How many numbers end in the four digits 1995 and become an integer number of times smaller when these digits are erased?

Question

turingtest · Answer

I guess we at least need the factors of 1995
3*5*7*19
that's a start...

anonymous · Answer

of the form 1995n/(n-1)where n is integer...and the no ends in 1995..trial

turingtest · Answer

It is as I posted it
It seems to mean something like how many numbers when changed from $$abcd1995 \to abcd$$ decrease by a factor that is an integer.

turingtest · Answer

No, an old magazine called "quantum" has these. They don't sell that magazine anymore :(

turingtest · Answer

it is sitting in the meta-math section in case nobody has the time to solve it here...

turingtest · Answer

not as far as I can see

zarkon · Answer

like this $19951995\to1995$

turingtest · Answer

so it is! 10001
how did zarkon know?

zarkon · Answer

10001,100001,.......

zarkon · Answer

an educated guess

turingtest · Answer

Oh I think I see where this is going...

turingtest · Answer

but that sequence is infinite

zarkon · Answer

yes

turingtest · Answer

I'm pretty sure the answer should be finite just by tendency of this magazine not to have trick questions like that.

turingtest · Answer

what's that?

turingtest · Answer

oh yeah, heard of it but never tried it
I'll be patient and let it sit in meta-math though

zarkon · Answer

so we have one so far :)

I believe I now understand the full problem

turingtest · Answer

I had a thought and lost it...
I'd appreciate it if you could walk me through your thought process zarkon, if you come up with something.

anonymous · Answer

abcd1995 --> abcd

Let x be the number abcd.
Let y be the integer number of times

10000x + 1995 = abcd1995

10000x + 1995 = xy

y = (10000x + 1995)/x

 (10000x + 1995) must be divisible by x

mathmate · Answer

Would 19951995 be an answer, since
19951995/1995=10001

anonymous · Answer

x must be one of the factors of 1995

anonymous · Answer

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