Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote -2 down vote favorite

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.

share|improve this question
1  
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
1  
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
show 3 more comments

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.

share|improve this answer
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.