The number 92 can be written as the sum of cubes of three different

Question

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

Vi09 · Answer

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

luisdicampo · Answer

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

dhruva09 · Answer

1. x^3+y^3+z^3=92

2. Possible values can be only be among {1,2,3,4}, since 5^3=125; 125>92

3. Start from the highest possible value and consequently descend

4. Let x=4, y=3, z=2; x^3+y^3+z^3= 99 not possible.

5. Let x=4, y=3, z=1; x^3+y^3+z^3= 92

6. 4+3+1= 8

Bunuel · Answer

Experts' Global Video Explanation:

https://www.youtube.com/watch?v=W0R3dAFuOYA

The number 92 can be written as the sum of cubes of three different

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