Question

What is the smallest integer $k>2000$ such that both $\dfrac{17k}{66}$ and $\dfrac{13k}{105}$ are terminating decimals?.

Answer 1

The smallest integer greater than 2000 for both the fractions $\frac{17k}{66} \ and \ \frac{13k}{105}$ to be terminating decimals is 2079.

Given

K is greater than 2000.

$k> 2000$.

Given fractions are

$\dfrac{17k}{66} \ and \ \dfrac{13k}{105}$.

How to find the smallest integer greater than 2000 for the fractions to be terminating decimals?

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 and 105 respectively.

Now factors of 66 will be 2,3,11.

And the factors of 105 will be 3,5,7.

So, the value of k must be multiples of 3, 7, and 11. The LCM of these numbers will be,

$3\times 7\times 11=231$

Now the value of K must be greater than 2000, so the multiple of 231 which is greater than 2000 is its $9^{th}$ multiple,

$231\times9=2079$

Hence 2079 is the smallest integer greater than 2000 for both the fractions $\frac{17k}{66} \ and \ \frac{13k}{105}$ to be terminating decimals.

For more details on terminating decimals follow the link:

brainly.com/question/5286788