Answer:
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
