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Math Help - Problem Solving

  1. #1
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    Problem Solving

    A and B are positive integers such that (i) A=7B and (ii) the sum of the digits of A is
    twice the sum of the digits of B. If C is the number formed by writing the digits of B
    immediately after the digits of A, then prove that C is a multiple of 9.
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  2. #2
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    Quote Originally Posted by matgrl View Post
    A and B are positive integers such that (i) A=7B and (ii) the sum of the digits of A is
    twice the sum of the digits of B. If C is the number formed by writing the digits of B
    immediately after the digits of A, then prove that C is a multiple of 9.
    "If C is the number formed by writing the digits of B immediately after the digits of A,..." Please explain this part further.
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  3. #3
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    A= 6 B= 3 C = BA or 36 which is a multiple of 9 we need to prove this somehow
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  4. #4
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    My thoughts

    So you're saying that C is a concatenation of BA whereby A = 7B and the sum of the digits of A is twice the sum of the digits of B.

    You would have to express A and B in terms of series and you can use mathematical induction to prove. Have you gotten that
    far?
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  5. #5
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    No i havent....but that is a start
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  6. #6
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    Sum of digits.

    Quote Originally Posted by matgrl View Post
    A and B are positive integers such that (i) A=7B and (ii) the sum of the digits of A is
    twice the sum of the digits of B. If C is the number formed by writing the digits of B
    immediately after the digits of A, then prove that C is a multiple of 9.

    (I assume these numbers are written base ten.)

    The sum of a number's digits is congruent to the number, mod 9.

    \text{Sum of A's digits }\equiv A (\bmod 9). A similar result holds for B or any other positive integer.

    \text{Sum of A's digits }\equiv 2\times (\text{Sum of B's digits })(\bmod 9)

    So, A \equiv 2B\ (\bmod 9).

    Also, we have that A=7B, so, A \equiv 7B\ (\bmod 9).

    Adding these gives: A \equiv 2B\ (\bmod 9).

    Since 2 is relatively prime to 9, A \equiv 0 \ (\bmod 9).

    Previously, we saw that A \equiv 2B\ (\bmod 9). Therefore, 0 \equiv 2B\ (\bmod 9), so that B \equiv 0 \ (\bmod 9)

    C is the number formed by writing the digits of B immediately after the digits of A, so \text{Sum of C's digits }= (\text{Sum of A's digits })+(\text{Sum of B's digits }).

    C \equiv A+B\ (\bmod 9) \ \ \implies\ \ C \equiv 0+0\ (\bmod 9)\ \ \implies\ \ C \equiv 0\ (\bmod 9).

    Therefore, C is divisible by 9.

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  7. #7
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    Hello, matgrl!

    Another approach . . . a bit fuzzy, though.


    A\text{ and }B\text{ are positive integers such that (i) }A=7B
    \text{ and (ii) the digit-sum of }A\text{ is twice the digit-sum of }B.

    \text{If }C\text{ is the number formed by writing the digits of }B\text{ immediately}
    \text{after the digits of }A\text{, then prove that }C\text{ is a multiple of 9.}

    Fact: If the digit-sum of a number is 0 or 9, the number is a multiple of 9.


    Let: . \begin{Bmatrix}d(A) &=& \text{digit-sum of }A \\<br />
d(B) &=& \text{digit-sum of }B \end{Bmatrix}


    Since A \,=\,7B, then: . d(A) \:=\:7\!\cdot\!d(B)

    But we are told that: . d(A) \:=\:2\!\cdot\! d(B)

    . . Hence: . 7\!\cdot d(B) \:=\:2\!\cdot d(B) \quad\Rightarrow\quad d(B) \:=\:0

    . . Then: . d(A) \:=\:0

    That is, both \,A and \,B are multiples of 9.


    \text{Since }C \:=\:\{AB\},\,\text{ then: }\,d(C) \:=\:d(A) + D(B) \:=\:0

    Therefore, \,C is a multiple of 9.

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