[ [Decimal numeral system ]

The numbers $2^{2000}$ and $5^{2000}$ are written in sequence. How many digits are written in total?

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If the number $2^{2000}$ contains m digits, then $10^{\mathrm{m}-1}<2^{2000}<10^{\mathrm{m}}$.

## Solution

Let the number $2^{2000}$ contain $\mathrm{m}$ digits, and the number $5^{2000}$ contain $\mathrm{n}$ digits. Then the following inequalities hold: $10^{\mathrm{m}-1}<2^{2000}<10^{\mathrm{m}}, 10^{\mathrm{n}-1}<5^{2000}<10^{\mathrm{n}}$ (the inequalities are strict because the power of two or five is not equal to the power of ten). Multiplying these inequalities, we get: $10^{\mathrm{m}+\mathrm{n}-2}<10^{2000}<10^{\mathrm{m}+\mathrm{n}}$. From this, it follows that the exponent 2000 is between $\mathrm{m}+\mathrm{n}-2$ and $\mathrm{m}+\mathrm{n}$, so $2000=\mathrm{m}+\mathrm{n}-1$ and $\mathrm{m}+\mathrm{n}=2001$. This means that a total of 2001 digits are written.

## Answer

2001.00