Given that $2.004 < \log_{10} 101 < 2.005$ , how many digits are there in the decimal representation of $101^{101}$ ?

**
Details and assumptions
**

Clarification: The decimal representation of $2^{10}$ is $2^{10}=1024$ , which has 4 digits.

The answer is 203.

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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.

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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.

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```
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.

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We write log for log base 10

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

So this has 203 digits

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*
101<202.505
therefore, 203.404<(log10(101))
*
101+1<203.505.
hence, the answer is 203

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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

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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

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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]