The smallest integer greater than 2000 for both the fractions [tex]\frac{17k}{66} \ and \ \frac{13k}{105}[/tex] to be terminating decimals is 2079.
K is greater than 2000.
[tex]k> 2000\\[/tex].
Given fractions are
[tex]\dfrac{17k}{66} \ and \ \dfrac{13k}{105}[/tex].
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,
[tex]3\times 7\times 11=231[/tex]
Now the value of K must be greater than 2000, so the multiple of 231 which is greater than 2000 is its [tex]9^{th}[/tex] multiple,
[tex]231\times9=2079[/tex]
Hence 2079 is the smallest integer greater than 2000 for both the fractions [tex]\frac{17k}{66} \ and \ \frac{13k}{105}[/tex] to be terminating decimals.
For more details on terminating decimals follow the link:
https://brainly.com/question/5286788
