Never Ends in 2015

We can write $nx=10000c+2015$ , where $c \geq 0$ and $100 \leq n \leq 999$ .

Writing this in the language of congruence, we have $nx \equiv 2015 \pmod{10000}$ . If $n$ is a $N_{3}$ -number, we will never be able to find a $x$ which satisfies the congruence.

We invoke the following well known result from elementary number theory:

The congruence $ax \equiv b \pmod{c}$ has a solution in $x$ if $\gcd(a,c)=1$ .

Take a look at the congruence $nx \equiv 2015 \pmod{10000}$ . If $n$ does not share a common factor with $10000$ , then we will always be able to find a $x$ which satisfies the congruence, hence such $n$ cannot be $N_{3}$ -numbers.

We focus now on cases where $n$ shares common factors larger than $1$ with $10000$ . We know that such common factors must take the form of $d=2^{a_{1}}5^{a_{2}}$ , where $0 \leq a_{1},a_{2} \leq4$ , with $a_{1}$ and $a_{2}$ not both $0$ .

We now argue that for $a_{1} \neq 0$ , $n$ is a $N_{3}$ number. To prove this, write $n=2k$ and substitute it into the original equation. We have $2kx-10000c=2015$ , but this is absurd since the L.H.S is divisible by $2$ while the R.H.S. isn't. We therefore conclude that no such $x$ exists.

Now, we focus on cases where $a_{1} =0$ . In other words, the only common factors shared between $n$ and $10000$ are powers of $5$ .

We now prove that for $a_{2} \geq 2$ , $n$ is a $N_{3}$ number. To prove this, write $n=25k$ and substitute it into the original equation. We have $25kx-10000c=2015$ , which simplifies to $5kx-2000c=403$ , which is absurd since the L.H.S. is divisible by $5$ while the R.H.S. isn't. We therefore conclude that no such $x$ exists.

All that is left for us to consider are cases where $a_{1}=0, a_{2}=1$ , or in other words, $\gcd(n,10000)=5$ . Write $n=5k$ and substitute it into our original equation. We get $5kx-10000c=2015$ , which simplifies to $kx-2000c=403$ . In the language of congruence, this is $kx \equiv 403 \pmod{2000}$ . However, since $\gcd(k,2000)=1$ , invoking the result from above, we can find a $x$ which satisfies the congruence. Therefore, such $n$ are not $N_{3}$ numbers.

To recap, all $N_{3}$ numbers must either be divisible by $2$ or/and divisible by $25$ . A quick computation reveals that there are $468$ such numbers.

Since 2015 is an odd number, hence all even 3-digit numbers are $N_3$ - numbers . There are 450 of them.

Next, it is clear that if the 3-digit numbers end in 25 or 75, then its multiples end in 25 or 75. Thus $\overline{X25}$ or $\overline{Y75}$ are $N_3$ - numbers . There are 18 of them.

So far, we have 468 $N_3$ - numbers . We claim that there are no other $N_3$ - numbers . The proof is as follows.

Suppose it is an odd 3-digit number not end in 5, says $n$ . Then there exist a unique 1-digit number $m$ such that the last digit of $mn \in \{0,1,2, \ldots,9\}$ . If $nk=\overline{A2015}$ , working backward from the last digit of $k$ (which obviously is 5), after obtain the value of $5n$ , check what is the second last digit of $k$ using the idea of getting the value of the 1-digit $m$ , repeat it again until we obtain $\overline{A2015}$ (you may see the solution of the End in 2015 to understand more). This shows that all odd 3-digit numbers not end in 5 are not $N_3$ - numbers .

Suppose it is an 3-digit number of the form $n=\overline{PQ5}$ , where $P,Q$ are single digit and $Q \neq 2,7$ . Then finding $k$ such that $nk=\overline{A2015}$ is equivalent to finding $k$ such that $2nk= \overline{A4030}$ . Note that the last two digit of $2n$ is $\overline{CD}$ , where $D=0$ and $C \in \{1,3,7,9\}$ . This means that it is similar to find the value of $k$ where $n'k= \overline{A403}$ with last digit of $n'$ is either 1, 3, 7 or 9. Using the previous argument, we see that such $k$ exists. So these 3-digit numbers are not $N_3$ - numbers .