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27 Nov 2025, 22:05
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
84% (01:27) correct
16%
(02:41)
wrong
based on 101
sessions
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The number 92 can be written as the sum of cubes of three different positive integers. What is the sum of these three positive integers?
A. 6
B. 7
C. 8
D. 9
E. 12
A. 6
B. 7
C. 8
D. 9
E. 12
28 Nov 2025, 01:36
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Ans is C as 3 cube + 4 cube+ 1 cube = 27+ 64 +1 = 92
therefore sum of digits = 3+4+1= 8
also to think for the numbers we know that none of the number would be beyond 4 as the cube of 5 is 125 which is greater than 92 , so we can directly think by taking the cubes of numbers from 1 to 4
therefore sum of digits = 3+4+1= 8
also to think for the numbers we know that none of the number would be beyond 4 as the cube of 5 is 125 which is greater than 92 , so we can directly think by taking the cubes of numbers from 1 to 4
28 Nov 2025, 06:57
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ExpertsGlobal5
To find the three integers, we first list the perfect cubes of positive integers that are less than 92.
Step 1: List Possible Cubes
We need \(x^3 + y^3 + z^3 = 92\).
Let's list the first few cubes:
\(1^3 = 1\)
\(2^3 = 8\)
\(3^3 = 27\)
\(4^3 = 64\)
\(5^3 = 125\) (Too large, so the integers must be chosen from the set {1, 2, 3, 4})
Step 2: Find the Combination
We need to pick three different numbers from the set of cubes \(\{1, 8, 27, 64\}\) that add up to 92.
Let's start with the largest possible value to cover the most "ground":
Try \(64\) (which is \(4^3\)).
Remaining sum needed: \(92 - 64 = 28\).
Now we need two distinct cubes from the remaining set \(\{1, 8, 27\}\) that sum to 28.
It is easy to see that:
\(27 + 1 = 28\)
So the cubes are \(64\), \(27\), and \(1\).
Let's verify:
\(4^3 + 3^3 + 1^3 = 64 + 27 + 1 = 92\).
The three distinct positive integers are \(4\), \(3\), and \(1\).
Step 3: Calculate the Sum
The question asks for the sum of these three positive integers:
\(Sum = 4 + 3 + 1 = 8\)
This matches Option (C).
Answer: C
