Question

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?

Answer 1

Answer:

  2079

Step-by-step explanation:

In order for the decimal equivalents to be terminating, the only factors that can remain in the denominators are 2 and 5.

Here, the given denominators are

  66 = 2·3·11

  105 = 5·3·7

So, the values of k must be multiples of 3, 7, and 11. The least common multiple of these numbers is 3·7·11 = 231, so ...

  S = {0, ±231, ±462, ±693, ...}

The smallest member of S that is greater than 2010 is ...

  ceiling(2010/231)·231 = 9·231 = 2079

_____

For k=2079, the terminating decimals are ...

  $\dfrac{17\cdot 2079}{66}=535.5\\\\\dfrac{13\cdot 2079}{105}=257.4$