What is the smallest integer $k>2000$ such that both $\dfrac{17k}{66}$ and $\dfrac{13k}{105}$ are terminating decimals? please help ty（＾-＾）

2079

Step-by-step explanation:

Let's look at the factorization of the denominators first. (I'm doing this to choose k so that everything cancels in the denominator that isn't a power of 2 or power of 5.)

66=2(3)(11) or 2(33)

105=5(3)(7) or 5(21)

Now, the reason I grouped the factors 3 and 11 together is because I don't recall anything divided by 3 and 11 giving me a decimal that terminates other than 3 and 11 itself.

I grouped 3 and 7 in the second factorization for similar reasoning.

So we want k to possibly be a multiple of

3(7)(11)=3(77)=231.

I didn't need two factors of 3 because each fraction contains k in numerator and each fraction as exactly one factor of 3 to cancel.

Let's try the first multiple.

17(231)/66=59.5

13(231)/105=28.6

The 1st multiple was a success.

We also want our number choice to be greater than 2000.

So we want to find m so that 231m>2000..

. m>2000/231

We want m to be an integer so choose smallest integer number in this set for m..

That's 9 since 2000/231 is approximately 8.7.

231(9)=2079

44

step-by-step explanation:

536

step-by-step explanation: