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Algebra Level 2

Find the number of digits in $\Large 9^{9^{2}}$ .

Note : You may use the fact that $\log_{10} 3 = 0.4771$ correct up to 4 decimal places.

The answer is 78.

3 solutions

Arulx Z
Feb 7, 2016

${ 9 }^{ { 9 }^{ 2 } }={ 9 }^{ 81 }$

Number of digits -

$=\left\lfloor \log { { 9 }^{ 81 } } \right\rfloor +1\\ =\left\lfloor 81\log { 9 } \right\rfloor +1\\ =77+1=78$

Okay , If instead of 9^9^2 , we have 2^2^222 , what would the process be ? Because in this problem we can proceed by first solving 9^2 from the top according to the tower rule and then compute log (9^81) but in 2^2^222 , if we go by tower rule , 2^222 is difficult to find.. so , How can I find the number of digits in 2^2^222 ?

Aniruddha Bagchi - 3 years, 8 months ago

Akshay Yadav
Feb 7, 2016

Let,

$9^{9^{2}}=x$

Taking $\log$ on both sides,

$\log 9^{81}=\log x$

$81\times2\times0.4771=\log x$

$77.2936=\log x$

Hence,

No. of digits= $77+1=\boxed{78}$

I don't get it why add 1

Colene Gammad - 5 years, 4 months ago

Let's say that number $n$ has $x$ digits.
$10^{x-1} \le n \lt 10^x$
$x-1\le \log_{10}n \lt x$

From the definition of floor/greatest integer function, $\lfloor{n}\rfloor \le n\lt \lfloor{n}\rfloor +1$
Comparing, we get $x=\lfloor{n}\rfloor +1$

shaurya gupta - 5 years, 4 months ago

Well, $\lfloor log_{10} x \rfloor$ is always 1 less than the number of digits. For example: $\lfloor log_{10} 10 \rfloor = 1$ , $\lfloor log_{10} 100 \rfloor = 2$

Jonathan Yang - 5 years, 4 months ago

The $\left\lfloor \log _{ 10 }{ x} \right\rfloor$ basically approximates how much digits of "zeros" are there in a positive number $x$ e.g $\left\lfloor \log _{ 10 }{ 1000 } \right\rfloor=3$ , $\left\lfloor \log _{ 10 }{ 54 } \right\rfloor =1$ . However, as it only counts the zeros, it forgets the leftmost digit, which must not be $0$ . So we plus $1$ because of that.

Vu Vincent - 3 years, 9 months ago

Surya Sharma
Feb 7, 2016

Let the number of digits in {9^9^2} be x. Then for some natural number n, {10^(n-1)} will also have x digits. (There are n+1 digit in {10^2} So, Log{10^(n-1)} = Log {9^9^2} (n-1) Log 10 = {9^2} Log 9 Log 10=1 and Log 9 =0.95 This implies that n-1= 81X(0.95) n-1= 76.95 n=76.95+1=77.95 As the number if digits cannot be a fraction and can only be a whole number So number of digits are [n]=77

The answer is 77 right?

I entered 77 and it showed that the answer is wrong :(

Then I entered 78 :P and got it correct

Mehul Arora - 5 years, 4 months ago

I believe it's 78 because if you remember characteristic+1

Department 8 - 5 years, 4 months ago

What did you mean by characteristic +1? Do you know when should we applied the +1? Thx

Victor Zhang - 5 years, 4 months ago

@Victor Zhang – When you take log the integer part is known as characteristic and we do +1 because logs irks values less than there no. of digits. Eg. log(2)=0.3010 but we know $2^1$ has one digit so we do +1. This is also known as ceiling function or smallest integer function. You may check the wiki.

Department 8 - 5 years, 4 months ago

@Department 8 – Thx much! Anyway, Beside this problem, Could you please tell me other problem that require me to +1? Can you list it all?

Victor Zhang - 5 years, 4 months ago

@Victor Zhang – Try to write ceiling in search and filter for problem you will get many question.

Department 8 - 5 years, 4 months ago

@Victor Zhang – You will always need to add 1. Because when you find the characteristic of a no.( to calculate its log) , it is the digits before the decimal point -1.

anweshan bor - 5 years, 4 months ago

certainly 78. as lakshay said characterstic+1

Shreyash Rai - 5 years, 4 months ago

Sakshi is me

aishwarya dadhich - 5 years, 3 months ago

As usual u fluked the answer $\Large \boxed{\boxed{\boxed{\boxed{\boxed{:3}}}}}$

Aditya Kumar - 5 years, 4 months ago

Copied

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Mehul Arora - 5 years, 4 months ago

Let's check how much you know

The answer is 78.

3 solutions

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