Set of Terminating Decimals
Let $ S $ be the set of all integers $ k $ such that, if $ k $ is in $ S $, then $ \frac{17k}{66} $ and $ \frac{13k}{105} $ are terminating decimals. What is the smallest integer in $ S $ that is greater than 2010?
- 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 us first analyze the fraction $ \frac{17k}{66} $. We can rewrite this fraction as $ \frac{17k}{2 \cdot 3 \cdot 11} $. Since the denominator can only contain powers of 2 and 5, we have that $ k $ must be a multiple of 33. We now continue to analyze the fraction $ \frac{13k}{105} $. We rewrite this fraction as $ \frac{13k}{3 \cdot 5 \cdot 7} $, and therefore deduce using similar logic that $ k $ must be a multiple of 21. From here, we proceed to find the least common multiple of 21 and 33. Since $ 21 = 3 \cdot 7 $ and $ 33 = 3 \cdot 11 $, we conclude that the least common multiple of 21 and 33 is $ 3 \cdot 7 \cdot 11 = 231 $.
We now know that $ S $ contains exactly the multiples of 231. The smallest multiple of 231 that is greater than 2010 is $ 231 \cdot 9 = \boxed{2079} $.
Freeflows