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Let S be the set of all integers k such that, if k is in S, then
(17k)/(66) and
(13k)/(105) are terminating decimals. What is the smallest integer in S that is greater than 2010?

User Max Tilley
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1 Answer

3 votes

Answer:

2079

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 ...


(17\cdot 2079)/(66)=535.5\\\\(13\cdot 2079)/(105)=257.4

User Galandil
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