Question

142n =47 using number base

Answer 1

To solve the equation \( 142_n = 47 \) for the base \( n \), we first need to convert \( 142_n \) to its decimal (base 10) equivalent. The expression \( 142_n \) means:

\[
1 \cdot n^2 + 4 \cdot n^1 + 2 \cdot n^0
\]

This can be simplified to:

\[
1n^2 + 4n + 2
\]

Setting this equal to \( 47 \), the equation becomes:

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

Now, we will simplify this equation:

\[
n^2 + 4n + 2 - 47 = 0
\]
\[
n^2 + 4n - 45 = 0
\]

Next, we can solve this quadratic equation using the quadratic formula:

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

Where \( a = 1 \), \( b = 4 \), and \( c = -45 \). Plugging in these values:

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

This gives us two potential solutions:

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

Thus, the only valid solution is:

\[
\boxed{5}
\]

To confirm, we can check if \( 142_5 \) equals \( 47 \) in base 10:

\[
142_5 = 1 \cdot 5^2 + 4 \cdot 5 + 2 = 1 \cdot 25 + 20 + 2 = 25 + 20 + 2 = 47
\]

So our calculations are correct, and the base \( n \) that satisfies the equation is \( n = 5 \).