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

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

2 Answers 2

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.

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Why $2^{1971}\times5^{1971}$????? –  bigli Jul 17 at 21:40
    
@JonasMeyer Thanks mate! corrected. –  cirpis Jul 17 at 21:41
    
@bigli: Because it works! Check out the neat argument showing why. –  Jonas Meyer Jul 17 at 21:41
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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. –  symplectomorphic Jul 17 at 21:42
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Just wanted to comment that I really like this answer. Nice and slick. –  mweiss Jul 17 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

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

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

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