An algebra problem by Hritesh Mourya

Algebra Level 3

What is the number of digits in 200 0 2000 2000^{2000} ?


The answer is 6603.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Hung Woei Neoh
Jul 5, 2016

Notice that:

log 10 10 = 1 , log 10 99 1.996 log 10 100 = 2 , log 10 999 2.9996 log 10 1000 = 3 , log 10 9999 3.99996 \log_{10} 10 = 1,\quad\log_{10} 99 \approx 1.996\\ \log_{10} 100 = 2,\quad\log_{10} 999 \approx 2.9996\\ \log_{10} 1000 = 3,\quad \log_{10} 9999 \approx3.99996

From here, we can see that the number of digits, n n in a number x x is represented as

n ( x ) = log 10 x + 1 n(x) = \lfloor \log_{10}x \rfloor + 1

Therefore, the number of digits in 200 0 2000 2000^{2000} :

n ( 200 0 2000 ) = log 10 200 0 2000 + 1 = 2000 log 10 2000 + 1 = 6602.06 + 1 = 6602 + 1 = 6603 n(2000^{2000})\\ = \lfloor \log_{10} 2000^{2000} \rfloor + 1\\ = \lfloor 2000 \log_{10} 2000 \rfloor + 1\\ =\lfloor 6602.06 \rfloor + 1\\ =6602+1\\=\boxed{6603}

Thanks for the correction

Hritesh Mourya - 4 years, 11 months ago
Manish Jha
Aug 26, 2016

log2000^2000 = 2000log2000 = 2000×3.301029 =6602.058 here characteristic is 6602 so the no of digits is (characteristic+1) =(6602+1) =6603 .

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...