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This is just a funny question that I was elaborating... I know one way to solve (or maybe it's wrong...), but I want know if there is another way to solve this (when we keep adding conditions, there is an possibility of letting the exercise trivial and don't see this, I think...)

The month and day of my birthday are perfect squares and its product is an power of an prime, with positive exponent. Denote by $x$ the square root of the month and $y$ the square root of the day. Then

  1. If you know the value of the product $xy$, you can certainly deduce the sum $x+y$.
  2. Knowing the sum $x+y$, even knowing that condition 1 holds, doesn't exists the possibility of deduce the product $xy$.

The sum $x+y$ is relatively prime with the product of the month by the day.

(This isn't my real birthday, but I accept gifts =p)

Thanks in advance!
And excuse my English...

  • 0
    Ah OK, now I got it! :)2012-08-15

1 Answers 1

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The simpliest way to solve is to list all possibilities and just remove those not fitting.

  • The month and day of my birthday are perfect squares

$\small \begin{matrix}1.01 & 4.01 & 9.01 & 16.01 & 25.01 & 1.04 & 4.04 & 9.04 & 16.04 & 25.04 & 1.09 & 4.09 & 9.09 & 16.09 & 25.09\end{matrix}$

  • Product is a power of a prime

$\small \begin{matrix} \textrm{day} & 4.01 & 9.01 & 16.01 & 25.01 & 1.04 & 4.04 &16.04 & 1.09 & 9.09 \\ \textrm{product} & 4 & 9 & 16 & 25 & 4 & 16 & 64 & 9 & 81 & \\ \textrm{prime power}& 2^2 & 3^2 & 2^4 & 5^2 & 2^2 & 2^4 & 2^6 & 3^2 & 3^4 \\ \end{matrix}$

  • Denote by x the square root of the month and y the square root of the day.

$\small \begin{matrix} \textrm{day} & 4.01 & 9.01 & 16.01 & 25.01 & 1.04 & 4.04 &16.04 & 1.09 & 9.09 \\ \textrm{x} & 1 & 1 & 1 & 1 & 2 & 2 & 2 & 3 & 3 & \\ \textrm{y}& 2 & 3 & 4 & 5 & 1 & 2 & 4 & 1 & 3 \\ \end{matrix}$

  • If you know the value of the product $xy$, you can certainly deduce the sum $x+y$.

(which is: for given $xy$ there is only one value of $x+y$) $\small \begin{matrix} \textrm{day} & 4.01 & 9.01 & 25.01 & 1.04 & 16.04 & 1.09 & 9.09 \\ xy & 2 & 3 & 5 & 2 & 8 & 3 & 9 & \\ x+y& 3 & 4 & 6 & 3 & 6 & 4 & 6 \\ \end{matrix}$

  • Knowing the sum $x+y$, even knowing that condition 1 holds, it isn't possibile to deduce the product $xy$.

(which is: for given $x+y$ there is more then one value of $xy$) $\small \begin{matrix} \textrm{day} & 25.01 & 16.04 & 9.09 \\ x+y&6 & 6 & 6 \\ xy & 5 & 8 & 9 & \\ \end{matrix}$

  • The sum $x+y$ is relatively prime with the product of the month by the day.

$\small \begin{matrix} \textrm{day} & 25.01\\ x+y&6 & \\ product & 5 \\ \textrm{GCD}& 1 \\ \end{matrix}$

So the answer is January 25.