How Many Digits? A Headache

As ${ 2 }^{ 1000 }$ is not a multiple of 10, it follows that $10^m < 2^{1000} < 10^{m+1},$ where $10^m$ and thus $2^{1000}$ contain $m+1$ digits.

Since $1000\log 2 \approx 301.02999,$ $m= \lfloor 301.02999 \rfloor = 301,$ so $2^{1000}$ contains 302 digits.

Before, note that $10^{n}$ has $n+1$ digits. Taking the common logarithm of $2^{1000}$ , $log2^{1000}$ $=1000log2$ $=1000*0.30103...$ (Sorry, approximating here) $=301.03...$ Therefore, $2^{1000}=10^{301.03....}$ Since: $10^{301}<10^{301.03}...<10^{302}$ Therefore, the number has 302 digits.

I did it with logs too but it's more fun to think about a solution with no logs. 2^10 is 1024 or 1000X1.024. Therefore 2^1000 = 2^10^100 = (1000X1.024)^100 = (1000^100)X(1.024^100). The least significant figure is 3 decimal spaces to the right... which means every time you multiply it by 1.024 the least significant decimal moves 3 more to the right. We're raising 1.024 to the 100th power which means the least significant figure is 300 spaces to the right. Raising 1.024 to the 100th power on my calculator is 10 plus a fractional remainder. Therefore we have two decimals to the left and 300 to the right... 302!

Using the simple formula to find number if digits in $a^{b}$ = $b \ log \ a + 1$

IN THIS CASE, the number of digits in $2^{1000} = 1000* log_{10}2 +1 = 1000 * 0.3010 +1 = \boxed{302}$

Let $2^{1000}=10^x$ then $int(x) +1 =$ the number of digits.

$1000log2=xlog10$

$x=1000\frac{log2}{log10}$

$x=1000log_{10}2$

$x=1000 \times 0.3010 =301.0$

$int(x)+1=301+1=302$

$Number\quad of\quad Digits\quad in\quad { 2 }^{ 1000 }=\left\lceil \log _{ 10 }{ { 2 }^{ 1000 } } \right\rceil \\ =\left\lceil 1000(\log _{ 10 }{ 2 }) \right\rceil \\ =\left\lceil 1000\times 0.3010299 \right\rceil \\ =\left\lceil 301.02 \right\rceil \\ =302$

NOT A SOLUTION!

BEAUTY OF LOGARITHMS:

Becomes quite simple, if you have a little knowledge of logarithms and understand why such a function evolved. Logarithm is thus one of the most beautiful functions of maths. It not only eases your calculations but also makes big and complicated numbers small ( so as they are readable) and makes very small numbers big enough so that it is easy to handle them, and obviously we can take antilog to get desired result. Moreover, as in this question it tells the number of digits in such a large number.

Beautiful function, isn't it? Love Math, Love Log!!

2^10 = 1024 (4 digits)
10^x = 2^10
x log 10 = 10 log 2
x = 10 * 0.30103
x = 3.01 ~ 4 digits
2^100 =1267650600228230000000000000000 (31 digits)

10^x = 2^100
x log 10 = 100 log 2
x = 100 * 0.30103
x = 30.103 ~ 31 digits
2^1000 =1267650600228230000000000000000....... (302 digits) 10^x = 2^1000
x log 10 = 1000 log 2
x = 1000 * 0.30103
x = 301.03 ~ 302 digits

Not going to lie. At first I calculated it to be 3011 digits...

total digits= 100Log(2)=301.02999≈302

10715086071862673209484250490600018105614048117055336074437503883703510511249361‌​224931983788156958581275946729175531468251871452856923140435984577574698574803934‌​567774824230985421074605062371141877954182153046474983581941267398767559165543946‌​077062914571196477686542167660429831652624386837205668069376. Count the number of digits, you will get the answer.Instead you can always use logarithms.

2^n have the number of digits is (3n+1)

where

n is multiple of 10

tried this way, I only need to know if this is correct or if there are better ways to solve this:

21000 does not have a factor of 5 obviously therefore we can assume

10m<21000<10m+1 for some m

Assume k=21000, then take log on both sides logk=1000log2≈301.02999>301 Therefore 21000 has 302 digits.

