Let N be of the form $\overline{abc}$ ( valid for 3 , 2 and one digit numbers )

Therefore the given condition can be written as : $\ \dfrac { \dfrac{ \begin{matrix} 0 & 1 & 2 & 3 \\ \times & a & b & c \end{matrix} } { \begin{matrix} 0 & 0 & c & 2c & 3c \\ 0 & b & 2b & 3b & X \\ a & 2a & 3a & X & X \end{matrix} } } { \begin{matrix} ?? & ?? & 0 & 0 & 1 \end{matrix} }$

We observe that :

$c = 7$ will be the least number possible to give us unit digit as $1$ .

$$

Now , carrying 2 to the tens place : [As $3 \times 7 = 21$ ] .

$2c + 3b + 2 = 3b + 16 \rightarrow 0$

$$

Here ,

$b = 8$ will be the * least number possible * to give us tens digit as $0$ .

$$

Now ,

carrying 4 to the hundreds place : [As $3 \times 8 + 16 = 40$ ]

$c + 2b + 3a + 4 = 3a + 27 \rightarrow 0$

$$

Here ,

$a = 1$ will be the * least number possible * to give us hundreds digit as $0$

$$

Hence. $\overline{abc} = 187$ .

$$

Therefore , $\boxed{187}$ is the required number .

$123N \equiv 1 (mod 1000)$ . The multiplicative inverse of 123 mod 1000 is 187, so $N \equiv 187 (mod 1000)$

Assume $N=\frac{}{abc}$ , where a is the hundred's digit, b is the ten's digit, and c is the one's digit.

We know that the ones digit has to be 1. The ones digit of the final product is the is the ones digit of the product of the ones digits of 123 and N.

So $3N=\frac{}{f1}$ , where f is a single digit number. The only number that works is 7, because $3*7=21$ .

The ten's digit is the sum of $((2*7+2)+3c$ So $16+3c$ has to have a one's digit of zero. The only number that adds with 6 to get some number of zero is 4. So $3c=\frac{}{d4}$ , where d is another single digit number. 3*8=24, so the ten's digit is 8.

At this point it would be easiest to simply do guess and check, since we narrowed it down to 10 numbers.

$123*87=10701$

$123*187=23001$

187 gives us a product of 23001, so the answer is $\boxed{87}$

The format of this solution doesn't recognized by the system.

Only 7 x 3 can get 1 as last digit. So N's last digit is 7. We get 7 x 123=861. To get 0 in 2nd last digit we must +4. The only way to get 4 as last digit is to multiply 3 by 8. now we get 10701. To get 0 as 3rd last digit we must +3. The only way is to multiply 3 by 1. Thus N is 187

Clearly, $N$ must end in $7$ . Therefore, $N$ must be of the form $(7+x)$ , where $x$ is some positive number. $123 \times N=123 \times (7+x) = 861+123x$ . We observe, last three digits of $123x$ must be $140$ to get the last $3$ digits of $N$ as $001.$ So, last $2$ digits of $x$ must be $80$ . $123 \times 80=9840$ ,which doesn't satisfy. So,when we try $x=180$ ,which may the next possible number, we get last $3$ digits as $140$ , which we require.Therefore, $N=7+180=187.$

See ..we go by the intuitive approach.. first step is to multiply with 7(as to get 1 in the last pos) then see which next number gives you a sum 0... continue till u get the last 3.. :)

Compile this code (Java) and look at the output. There are two methods you can use below, for those who are programmers and those who aren't.

public static void main(String[] args) {
    Scanner input = new Scanner(System.in);

    int x;
    int y;

    for (x=0;x<=999;x++){

        y = 123 * x;


       System.out.println(x + ".  -  " + y);


        }
    } 


}

Copy the output put it in word and search for 001 or just insert a parse.int command and dissect 001 with the statement and insert a condition if the value of y contains 001 print the remaining, x.

If we want N * 123 ending with 1, then N has to be a number of the form 7 + 10x. So we have 123(7 + 10x) = 1230x + 861. Now, if we want 1230x + 861 ending with 01, then x has to be of the form 8 + 10k. So we get 1230(8 + 10k) + 861 = 12300k + 10701. If we want 12300k + 10701 ending with 001, then k has to be a number of the form 1 + 10r. But we are looking for the smaller value, so r = 0, k = 1, x = 18 and N = 7 + 10x = 187.

We know that 3*(last digit of N) has last digit of 1. The only number from 0-9 that satisfies this condition is 7. Therefore we know that the unit digit of N = 1.

Let the number be xy7 123*7 = 861

        123
       *xy7

---

        8 6 1
 y(2y)(3y)
   (3x)

By the good old multiplication by hand method, last digit of (3 y+6) = 0 *(or (3 y+6)%10 = 0)* This implies that (3*y)%10 = 4, smallest y = 8. 6+24 = 30, bring over 3 to left side.

(3+8+2 y+3 x)%10 = 0 (3+8+16+3 x)%10 = 0 (27+3 x)%10 = 0 (3*x)%10 = 3 x = 1.

Thus, smallest N = 187