I was trying to go through some basic number theory and there I saw a lot of questions based on calculating the number of digits of a number raised to some big power. Many a time the expression gets reduced to multiples of $10$ yielding many zeroes in the end but I am unable to deal with the ones where this doesn't happen.

For e.g.: All the digits of $2^{1989}$ and $5^{1989}$ are written side by side and we need to calculate the total number of digits thus obtained.

Can anybody please help by telling me about some kind of method that I can follow to tackle such problems?


Note that if $x$ has $n$ digits, then $10^{n-1} \leq x \leq 10^n-1$. Taking the $log_{10}$ on all sides of the inequality and rounding down should yield the answer.


Well, the number of digits of a number $\text{N}\in\mathbb{N}^+$ is:


So, when $\text{N}$ is in the form of $\alpha^\beta$:


Using Hermite's identity:


  • $\begingroup$ I found this question in an olympiad textbook and though your answer gives a very simple way to solve the problem, yet I believe that problem was meant for some other approach. Is there any other approach possible without using the logarithmic one? $\endgroup$ – Harsh Sharma Mar 1 '18 at 19:38

$\log_{10} 2 = 0.30102999566398114$ (to plenty of significant digits)

$\log_{10} 2^{1989} = 1989\cdot \log_{10} 2 = 598.748661375658$

$\implies 10^{598} < 2^{1989} < 10^{599}$, so $2^{1989}$ has $599$ digits.

Similarly $\log_{10} 5 = 0.6989700043360187$, again with sufficient precision for solution, etc.

Any other method of showing that the results are between two adjacent powers of $10$ would also work but logarithms give the direct method.

However there is also a "cheat". Note that $\log_{10} 2 + \log_{10} 5 = 1$ (since $2\times 5 =10$). Thus $1989\log_{10} 2 + 1989\log_{10} 5 = 1989$, but we know that neither $ 2^{1989}$ nor $5^{1989}$ will be exact powers of $10$ so the combined number of digits will be $1989+1 = 1990$. (Try it out with smaller powers!)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.