2.
How many zeroes do we write when we write all the integers from 1 to243 in base 3?
\(\text{Let base $b = 3$} \\ \text{Let digit numbers $= n$ }\)
1. The one digit numbers \((n=1)\) don't have any zeroes.
\((n-1)\times \dfrac{(b-1)\cdot b^{(n-1)} }{b} = 0\times \dfrac{2\cdot 3^0}{3} = 0 \text{ zeroes}\)
2. The two digit numbers \((n=2)\) have:
\((n-1)\times \dfrac{(b-1)\cdot b^{(n-1)} }{b} = 1\times \dfrac{2\cdot 3}{3} = 2 \text{ zeroes}\)
3. The three digit numbers \((n=3) \) have:
\((n-1)\times \dfrac{(b-1)\cdot b^{(n-1)} }{b} = 2\times \dfrac{2\cdot 3^2}{3} = 12 \text{ zeroes}\)
4. The four digit numbers \((n=4)\) have:
\((n-1)\times \dfrac{(b-1)\cdot b^{(n-1)} }{b} = 3\times \dfrac{2\cdot 3^3}{3} = 54 \text{ zeroes}\)
5. The five digit numbers \((n=5)\) have:
\((n-1)\times \dfrac{(b-1)\cdot b^{(n-1)} }{b} = 4\times \dfrac{2\cdot 3^4}{3} = 216 \text{ zeroes}\)
6. The six digit numbers \((n=6) \) have:
\(243 = 100000_3 = 5 \text{ zeroes}\)
\(\text{Ths sum $= 0 + 2 + 12 + 54 + 216 + 5 = 289$ zeroes } \)
We write 289 zeroes, when we write all the integers from 1 to 243 in base 3