Accept The Situation And Move On

Well, first of all overrated!(it is level 5)

Secondly, 1 is also a positive integer. @Satvik Golechha

Now, let's try to find what we require.

${3}^{n} = 3 mod 1000$

${3}^{n-1} = 1 mod 1000 \Rightarrow {3}^{x} = 1 mod 1000$

Now, of course we know by Euler's formula that $3**400 = 1 mod 1000$ as $\phi(1000) = 40$ but we don't know if this is the order of 3 modulo 1000.

Hence, we need to find the order of 3 modulo 1000.

With the help of some calculations, we find that

${3}^{10} = 49 mod 1000 \Rightarrow {3}^{10} = {7}^{2} mod 1000$

Therefore, we actually need to find the order of 49 modulo 1000.

Let us have a look at $B = {7}^{y} mod 1000$ for some values of y.

$y = 1, B = 7$

$y = 2, B = 49$

$y = 3, B = 343$

$y = 4, B = 401$

$y = 5, B = 807$

$y = 6, B = 649$

$y = 7, B = 543$

$y = 8, B = 801$ and so on.

Hence, we can see a pattern following here that 4 is added on every 4th number from the preceding term. Like ${B}_{4} = 401, {B}_{8} = 801, {B}_{12} = 201, {B}_{16} = 601, {B}_{20} = 001$

Hence, 20 is the order of the 7 modulo y, or 10 is the order of 49 modulo y.

As a result $10*10 = 100$ is the order of 3 modulo 1000.

or ${3}^{101} = 3 mod 1000$ , n = 101(for n being all positive integers >1)

If allowed to use calculator as in other solutions, i have more easy solution:-

first of all, we'll get the pattern of last digit of $3^n$ ,

$3^1=3$

$3^2=9$

$3^3=27$

$3^4=81$

$3^5=243$

So the pattern is $3,9,7,1$ , so to get $3^n-3$ divisible by 1000 we must have 003 as last digit,

hence $n mod 4=1$ , or n can be 5 or 9 or 13 or 17 or 21.............

on calculation we get that when n is $21 or 41 or 61$ we get the number as _ _ _ 203(for n=21) or _ _ _ 403(for n=41) or .......... hence to make the last digits _ __ 003 n minimum n should be $\boxed{101}$ . Author, please fix the question as 1 can be also possible.