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I'm new to binary numbers and number bases in general, there's a question in a book I'm looking to answer, I infer that it has $ 100 + 1 $ digits.

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    $\begingroup$ It does indeed have $101$ binary digits. Can you describe what it looks like? $\endgroup$ – Brian M. Scott Oct 13 '16 at 7:42
  • $\begingroup$ The digits represent $2^{100}, 2^{99} \dots 2^1, 2^0$. $0 \to 100$ is 101. But if it handles negative numbers (2s complement) then it has one more bit for the sign bit, so 102 digits. $\endgroup$ – arthur Oct 13 '16 at 7:45
  • $\begingroup$ @AlbertMasclans "$2^{100}$" has $3$ distinct symbols, and hence it cannot be a binary representation. $\endgroup$ – GoodDeeds Oct 13 '16 at 9:18
  • $\begingroup$ @arthur i disagree that a (signing) bit should be considered a digit, nor that a byte by definition should be considered a character $\endgroup$ – nl-x Oct 13 '16 at 9:38
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Not sure how to give a hint without blatantly giving the answer but...

Consider an n - digit number in base b.

That is $N = a_{n-1}a_{n-2}.....a_0= \sum_{k=0}a_kb^k $

Note $a_k < b$ so we can easily show $N <b^n $ (may have to repeat and argue inductively.

And presumably to be n - digit than $a_{n-1}\ne 0$ so $N \ge b^{n-1} $.

So we have: every n digit number is between $b^{n-1} $ inclusively and $b^n $ exclusively. This should be blindingly obvious to us if $b=10$.

So... that's a really important and fundamental result. Remember and use it.

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    $\begingroup$ Wow thanks , i'll go through the result , i'll ask questions if needed. $\endgroup$ – Gal Zafar Oct 13 '16 at 10:03
  • $\begingroup$ It's a basic result that often goes without being stated explicitely. Each digit represents a power of b. so 2^100 would be written as 1 with 100 zeros so there should be 101 digits. It should be self evident. $\endgroup$ – fleablood Oct 13 '16 at 17:00
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$2^{100}$ in base $2$ is $1$ followed by $100$ zeroes. Thus, the number of digits is $101$.

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How many digits does $2^{100}$ have if written in base $2?$

$1100101$

(That's in base $2,$ which is what the question asked for!)

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  • $\begingroup$ But as 2 is and 100 are base ten the answer should actually be 11111 as that's how many base 10 digits it'd have if the answer is written in base 2. As long as we are making jokes. Which we probably shouldn't. $\endgroup$ – fleablood Oct 13 '16 at 8:39

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