1
$\begingroup$

I have a homework problem and I don't understand it. Here is the problem: The base two number 11111(base 2) has the same digit in all places. The same number can be written in different bases. Find two other bases where the number has the same digit in all it's places.

I get that 11111 base 2 = 31, but I don't really understand what the question is asking me or how to "solve" it.

$\endgroup$
  • $\begingroup$ Want to know why $11111_2=31,$if want to know then you have the solution below $\endgroup$ – Madrit Zhaku Nov 12 '13 at 19:13
1
$\begingroup$

$$11111_2=1\cdot 2^0+1\cdot 2^1+1\cdot 2^2+1\cdot 2^3+1\cdot 2^4=1\cdot 1+1\cdot 2+1\cdot 4+1\cdot 8+1\cdot 16$$ $$=1+2+4+8+16=31$$

$\endgroup$
0
$\begingroup$

A number of the form $aaa$ in base $b$ represents $ab^2+ab+a$, with the number of terms matching the number of digits. You are being asked to find two pairs $a,b$ (with some number of digits) so that it can represent $31_{10}$

$\endgroup$
  • $\begingroup$ So if I understand, a valid answer might be 1011 in base 3? And I would need to find one more, correct? $\endgroup$ – bert Nov 12 '13 at 19:15
  • $\begingroup$ You are correct that $1011_3=31_{10}$, but it doesn't meet the requirement that all the digits be the same. $\endgroup$ – Ross Millikan Nov 12 '13 at 19:16
  • $\begingroup$ Oh ok. Would 111 base 5 work? But I can't even think of what the next one might be. $\endgroup$ – bert Nov 12 '13 at 19:17
  • $\begingroup$ Yes, $111_5$ is one of the solutions. There aren't too many choices left for the number of digits. Bases can be larger than $10$ $\endgroup$ – Ross Millikan Nov 12 '13 at 19:18
  • $\begingroup$ Ok I figured it out. I got 11 base 30. Thank you very much. $\endgroup$ – bert Nov 12 '13 at 19:28
0
$\begingroup$

The number is $31$. If $dd\dots dd_b = 31_{10}$, then $d$ divides $31$. Only possibilities are $d= 1$ and $d=31$. So for all bases $b \gt 31$ you have a single digit which clearly passes the requirement. Now check all $11\dots 11_b$ for different bases $b$ and you have the rest.

$\endgroup$
  • $\begingroup$ I don't think a single digit solution was intended, though technically all the digits are the same. There are two multidigit solutions. $\endgroup$ – Ross Millikan Nov 12 '13 at 19:17
0
$\begingroup$

The number in decimal is 31. I think we've established that.

It appears that you're being asked for two other bases in which 31 is written with all the same digits.

Any base 32 or larger will work. In this case, decimal 31 will be a single digit. I guess they're all the same if there's only one, right?

Base 30 will work:

$$11_{30}=1\cdot 30^1+1\cdot 30^0$$

Since 31 is odd, the next possibility would be $33$, but that won't work because 28 isn't divisible by 3.

$55$ and higher won't work because the number would need to be at least base 6, and $55_6$ exceeds 31.

Let's go to $111$ then. Is there a value for $n$ such that $n^2 + n = 30$ ? Why, yes there is:

$$111_{5}=1\cdot 5^2+1\cdot 5^1+1\cdot 5^0$$

So there are the two (non-trivial single-digit) solutions.

$\endgroup$
  • 1
    $\begingroup$ Because $31$ is prime, the digit must be $1$ or $31$ $\endgroup$ – Ross Millikan Nov 12 '13 at 19:45
  • $\begingroup$ Good point. That would have made things easier. :) $\endgroup$ – John Nov 12 '13 at 19:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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