$log_{10} 2 = 0.301 → 2 = 10^{0.301}$

$2^{1000} = (10^{0.301})^{1000} = 10^{301}$

$10^{301}$ is $1$ followed by $301$ $0$ 's

$301 + 1 = 302$ digits

Given that $log_{10} 2 \approx 0.3010$ , we can show that $2 \approx 10^{0.3010}$ using the definition of a logarithm. So now we have $(10^{0.3010})^{1000}$ , which is equivalent to $10^{301.0}$ , which has 302 decimal places.

Given that log(10) 2≈0.3010, we can rewrite

this as 10^(301/1000)≈2. Since we are

attempting to find the number of digits in

2^1000, we can simply use the property of

substitution and obtain

10^((301/1000)×1000). The 1000's cancel

each other and we are left with 10^301. Now

at this point we may find that the total

number of digits in 10^301 is 302 because the

number of digits in 10^n is n+1. And since

10^301≈2^1000, we may conclude that the

number of digits in 2^1000 is 302. Thus we

obtain 302 as our final answer.

Bash shell one-liner:

1

dc <<< "2 1000 ^ p" | perl -pe 's/[\s\\]//g' | wc -c

and you get 302.

We have $2^{1000}$ $=10^{\log_{10}{2^{1000}}}$ $=10^{1000\log_{10}{2}}$

And using $\log_{10}{2} \approx 0.3010$ , we have

$2^{1000}=10^{1000\log_{10}{2}} \approx 10^{0.3010*1000}$ $=10^{301}$

Then, $2^{1000}$ is approximately equal to (greater than) $10^{301}$ , which have $302$ digits. And $2^{1000}=10^{1000\log_{10}{2}}$ is less than $10^{302}$ , which have $303$ digits. It is easy to see that $10^{1000\log_{10}{2}}=2^{1000}$ is nearer from $10^{301}$ than from $10^{302}$ , and therefore $2^{1000}$ must have $302$ digits, and not $303$ .

Given the logarithm: 10^(0.3010) = 2

hence: 2^1000 = 10^301 = 100000000.......... (301 zeros)

hence : the number of digits is 302

2^1000 1000(log2)=301.3 When u take antilog , the characteristic is 302 ... Therefore it would be of the order 10^302 .. So it has 302 digits ..

We can let $2^{ 1000 }$ = k, so ${ log }_{ 2 }$ k = 1000. Then, change the base, $\frac { { log }_{ 10 }{ k } }{ { log }_{ 10 }{ 2 } }$ =1000. Given ${ log }_{ 10 }{ 2 }$ =0.3010, so ${ log }_{ 10 }{ k }$ =301. Change to index form, k = ${ 10 }^{ 301 }$ . In this way, we know that there are 302 digits for this value.

Log10 (2^1000) =1000*Log10(2) =1000 * .301=301 So, 10^301 = 2^1000 And 10^n contains n+1 digits So, we get 301+1=302 digits

X=2^1000, LogX=1000Log2 where Log is in base 10.. LogX=1000×0.3010 (Given fact in the problem) LogX=301 X =10^301 The digits in 10^n term are n+1... Hence the number of digits in the answer are 302

2^n contains x+1 digits, where x = floor(n*log2)

(\log_10 2\ )= x

10^{x} = 2

if x = 0.3010

10^{0.3010} = 2

We know that log2=0.301. So, 10^0.301=2. So 2^1000=10^301. 10^301 has 302 digits, so the answer is 302 digits. (P.S. 10^301 is not equal to 2^1000, it's an approximation due to an approximate logarithmic value)

Given that log 2 = 0.3010

log 2 ^ 1000 = 1000 log 2 = 1000 (0.3010) = 301

2 ^ 1000 = 10 ^ 301

2 ^ 1000 contains 302 digits.

log $2^{1000}$ = 1000log 2 = 301.0299957 $\simeq$ 302

2^1000 is not a multiple of 10 therefore: 10^x < 2^1000 < 10^(x + 1) 10^x = 2^1000 xlog10 = 1000log2 x = 301.3 therefore 2^1000 is between 10^301 and 10^302 Therefore it has 302 Digits.

2^1000 = x . Taking Log on both sides :- 1000 log 2 = log x

1000 (.3010) = log x

301 = log x Taking antilog :- 10^301 = x

Since , raising 10 to any power gives a number having number of digits (1+value of power raised ). So x = 2^1000 will have 301+1=302 digits :)