How many digits does the number 1000 need when it is written in base three?

The place values in base three are 1, 3, 9, 27, 81, 243, 729, 2187, … . Since 729 ≤ 1000 < 2187, the leading digit sits in the 729 column, so the columns run from $3^6$ down to $3^0$ — seven of them. Written out, 1000 = 729 + 243 + 27 + 1 = 1101001 in base three. Stopping at $3^5=243$ gives only 6 columns, and six columns reach no further than 728. The trap 10 is the digit count in base TWO, where 1000 = 1111101000 needs ten places because $2^9=512$ ≤ 1000 < 1024. Answering 8 overshoots by one column.