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Thread: Find a three-digit number N = abc

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by Statsnoob2718 View Post
    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.
    $\displaystyle N=abc$

    hence

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

    from condition ii)

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

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

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

    now from (1) and (2)

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

    ok now take (mod 2)

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

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

    now (*) mod 3

    $\displaystyle 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)

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

    if you take (mod 4)

    you will have

    $\displaystyle 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

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

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

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

    so

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

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

    we have

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

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

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

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

    I attempt to find out $\displaystyle (a,b,c) $ by putting $\displaystyle a=1,2,...,9$ and finally I obtain $\displaystyle (a,b) = (5,9) $ or $\displaystyle (8,8)$ but given that they are distinct so $\displaystyle a=5 , b=9 $ is the solution . Moreover , since $\displaystyle a+b+c = 9k $ , $\displaystyle c = -(5+9) mod(9) = 4 mod(9) $ , $\displaystyle c=4 $ .Therefore , the number is $\displaystyle 594$
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  4. #4
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    Hello, Statsnoob2718!

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


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

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

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

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

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


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

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

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


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

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


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

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


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


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


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


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


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

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