Base Conversion Puzzle 3
A particular number written in base 3 requires three digits ($ {\_ \_ \_}_3 $). When the number is written in base 3 and 4, the digits are the reverse of each other. What is this number expressed in base 10?
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
Let $ abc $ represent the three-digit number in base 3, where $ a $, $ b $ and $ c $ each represent a digit 0, 1 or 2. The place values in base 3 are 9, 3 and 1, so the base-ten value of $ abc $ is $ a \times 9 + b \times 3 + c \times 1 $, which can be written as $ 9a + 3b + c $. This same value is $ cba $ in base 4, which we can write as $ 16c + 4b + a $. Equating these two expressions, we get $ 9a + 3b + c = 16c + 4b + a $. We can simplify this to $ 8a = 15c + b $. Now, there are only three digits to try for each letter. It turns out that $ 8 \times 2 = 15 \times 1 + 1 $, so the base-three number is $ 211_3 $ and the base-four number is $ 112_4 $. The base-ten value is $ (2 \times 9) + (1 \times 3) + 1 = 18 + 3 + 1 = 22 $. To confirm this answer, we check the base-four value: $ 1 \times 16 + 1 \times 4 + 2 \times 1 = 16 + 4 + 2 = \boxed{22} .$
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