Convert the decimal number 128 to base 3 number system.
My attempt to solution:
\(\frac{128}{3^4}=1 \text{r} 47\)
\(\frac{47}{3^3}= 1 \text{r} 20\)
\(\frac{20}{3^2}=1\text{r}2\)
\(\frac{2}{3^1}=0\text{r}2\)
\(\frac{2}{3^0}=2\text{r}0\)
\(\therefore 128_{10}=11202_{3}\)
How come the text book is saying my answer is wrong. The answer in the text book is :
\(1122_{3}\)
My attempt to solution:
\(\frac{128}{3^4}=1 \text{r} 47\)
\(\frac{47}{3^3}= 1 \text{r} 20\)
\(\frac{20}{3^2}=1\text{r}2\)
\(\frac{2}{3^1}=0\text{r}2\)
\(\frac{2}{3^0}=2\text{r}0\)
\(\therefore 128_{10}=11202_{3}\)
How come the text book is saying my answer is wrong. The answer in the text book is :
\(1122_{3}\)
