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I noticed an interesting pattern the other day. Let's take a look at the powers of 10 in binary:

  • $10^0$ = 1 = 1 b
  • $10^1$ = 10 = 10 10 b
  • $10^2$ = 100 = 1100 100 b
  • $10^3$ = 1000 = 111110 1000 b

Basically, it seems that $10^n$ for any non-negative integer $n$ written out in base 2 ends with its base 10 representation.

Does this pattern go on forever, and if so, can anyone provide me with a satisfactory explanation as to why this happens?

BambooleanLogic
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1 Answers1

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Multiplication of a number in binary by $2^{n}$ adds $n$ zeroes to the expression. $10^{n}=2^{n}5^{n}$, so as $2^{n}$ divides $10^{n}$, when expressed in binary you are adding $n$ zeroes to the number $5^{n}$ expressed in binary form. This is directly analogous to the base $10$ case.

Edward ffitch
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    Ah, of course! As an addendum to your argument, I'd just like to point out that $5^n$ is always odd and therefore ensures that it never contributes to any additional trailing zeroes. – BambooleanLogic Jan 19 '14 at 11:44