# Thread: Find a three-digit number N = abc

1. ## Find a three-digit number N = abc

Find a three-digit number N = abc satisfying the
following conditions:
(i) the digits a, b, c are distinct;
(ii) N is equal to the sum of all two-digit numbers formed from the three
digits a, b, c.

2. Originally Posted by Statsnoob2718
Find a three-digit number N = abc satisfying the
following conditions:
(i) the digits a, b, c are distinct;
(ii) N is equal to the sum of all two-digit numbers formed from the three
digits a, b, c.
$N=abc$

hence

$N = 100a+10b+c$ ...(1)

from condition ii)

$N = aa+ab+ac+bb+ba+bc+cc+ca+cb$

$N= 10a+a+10a+b+10a+c+10b+b+10b+a+10b+c+10c+c+10c+a+10 c+b$

$N = 33a + 33 b + 33c$ ...(2)

now from (1) and (2)

$100a+10b+c = 33a+33b+33c$ ...(*)

ok now take (mod 2)

$c \equiv a+b+c \;\;(mod2) \Rightarrow a+b\equiv 0 \;\; (mod 2)$

that means a,b both even or both odd since their sum is even

now (*) mod 3

$a+b+c \equiv 0 \;\; (mod 3)$

that means a+b+c is a multiple of 3 so N is a multiple of 3

now take (*) (mod 11)

$a - b +c \equiv 0 (mod 11) \Rightarrow b \equiv a+c \;\; (mod 11)$

if you take (mod 4)

you will have

$b-a \equiv 0 \;\; (mod 4)$

a and b are distinct numbers so the difference between a and b is a multiple of 4,

mod 5

$c \equiv 3a+3b+3c \;\; (mod 5)$

$-2c \equiv 3a+3b \;\; (mod 5)$

$3c = 3a+3b \;\; (mod 5)$ but 3 is primitive with 5 not common divisors between them

so

$c \equiv a+b \;\; (mod 5)$

let's try some numbers let a be "1", b have two choices 5 or 9 false in the each probability
if a "2" , b = 6 , c=3 these fail in equation (mod 3)

if a "3" , b =7 , c =5 fail in (mod 11)

if a "4" , b =8 , c =7 fail in (mod 3)

if a "5" , b =9 , c =4 works in (mod 2,3,4,5,11) now justify the answer
is this true
594 = 33(5)+33(9)+33(4)

33*18 = 594 true

nice question

3. Starting from here $33(a+b+c) = 100a + 10b + c$

I find that if we put $100a+10b+c = 99a +9b+ (a+b+c)$

we have

$33(a+b+c) = 9(11a+b) + (a+b+c)$

$32(a+b+c) = 9(11a+b)$

Therefore , $a+b+c$ is the multiple of $9$

and $11a+ b$ is the multiple of $32$ .

I attempt to find out $(a,b,c)$ by putting $a=1,2,...,9$ and finally I obtain $(a,b) = (5,9)$ or $(8,8)$ but given that they are distinct so $a=5 , b=9$ is the solution . Moreover , since $a+b+c = 9k$ , $c = -(5+9) mod(9) = 4 mod(9)$ , $c=4$ .Therefore , the number is $594$

4. Hello, Statsnoob2718!

I used a straight algebra approach.
It takes longer, but requires neither modulo arithmetic nor guessing.

Find a three-digit number $N = abc$ satisfying the
following conditions:

(i) the digits $a, b, c$ are distinct,
(ii) $N$ is equal to the sum of all two-digit numbers formed from the digits $a, b, c$.

The sum of the possible two-digits numbers is: . $33(a+b+c)$

This sum equals $N\!:\quad 33a + 33b + 33c \;=\;100a + 10b + c \quad\Rightarrow\quad 23b + 32c \:=\:67a$

We have: . $b \;=\;\frac{67a-32x}{23} \;=\;2a - c + \frac{21a - 9c}{23}$ .[1]

Since $b$ is a digit, then $\frac{21a-9c}{23}$ is an integer $p.$

. . $\frac{21a-9c}{23} \:=\:p \quad\Rightarrow\quad 21a - 9c \:=\:23p$

. . $c \:=\:\frac{21a-23p}{9} \;=\;2a - 2p + \frac{3a-5p}{9}$ .[2]

Since $c$ is a digit, then $\frac{3a-5p}{9}$ is an integer $q.$

. . $\frac{3a-5p}{9} \:=\:q \quad\Rightarrow\quad 3a-5p \:=\:9q \quad\Rightarrow\quad a \:=\:\frac{9q+5p}{3} \:=\:3q+\frac{5p}{3}$

Since $a$ is an integer, then $\frac{5p}{3}$ is an integer.

. . Let $p = 3 \quad\hdots\quad \text{ then: }\,\boxed{a \:=\: 3q+5}$

Substitute into [2]: . $c \;=\;\frac{21(3q+5) - 23(3)}{9} \;=\;\frac{63q + 36}{9} \quad\Rightarrow\quad \boxed{c \:=\:7q + 4}$

Substitute into [1]: . $b \:=\:\frac{67(3q+5) - 32(7q+4)}{23} \;=\;\frac{207 - 23q}{23} \quad\Rightarrow\quad \boxed{b \:=\:9-q}$

Hence, we have: . $\begin{Bmatrix}a &=& 3q+5 \\ b&=& 9-q \\ c &=& 7q+4\end{Bmatrix}$

Since $a,b,c$ are digits, $q$ must be 0.

. . Therefore: . $N \;=\;594$