How Many Digits in the Number $2^{100}$?

[kaliprasad]

without using logs and calculator find the number of digits in $2^{100}$

 

[Nono713]

Re: number of digit

Spoiler

(EDITED)

$$2^{100} = \left ( 2^{10} \right )^{10} = 1024^{10} = \left ( 1000 + 24 \right )^{10} = 1000^{10} + \sum_{i = 0}^{9} c_i 1000^{i} 24^{10 - i}$$
Where the $c_i$ are the binomial coefficients for this binomial term:
$$c_i = \left ( 1, 10, 45, 120, 210, 252, 210, 120, 45, 10 \right )$$
Now observe that for $0 \leq i \leq 9$ we have that:
$$c 1000^i 24^{10 - i} < \frac{9}{10} 1000^{10} ~ ~ \iff ~ ~ c 24^{10 - i} < \frac{9}{10} 1000^{10 - i} ~ ~ \iff ~ ~ c < \frac{9}{10} 40^{10 - i} < \frac{9}{10} \left ( \frac{1000}{24} \right )^{10 - i}$$
It is easy to see that this inequality holds true for all the binomial coefficients:
$$c_0 = 1 < \frac{9}{10} 40^{10}$$
$$c_1 = 10 < \frac{9}{10} 40^9$$
$$\cdots$$
$$c_7 = 120 < \frac{9}{10} 40^3$$
$$c_8 = 45 < \frac{9}{10} 40^2$$
$$c_9 = 10 < \frac{9}{10} 40^1$$
So we can say that:
$$c_i 1000^i 24^{10 - i} < \frac{9}{10} 1000^{10} ~ ~ ~ \left ( 0 \leq i \leq 9 \right )$$
And so:
$$\sum_{i = 0}^{9} c_i 1000^i 24^{10 - i} < 10 \cdot \frac{9}{10} \cdot 1000^{10} = 9 \cdot 1000^{10}$$
We therefore conclude that:
$$1000^{10} = 10^{30} < 2^{100} < 1000^{10} + 9 \cdot 1000^{10} = 10 \cdot 1000^{10} = 10^{31}$$
And so $2^{100}$ has $31$ digits.

 

[kaliprasad]

Re: number of digit

Spoiler

We therefore conclude that:
$$1000^{10} < 2^{100} < 1000^{10} + 9 \cdot 1000^{10} = 1000^{11}$$

Spoiler

it should be
$1000^{10} < 2^{100} < 1000^{10} + 9 \cdot 1000^{10} = 10 *1000^{10}$

 

[Greg]

Re: number of digit

Spoiler

\(\displaystyle 2^{100}=1024^{10}=(1024^5)^2\)

\(\displaystyle 1024^2\) has 4 + 4 - 1 = 7 digits, \(\displaystyle 1024^4\) has 7 + 7 - 1 = 13 digits,

\(\displaystyle 1024^5\) has 13 + 4 - 1 = 16 digits, \(\displaystyle 1024^{10}\) has 16 + 16 - 1 = 31 digits.

Speculation at this point; $1024^{100}$ has 302 digits. :)

 

[kaliprasad]

Re: number of digit

Spoiler

\(\displaystyle 2^{100}=1024^{10}=(1024^5)^2\)

\(\displaystyle 1024^2\) has 4 + 4 - 1 = 7 digits, \(\displaystyle 1024^4\) has 7 + 7 - 1 = 13 digits,

\(\displaystyle 1024^5\) has 13 + 4 - 1 = 16 digits, \(\displaystyle 1024^{10}\) has 16 + 16 - 1 = 31 digits.

Speculation at this point; $1024^{100}$ has 302 digits. :)

not too convincing

Spoiler

2 digit * 2 digit can be 4 digit as 32 * 32 = 1024

 

[Greg]

Re: number of digit

not too convincing

Spoiler

2 digit * 2 digit can be 4 digit as 32 * 32 = 1024

Spoiler

Leading 10...

 

[kaliprasad]

Re: number of digit

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Leading 10...

Spoiler

explanation was missing . now is is 100% correct

 

[kaliprasad]

I should close it before the year end.
both the ans are correct and greg1313's ans is elegant.
here is my ans

Spoiler

$2^{10} = 1024 > 1000 = 10^3$
so $2^{100} > 10^{30}$

further
$2^{10} < 1025 < 10^3 * \frac{41}{40}= 10^3 *( 1+ \frac{1}{40}) $
so $2^{100} < 10^{30} * ( 1+ \frac{1}{40})^{10} < 10^{30} * ( 1+ \frac{1}{40})^{40}< 10^{30} * e < 10^{30} * 3$
so $10^{30} < 2^{100} < 3 * 10^{30}$
hence 31 digits

 

[kaliprasad]

Re: number of digit

Spoiler

Speculation at this point; $1024^{100}$ has 302 digits. :)

your speculation is right as per http://mathhelpboards.com/challenge-questions-puzzles-28/how-many-digits-5251.html