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Teacher asked the students to find the cube root of a natural number but she did not mention the base. Students assumed the base found the cube root. Each student got an integer. Find the sum of digits of that number.

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Your question doesn't make sense, I'm afraid. How many different bases? If the teacher has a countable infinity of students, each assuming a different base, then I suppose the answer must be 1. –  TonyK Aug 22 '11 at 21:15
    
See Ross's answer. Also, not all numerals have enough digits such that the digits can get summed. For instance, consider 6 in base ten. The sum of the digits here does not exist here where "the sum" means an addition, or a sequence of additions. Addition consists of a binary operation, and when people talk about "the sum" of more than two numbers, they (usually implicitly) mean a sequence of additions of all those numbers, where addition still qualifies as a binary operation. +6 has no meaning as a sum, and consequently the problem might not even have an ambiguous set of answers. –  Doug Spoonwood Aug 22 '11 at 22:03
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What I want to know is, how do you take the cubed root of a number without knowing the number? :/ –  anon Aug 23 '11 at 0:51
    
This problem appears in various places on the web, e.g., rangrut.com/share-post/selection-procedure/r1/Mzc1/in/Yahoo where it is given as a multiple choice question with the options being 0, 1, 6, 7, and 8. Stated that way, I guess the correct answer is 8. –  Gerry Myerson Aug 23 '11 at 3:18
    
@gerry can you please explain how you calculate this answer 8 –  rohit Aug 24 '11 at 15:28
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1 Answer

up vote 4 down vote accepted

This is not a well-formed question, as $1^3=1$ with digit sum $1$, $11^3=1331$ (in any base higher than $3$) with digit sum $2$, and $111^3=1367631$ (in any base higher than $7$) with digit sum $3$ appear to satisfy the requirement, among others.

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why dont you take 1111^4 ? –  rohit Aug 22 '11 at 20:20
    
The question referred to cubes. When $1111$ is cubed in base $10$, it carries. You can do it without carries in base $13$ or above. But I thought the pattern was clear. In fact any number is a candidate for this treatment if you choose a base high enough that there are no carries. –  Ross Millikan Aug 22 '11 at 22:19
    
@Ross: Don't you mean "any base higher than $3$" rather than "any base higher than 2$"? –  Michael Hardy Aug 23 '11 at 0:24
    
@Michael Hardy: Right. Fixed. I saw the need for a digit $3$ and thought I got one in base $3$. –  Ross Millikan Aug 23 '11 at 0:40
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