Prove that $2002^{2001}$ and $2002^{2001}+2^{2001}$ have the same number of digits. I have an arithmetic exercise as follows: Prove that $2002^{2001}$ and $2002^{2001}+2^{2001}$ have a same number of digits. It seems easy but I don't know how to do. I can do it if $2002$ is replaced by $2000$. Any hints are appreciated.
Thanks a lot!
 A: For those who learned key facts about logarithms in their youth, here is a way through which may not appeal to OP (though if decent log tables to five digits are available this will do - I did it using facts I remember).
The first number is $$M=2^{2001}\cdot1001^{2001}$$
The second is $$N=2^{2001}\cdot\left(1001^{2001}+1\right)$$
Now if we take logs to base $10$ $$\log M=2001 \log 2+2001 (3+\log 1.001)$$
We want to estimate the leading digit (or few) of $M$ and hence the decimal part of the logarithm to a significant figure or two.
Now $\log 2 =0.30103$ to the nearest $0.00001$ so $2001 \log 2$ is within the bounds $602.36\pm 0.02$
$\ln 10 \approx 2.303$ (within 0.05%) and $\ln 1.001 \approx 0.001$ (easily within 0.5%) so $\log 1.001 =\cfrac {\ln 1.001}{\ln 10} \approx 0.00043$ and $2001 \log 1.001 \approx 0.86$ certainly to within $2\%$ so say $\pm 0.02$
So the fractional part of the logarithm is $0.86+0.36\pm0.04=1+0.22\pm 0.04$ which makes the leading digit $1$.
If the leading digit is $1$, then increasing the number by less than 1% will not increase the number of digits.
A: Basically, the number of digits in a decimal number N is given by $\lfloor{\log_{10}N}\rfloor + 1$
They key is to find an estimate for the base-10 logarithm of the number.
Start with the Taylor expansion of $\ln(1+x)$ and convert that to a base-10 log.
Then try to figure out how to use that to compute a series for $\log_{10}(2002^{2001} + 2^{2001})$
Now see if you can manipulate that into proving that the result is very close to $\log_{10}2002^{2001}$. More precisely, you want to prove that the floor of the base-10 log of one number is equal to the floor of the base-10 log of the other.
A: Note that $2002^{2001} \gt 2^{2001}\cdot 1000^{2001}$, so there will not be a carry unless $2002^{2001}$ starts with over $6000\ \ 9$'s.  $\log_{10} {2002}^{2001}=2001\log 2002\approx 2001\cdot (3.30103+\log_{10} 1.001)$ and the decimal part is far from $0.9999$
A: $\ln (a + b) = \ln (a(1 + \dfrac{b}{a})) = \ln a + \ln (1 + \dfrac{b}{a})$.
The number of digits in the base-10 numeral for integer $n > 0$ is $\lceil \log_{10}n \rceil$.  $\log_{10} n = \dfrac{\ln n}{\ln 10}$
Let $a = 2002^{2001}$ and $b = 2^{2001}$
$\lceil \log_{10} a \rceil = \left\lceil \dfrac{\ln a}{\ln 10} \right\rceil = \left\lceil \dfrac{\ln 2002^{2001}}{\ln 10} \right\rceil=\left\lceil \dfrac{2001 \ln 2002}{\ln 10} \right\rceil = \lceil 6606.2296\dots\rceil = 6697$
$$\begin{align}
\\ \lceil \log_{10} (a + b)\rceil &= \left\lceil \dfrac{\ln (a+ b)}{\ln 10} \right\rceil 
\\ & =  \left\lceil \dfrac{\ln a + \ln (1 + \dfrac{b}{a})}{\ln 10} \right\rceil 
\\ & =  \left\lceil \dfrac{\ln a + \ln \left(1 + \left(\dfrac{2}{2002}\right)^{2001}\right)}{\ln 10} \right\rceil 
\\ & =  \left\lceil \dfrac{\ln a + O(10^{-6003})}{\ln 10} \right\rceil 
\\ & =  \left\lceil \dfrac{\ln a}{\ln 10} + O(10^{-6003}) \right\rceil
\\ & =  \left\lceil 6606.2296\dots + O(10^{-6003}) \right\rceil
\\ &= 6697
\end{align}$$
