Base Conversion Puzzle 1
In this problem, $ a $ and $ b $ are positive integers.
When $ a $ is written in base $ 9 $, its last digit is $ 5 $.
When $ b $ is written in base $ 6 $, its last two digits are $ 53 $.
When $ a-b $ is written in base $ 3 $, what are its last two digits? Assume $ a-b $ is positive.
- 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
In base $ 3 $, the place values are $ \ldots\ldots, 3^4, 3^3, 3^2, 3, 1 $. Notice that all of these except the last two are divisible by $ 3^2=9 $. Thus, the last two digits of $ a-b $ in base $ 3 $ are the base-$ 3 $ representation of the remainder when $ a-b $ is divided by $ 9 $. (This is just like how the last two digits of an integer in base $ 10 $ represent its remainder when divided by $ 100 $.)
For similar reasons, we know that $ a\equiv 5\pmod 9 $ and $ b\equiv 5\cdot 6+3=33\pmod{6^2} $. This last congruence tells us that $ b $ is $ 33 $ more than a multiple of $ 36 $, but any multiple of $ 36 $ is also a multiple of $ 9 $, so we can conclude that $ b\equiv 33\equiv 6\pmod 9 $.
Finally, we have $$a-b \equiv 5-6 \equiv -1 \equiv 8 \pmod 9.$$This remainder, $ 8 $, is written in base $ 3 $ as $ 22_3 $, so the last two digits of $ a-b $ in base $ 3 $ are $ \boxed{22} $.
Freeflows