Thousand Power Thousand

For positive integer $x$ , any expression $10^{x}$ ,simplified, has $x+1$ digits, e.g. $10^{0}=1$ , $10^{1}=10$ . Hence, $1000^{1000}=(10^{3})^{1000}=10^{3000}$ , so the simplified expression has $3000+1=3001$ digits.

1000^1000=10^3000 and answer is 3001 becuase 10^x has x+1 digits 10²=100

$1000^{1000}=1*10^{3000}=10^{3000}$ . This means there are 3,000 zeroes in the answer, however the part where it tricks you comes in when you need to realize that there is also a 1 at the beginning of the answer, therefore we get $3000+1=\boxed{3001}$ digits.

Guyz its just a pattern... we can make a formula to get the no. of digits in the solutions. It goes like this... the formula is x(n)+1 ... where "x" is the no. of zeroes in the base (limited to only 10,100,1000, etc of this type) and "n" is the exponent of the given number. this would give the total number of digits. Here, no. of zeroes=3 and exponent=1000, therefore, 3(1000)+1=3001 ... this is the answer...

Each time you multiply by 1000 you add 3 zeros, so you multiply 1000 times by 1000, then this is 3000 zeros, plus the number one in the beginning. This is 3000+1=3001

Hello, $1000^{1000}$ can be expressed as $(10^3)^{1000} = 10^{3000}$ . This number will look like this: $100000\cdots 000$ where there are exactly $3000$ zeroes. Thus, there are $\boxed{3001}$ digits (including the $1$ in front).

You can simply use log in this types of problem to count digits of big exponents. Just calculate the 10 based logarithm for any other exponent like log(5^56) for calculating the digits of the number 5^56. For knowing details you can visit This Algorithm . But , in this case this algorithm is not effective because it is too big exponent to calculate.

Let 1000^1000=x, taking log both side yhen finding x=10^3000. We have the solution as 10^a has always number of digits a+1 hence 3000+1=3001

An efficient way to find the amount of digits D in an exponent is to use the formula D = floor(1 + (b)Log10(a)), which yields the result of 3001

1000^1=1000, 3zeroes,4digits

1000^2=1000000. 6zeroes,7 digits

1000^3=1000000000, 9zeroes,10 digits.

so,1000^1000 has=(3x1000+1)digits=3001 digits

Every time you multiply something by 1000, you add three zeroes. We're already at four from 1000^1, so for the other 999 thousands you add three more. 3x999 + 4 gives you 3001

I am a big fan of simple solution to hairy looking problems .....so here goes 10^2 = 100 : 3 digits ( 2 zeros and 1) 100^2= 10000: 5 digits ( 4 zeros and 1) So you just have to calculate the number of zeros and increase it by one Number of zeros as can be seen by above examples = the numerical product of the power and the number of zeros in the base number So in 1000^1000 , number of zeros = 1000 x 3 ( no of zeros in the base number i.e. 1000) = 3000 and now add one to accommodate 1 So 3001 ....hope I have painted a picture

1000^1000=(10^3)^1000=10^3000. now 10^1 =2 digit 10^2=3 digit 10^3=4 digit 10^4= 5 digit that means 10^n contains (n+1) digits. so ,1000^1000 (=10^3000) will contain (3000+1=3001) digits

$1000^1$ has (1+3) digits. $\newline 1000^2$ has (1+3*2) digits. So $1000^{1000}$ has $1+3*1000 = 3001$ digits.

go to https://www.quora.com/How-many-digits-are-in-1000-1000

I don't know

1000 have 4 digits, 1000^2 have 7, ..10, 13,... Number of digits of 1000^1000: 4 + 999*3 equal to 3001 digits

What I did: It's (10^3)^(10^3) So that's 10^(3+3+3+3+3+... 10^3 times ) Therefore: 10^3000. Which is 3000+1 = 3001 digits. Since 10^x has x+1 digits.

No of terms in an exponential expression = $(log x) + 1$ , where $x$ is the exponential expression

So, here $x = 1000^{1000}$

$\text{No of terms} = log (1000^{1000}) + 1 = [1000 * log(1000)] + 1 \\ = [1000 * log(10^{3})] + 1 = [1000 * 3 * log (10)] + 1 = [1000 * 3 * 1] + 1 \\ = 3000 + 1 = 30001 \text{ terms}$

Do it with logic it will be easy

See, 1000^1=1000 so if power is 1 there are 3 zeros similarly if power is 2 then zeros are 6

applying unitary method u get that if power is 1000 no of zeros is 3000 and there os a 1 in 1000 so no. Of digits is 3000+1

10^1=10 so 1+1 THAT IS ONE IS THE NO. OF ZEROES THAT IS THE POWER AND REST 1 SO 1+1 SIMILARLY 1000^1000=(10^3)^1000 SO 3000 ZEROES AND 1 ONE SO 3000+1=30001

Not the best solution but how i figured it out: 1000 = 4 digits 1000000 = 7 digits 1000000000 = 10 digits

you form a pattern starting at 4 then you add 3 digits for every subsequent power therefore 1000^1000 would be 4+3(999) = 3000-3+4 = 3001

( 1000 x 3 ) + 1

(000+000....1000) times means 3*1000+1=3001

10^1 has 2 digits, 10^3 has 4. 10^4 has 5,..., 10^x has (x+1) digits. 1000^1000 = 10^3(1000) = 10^3000. Therefore 1000^1000 has 3001 digits.

Simple solution

(1000) $^{1}$ = 1000 contains 3 zeros

(1000) $^{2}$ = 1000000 contains 6 zeros

So (1000) $^{100}$ contains 300 zeros

And (1000) $^{1000}$ contains 3000 zeros and so total digits are 3001.

For each time you times it by 1000 you are adding 3 digits. You are adding 3 digits 1000 times therefore 3*1000=3000 plus the 1 at the start 30001

There are 3 zeroes in the no. So simply multiply 3 with power of the no. i.e., 1000 Now 3×1000= 3000 so, 3000+ (1)= 3001 1 came from the first digit which is "1"