Question

Find'n' if 142n=47ten

Answer 1

To solve the equation \( 142_n = 47_{10} \), we first need to convert \( 142_n \) (a number in base \( n \)) into base 10.

The value of \( 142_n \) in decimal form is computed as follows:

\[
142_n = 1 \cdot n^2 + 4 \cdot n^1 + 2 \cdot n^0
\]

This simplifies to:

\[
142_n = 1 \cdot n^2 + 4 \cdot n + 2
\]

So we need to set this equal to \( 47_{10} \):

\[
n^2 + 4n + 2 = 47
\]

Next, we move \( 47 \) over to the left side of the equation:

\[
n^2 + 4n + 2 - 47 = 0
\]

This simplifies to:

\[
n^2 + 4n - 45 = 0
\]

Now we can solve this quadratic equation using the quadratic formula, which is:

\[
n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 1 \), \( b = 4 \), and \( c = -45 \). Substituting these values into the formula:

\[
n = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-45)}}{2 \cdot 1}
\]
\[
n = \frac{-4 \pm \sqrt{16 + 180}}{2}
\]
\[
n = \frac{-4 \pm \sqrt{196}}{2}
\]
\[
n = \frac{-4 \pm 14}{2}
\]

Now we calculate the two possible values for \( n \):

1. \( n = \frac{10}{2} = 5 \)
2. \( n = \frac{-18}{2} = -9 \) (not a valid base)

Since base \( n \) must be a positive integer greater than 4 (because of the digit 4 in \( 142_n \)), we conclude:

\[
n = 5
\]

Thus, the solution to the equation \( 142_n = 47_{10} \) is:

\[
\boxed{5}
\]