-2
$\begingroup$

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.

  • 2
    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 1.2011-08-22
  • 0
    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.2011-08-22
  • 1
    What I want to know is, how do you take the cubed root of a number without knowing the number? :/2011-08-23
  • 0
    This problem appears in various places on the web, e.g., http://www.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 8.2011-08-23
  • 0
    @gerry can you please explain how you calculate this answer 82011-08-24
  • 0
    As has been noted elsewhere, 1331 is a possibility for the number given by the teacher, and $1+3+3+1=8$.2011-08-25
  • 0
    @Gerry: I looked at your link, and the test is so garbled that I can't believe it was set in earnest. One hopes it was the poster who mangled it! So the OP's question is definitely not to be taken at face value.2011-09-01
  • 0
    @TonyK, it's that garbled version that I found in various places on the web. I didn't find an ungarbled version though I think you and I agree there must be one. Presumably it runs something like this: A teacher asked her students to find the cube root of a natural number. She did not mention what base the number was written in. Different students assumed different bases, and each student found an integer answer. Which of these could be the sum of the digits of the original number: 0, 1, 6, 7, 8. Well, you'd need something to rule out the trivial answers 0 and 1.2011-09-02

1 Answers 1

5

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.

  • 0
    why dont you take 1111^4 ?2011-08-22
  • 0
    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.2011-08-22
  • 0
    @Ross: Don't you mean "any base higher than $3$" rather than "any base higher than 2$"?2011-08-23
  • 0
    @Michael Hardy: Right. Fixed. I saw the need for a digit $3$ and thought I got one in base $3$.2011-08-23