A school bus has to make three complete trips. The times, in minutes, for the three trips are $254_6$, $401_5$, and $1212_4$. Determine the time, in base ten, of the fastest complete trip to send the pupils to school.

1. Problem Description

A school bus has to make three complete trips. The times, in minutes, for the three trips are $254_6$, $401_5$, and $1212_4$. Determine the time, in base ten, of the fastest complete trip to send the pupils to school.

2. Solution Steps

First, convert each time from its given base to base

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For $254_6$:

$254_6 = (2 \times 6^2) + (5 \times 6^1) + (4 \times 6^0) = (2 \times 36) + (5 \times 6) + (4 \times 1) = 72 + 30 + 4 = 106$

For $401_5$:

$401_5 = (4 \times 5^2) + (0 \times 5^1) + (1 \times 5^0) = (4 \times 25) + (0 \times 5) + (1 \times 1) = 100 + 0 + 1 = 101$

For $1212_4$:

$1212_4 = (1 \times 4^3) + (2 \times 4^2) + (1 \times 4^1) + (2 \times 4^0) = (1 \times 64) + (2 \times 16) + (1 \times 4) + (2 \times 1) = 64 + 32 + 4 + 2 = 102$

The times in base 10 are 106, 101, and

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2. The fastest time is the smallest of these values, which is

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1.

3. Final Answer

101