# Counting the number of digits in a concatenation

Concatenate the numbers $2^{1971}$ and $5^{1971}$. How many digits are there in the new number? How do I count them?

• You mean total number of digits in the two numbers? What do you mean by new number? Jul 17, 2014 at 21:22
• Example: For $2^3, 5^3$ the new numberis $8125$ so the new number($8125$) has $4$ digits. Jul 17, 2014 at 21:32
• So you want the sum of digits of the two numbers. Jul 17, 2014 at 21:34
• He wants the number of digits in the new number, which is the sum of the numbers of digits in the two separate numbers which are being concatenated. Jul 17, 2014 at 21:41
• I edited the post to improve the meaning. Others can give it a try if you'd like something better. Jul 17, 2014 at 21:47

let $$10^m<2^{1971}<10^{m+1}$$ and $$10^n<5^{1971}<10^{n+1}$$ This inequality is true since every number that is not a power of ten is between two consecutive powers of ten. Now let us multiply both inequalities $$10^m*10^n=10^{n+m}<2^{1971}*5^{1971}=10^{1971}<10^{m+1}*10^{n+1}=10^{n+m+2}$$ thus $$m+n<1971<m+n+2$$ the only whole number between $m+n$ and $m+n+2$ is $m+n+1$,thus $$m+n+1=1971$$ $$m+n=1970$$ and since $m+1$ and $n+1$ are the number of digits of $2^{1971}$ and $5^{1971}$ respectiveley,then their sum is equal to the number of digits of the new number. Your new number will have $$m+n+2=1972$$ digits.

• Why $2^{1971}\times5^{1971}$????? Jul 17, 2014 at 21:40
• @JonasMeyer Thanks mate! corrected. Jul 17, 2014 at 21:41
• @bigli: Because it works! Check out the neat argument showing why. Jul 17, 2014 at 21:41
• maybe it's worth pointing out that this worked out cleverly because the bases were 2 and 5 and the exponents were the same, so that multiplying gives a power of ten? in general, you'd have to find $m$ and $n$ separately. Jul 17, 2014 at 21:42
• Just wanted to comment that I really like this answer. Nice and slick. Jul 17, 2014 at 22:06

This is only one way and probably will not be "nicest" way, but we can get the number of digits of a number if we just take log$_{10}|x|$ and then round up.

So for $2^{1971}$ we have

$$\text{number of digits } = \lceil 1971 \text{log}_{10} (2) \rceil = 594$$

• I'd just like to point out that this works for $x\geq1$ and $x\leq-1$ as it's undefined at 0.
– Jam
Jul 17, 2014 at 21:39