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Thread: Algebra solution to (unusual) digit reversal problem

  1. #1
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    Algebra solution to (unusual) digit reversal problem

    The problem is as follows:

    "By a mistake a number was doubled instead of raising it to the power of two. The result was a two digit number which was the reverse of what it should have been. What was the original number?"

    I realized rapidly that the solution must be a number between 5 and 9. Any less than 5 and doubling does not result in a two digit number, any more then 9 and squaring does not result in a two digit number.

    Clearly the answer is 9 since $29 = 18$ and $9^2 = 81$. [Simple, well below my pay grade.]

    But is there an algebra solution to this problem?

    I got as far as this simple system...

    $\displaystyle 2n = 10a + b \Longrightarrow n = 5a + \frac{b}{2}$
    $\displaystyle n^2 = 10b + a$

    Substituting $n$ from the top equation into the bottom one leads to nothing useful.

    $\displaystyle (5a + \frac{b}{2})^2 = 10b + a \Longrightarrow 100a^2 + 20ab + b^2 = 40b + 4a$

    Am I missing some clever way of solving the problem using algebra?

    Thanks.
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  2. #2
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    Re: Algebra solution to (unusual) digit reversal problem

    adding the two equations

    $2n=10a+b$ and $n^2=10b+a$ we get

    $n(n+2)=11(a+b)$

    so $11$ must divide $n+2$ since it cannot divide $n$

    using the condition $5\leq n\leq 9$
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  3. #3
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    Re: Algebra solution to (unusual) digit reversal problem

    I'm sorry but I do not follow the logic. Could you spell it out for me please?

    Note: I understand the equation addition, just not what follows.

    Thanks.
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  4. #4
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    Re: Algebra solution to (unusual) digit reversal problem

    if a prime number p divides the product xy then
    p divides x or p divides y

    we have that 11 divides n(n+2) so
    11 divides n or 11 divides n+2

    but 11 cannot divide n since n is smaller than 11
    so 11 must divide n+2 which means
    n=9
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  5. #5
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    Re: Algebra solution to (unusual) digit reversal problem

    Okay, got it. Thanks for your help.
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