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How many zeros are there in this equation?

10000^999

In my calculator it says infinity but that doesn't seem right.

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  • $\begingroup$ what have u tried? $\endgroup$ – Shobhit Oct 20 '13 at 18:57
  • $\begingroup$ should it be 1*10^4^999 = 10^3996 ? $\endgroup$ – user101946 Oct 20 '13 at 18:58
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    $\begingroup$ We know $(a^m)^n=a^{mn}$ Here $10000=10^4$ $\endgroup$ – lab bhattacharjee Oct 20 '13 at 19:01
  • $\begingroup$ Further, calculators aren't perfect. $\endgroup$ – Ian Coley Oct 20 '13 at 19:39
  • $\begingroup$ Calculators make a lot of approximations, so that they can do arithmetic with finite decimal numbers. You're probably familiar with approximations like writing $1/3 \sim 0.33333333$. If your calculator only uses 8 decimal places, this error is too small for the calculator to represent. Approximating sufficiently large numbers with $+\infty$ is the same idea. $\endgroup$ – user14972 Oct 20 '13 at 23:12
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The answer is: $3996$.

$10000^1 - 4$ zeroes $(4 \cdot 1)$

$10000^2 - 8$ zeroes $(4 \cdot 2)$

...

$10000^{999} - 3996$ zeroes $(4 \cdot 999)$

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Everytime we mulltiply a number with $10$ we get:

$$n \cdot 10 = \overline{n0}$$

This would hold for any power of $10$. So for $1000=10^4$, this would be:

$$n \cdot 10000 = \overline{n0000}$$

If we multiply again we have:

$$\overline{n0000} \cdot 10000 = \overline{n00000000}$$

To sum up everytime we multiply we add $k$ zeroes at the end of the number, where $k$ is the exponent in the prime power of $10$.

So in our specific case, take $n=1$ and multiply by $10^4$ that would add $4$ zeroes at the end of the number, repeat that $999$ times and you'll get that $10000^{999}$ will have $4\cdot 999 = 3996$ zeroes.

Note that this property is a consequence of our decimal numbercal system. If we were working in a numberical system with base $k$ then the same property would hold for every power of $k$.

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