2
$\begingroup$

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

$\endgroup$
10
  • 1
    $\begingroup$ You mean total number of digits in the two numbers? What do you mean by new number? $\endgroup$
    – Juanito
    Jul 17, 2014 at 21:22
  • $\begingroup$ Example: For $2^3, 5^3$ the new numberis $8125$ so the new number($8125$) has $4$ digits. $\endgroup$
    – bigli
    Jul 17, 2014 at 21:32
  • $\begingroup$ So you want the sum of digits of the two numbers. $\endgroup$
    – Juanito
    Jul 17, 2014 at 21:34
  • 1
    $\begingroup$ 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. $\endgroup$
    – mweiss
    Jul 17, 2014 at 21:41
  • 1
    $\begingroup$ I edited the post to improve the meaning. Others can give it a try if you'd like something better. $\endgroup$ Jul 17, 2014 at 21:47

2 Answers 2

9
$\begingroup$

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.

$\endgroup$
10
  • $\begingroup$ Why $2^{1971}\times5^{1971}$????? $\endgroup$
    – bigli
    Jul 17, 2014 at 21:40
  • $\begingroup$ @JonasMeyer Thanks mate! corrected. $\endgroup$
    – cirpis
    Jul 17, 2014 at 21:41
  • $\begingroup$ @bigli: Because it works! Check out the neat argument showing why. $\endgroup$ Jul 17, 2014 at 21:41
  • 1
    $\begingroup$ 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. $\endgroup$ Jul 17, 2014 at 21:42
  • 1
    $\begingroup$ Just wanted to comment that I really like this answer. Nice and slick. $\endgroup$
    – mweiss
    Jul 17, 2014 at 22:06
2
$\begingroup$

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

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

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

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .