101 to the 101

The number of digits in a number is given by $\lfloor \mbox{log of that number} \rfloor+1$ . Let $\log 101 = x$ . $\log (101^{101}) = 101\times \log 101$ by the laws of logarithm. Therefore, the number of digits in $101^101$ is $[101\times x] + 1$ . It is given that $2.004<x<2.005$ , so $202.404 < 101 x < 202.505$ and the no.of digits is $\lfloor 101 \times x \rfloor + 1 = 202 + 1 = 203$ .

Note: $\lfloor x \rfloor$ denotes the greatest integer function.

[LaTex edits. - Calvin]

log(base 10)101>2.004--------10^2.004<101

Similarly, log(base 10)101<2.005--------10^2.005>101

Thus (10^2.004)^101<101^101<(10^2.005)^101

We can then get 10^202.404<101^101<10^202.505

The decimal representation is the number of digits in 10^202 (10^0.404 is single-digit), which is 100000....(202 zeroes), or 203 digits.

2.004 < log 10 (101) < 2.005 --> 2.004*101 < log 10 (101^101) < 2.005*101 --> 202.404 < log_10 (101^101) < 202.505 --> 10^202.404 < 101^101 < 10^202.505

10^202 has 203 digits (1 followed by 202 zeroes), while 10^203 has 204 digits (1 followed by 203 zeroes), and 101^101 is between them, so 101^101 has 202+1=203 digits.

  2.004 < log_10 (101) < 2.005

or; 2.004 * 101 < (101)*( log 10 (101)) < 2.005 or, 202.404 < log 10 (101^101) < 202.505 or, 10^202.404 < 101^101 < 10^202.505

since, 10^202.404 = (10^202) (10^.404) = 2.53 (10^202)
so, the total no of digits in 10^202.404 is 203. again, 10^202.505 = (10^202) (10^.505) = 3.198 (10^202)
so, the total no of digits in 10^202.505 is also 203.

101^101 exists in between these two numbers. so, it wil also have 203 digits.

We write log for log base 10

log (101)^101=101*log(101) 202.404<101log(101)<202.505

So this has 203 digits

since 2.004<log10(101)<2.005, 202.404<(log10(101)) 101<202.505 therefore, 203.404<(log10(101)) 101+1<203.505. hence, the answer is 203

As log (101)^101

  = 101* log(101)

  =101*(2.0043)

  =202.4343

now

Number of digits in (101)^101 = [ log{(101)^101} ]

                                                = [ 202.4343 ]

                                                =  203

where [ ] denote greatest integer function

first get the value of 101^101;

log 101^101=log x; which results to

101(log 101)=log x ;using the properties of logarithm

since log 101>2.004; we substitute it ; we get

101(2.004)=log x

log x=202.404

x=10^202.404; which has 203 digits because 10^202 has 202 0's and one 1

WE KNOW 2^{10}=1024 HAS 4 DIGITS ALSO LOG2^{10}=3.010 SO INTEGRAL PART IS 3 SONO OF DIGIT OF 2^{10}=3+1 IN THE SAME WAY LOG101^{101} IS NEARLY EQUAL TO202.404 SO NO OF DIGIT=202+1

as you can see 10^3 has 4 digits, using log 10^3 >> 3 log 10 = 3, then just add 1, it's 4! Works for another, like 10^6 has 7 digits >> log 10^6 = 6 log 10 >> 6, add 1, it's 7! I don't know, intuition. LOL So 101^101 >> using log >> log 101^101 = 101 log 101, as you can see 2.004<log 101<2.005, so, 101x2 = 202, then add 1. LOL. YAY!

The number $N$ has $\lfloor \log_{10} N \rfloor +1$ digits. We have that $202 <202.4 < \log_{10} 101^{101} < 202.6 < 203$ . Hence, $101^{101}$ has $\lfloor \log_{10} 101^{101} \rfloor +1 = 202+1 = 203$ digits.

First take {100}^{100}. This is equal to {10}^2 to the power of {10}^2. This has 201 digits. Since 101 is 10 to the power of a number between 2.004 and 2.005, which is below 2.1, we have to add 2 more digits since it is {101}^{101}.

In this problem is use my logic because i'm in ninght grade and i dont learn The digits of 101^n is 2n+1(it makes half hour to find 2n+1)