What is the smallest positive integer $N$ such that the last three digits of $123 \times N$ are 001?

The answer is 187.

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Nice calculations.

However, I would like to point out that when finding the minimum possible value of $\overline{abc}$ , you want to minimize $a$ then $b$ then $c$ , as opposed to the other way around. E.g. To find the smallest 3-digit number that is a multiple of 3, you'd want $a = 1, b = 0, c = 2$ . As opposed to saying $c = 0, b = 0, a = 3$ .

This doesn't affect your solution, because there is a unique value of $c$ and $b$ . Hence, instead of saying

$c=7$ will be the

least number possibleto give us unit digit as 1.

it should instead be

$c = 7$ is the

only possible valueto give us unit digit as 1.

Likewise for the case of $b$ , and the case of $a$ is properly written.

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Oh yes sir! Thank you for pointing it out. Ill surely keep that in mind.

Priyansh Sangule
- 7 years, 7 months ago

I used the same way and this is the easiest solution possible without the knowledge of mod values..

Ayush Garg
- 7 years, 8 months ago

we had the same solution

Rindell Mabunga
- 7 years, 8 months ago

Hah, I used a Python program because I was tired of guessing values on my calculator. (I knew that the last digit was 7 though, so calcing wasn't too hard)

William Cui
- 7 years, 8 months ago

hi I dont understand how u done this plz tell me simlest and clear way t solve this .. tthanxxx

Purwa Ali
- 7 years, 7 months ago

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Basically, we are looking at the numbers that can be multiplied to 123 to produce the digits 001. For example, for the units digit, only 7 can be multiplied to 3 to obtain a number with a units digit of 1.

William Cui
- 7 years, 7 months ago

for me using modulo is hard hahaha

Rindell Mabunga
- 7 years, 8 months ago

nice skills you have

Devesh Rai
- 7 years, 8 months ago

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How did you deduce that 187 was the multiplicative inverse mod 1000 of 123?

William Mitchell
- 7 years, 8 months ago

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gcd(1000, 123) = 1, so we can apply the extended Euclidean algorithm: $1000 - 123 * 8 =16$ $123 - 16 * 7 = 11$ $16 - 11*1 = 5$ $11 -5*2 = 1$ Plugging back in backwards, we find that: $11-(16-11*1)*2 = 1\rightarrow 3*11-2*16 = 1$ $3*(123-16*7)-2*16 = 1\rightarrow 3*123-23*16=1$ $3*123-23*(1000-123*8)=1\rightarrow 187 * 123-23*1000=1$ So $187*123\equiv1(mod1000)$ .

Carl Denton
- 7 years, 8 months ago

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explain how did u get that through multiplicative inverse

rajachowdary raj
- 7 years, 7 months ago

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There's an explanation in the comments, apologies for that. Unfortunately I don't know any good books on number systems, sorry.

Carl Denton
- 7 years, 7 months ago

can u tell me any good book for number systems plz

rajachowdary raj
- 7 years, 7 months ago

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 $\boxed{87}$

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

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answer = 187

00861 09840

23001

to get 1st digit "1" in 001 , 3 in 123 should be multiplied by 7

so 7 is units digit of N

units digit of 123 * 7 = 861 6 need 4 to be 0 in 001 so the next number is 8 as 8*3 = 24

just swim in space :(

Moataz Mohamed
- 7 years, 5 months ago

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Sorry , forgot to mention that $x$ must end in $0$ in the $2nd$ line.

Bhargav Das
- 7 years, 8 months ago

nice one man!

Jun Das
- 7 years, 7 months ago

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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.

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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
**

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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 possibleto 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 .