This was not as difficult a puzzle as it appeared to be, though required some work! Was glad to see most of the regular puzzlers answer this, which included Karan, Abhinav Jain, Suman Saraf, Girish Tutakne and Ruchir Godura (welcome back!).
I am taking the liberty of reproducing Karan’s answer:
We’ll proceed after reducing it to the following form.
Now 9999 has the following prime factors 3,3,11,101 out of which possible 3 digit numbers with similar first and last digits are 101, 303 and 909. 101 can be ruled out as value of the fraction is less than 1 and 101 is the smallest number of the above mentioned form.
9999/909=11, when EVE*11 the last digit would be E which is not the case, hence 909 can also be rejected. Which leaves us with 303=DID.
The options for EVE are 1?1, 2?2 (?=[0,9])
1?1 can be rejected as 1?1*33 will leave K=3 which contradicts D=3.
We’re now left with 212, 242, 252, 262, 272, 282, 292.
We now know for sure that K=6 so all we have to check is 212,242,252,272,282,292.
hope you all enjoyed the puzzle!