + b + c = 31, wha 20 May 2020, 12:45
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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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If a, b, and c are distinct positive integers, and a + b + c = 31, wha 24 May 2020, 08:16
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Bunuel
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 24 May 2020, 16:16
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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.
{The Transformation-Nigeria}

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If a, b, and c are distinct positive integers, and a + b + c = 31, wha 17 Oct 2024, 05:13
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Given that a, b, and c are distinct positive integers, and a + b + c = 31, And we need to find the greatest possible value of a*b*c?

Theory: Given the sum of a set of number, their product will be maximum when they are equal or are close by

For a * b * c to be maximum all of them should be close to each other
Now, 31/3 ~ 10
=> Numbers can be 10, 10, 11 to make the sum as 31
But, they need to be distinct
=> we can try with 9, 10, 12 to get the product as 9 * 10 * 12 = 1080

So, Answer will be D
Hope it helps!

Watch the following video to MASTER Sequence problems

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If a, b, and c are distinct positive integers, and a + b + c = 31, wha 15 Jun 2025, 13:44
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Easy way to start:

Factor the highest number given (1200)

You find the three main factors (12, 10, and 10) are not only not distinct but also equate to 32

Then trickle down to the next option, factor 1080 to get down to (12, 9, 10)

Answer D :)
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