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Math Help - 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.
    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
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  3. #3
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    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
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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 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

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