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I pick two whole numbers $x$ and $y$ between $1$ and $10$ inclusive (not necessarily distinct). My friend picks two numbers $x -4$ and $2y-1$. If the product of my friend's numbers is one greater than the product of my numbers, then what is the product of my numbers?

gueesstt Apr 22, 2018

#1**+3 **

I think I have a way of figuring out this problem. It may not be the most efficient, but it gets the problem done.

Both ranging from 1 to 10, "x" and "y" are two numbers, and their product plus one equals the the number your friend got. I can create an equation with this.

\(xy+1=(x-4)(2y-1)\) | I am going to solve for y and see if I can make any more observations. First, let's expand. |

\(xy+1=2xy-x-8y+4\) | In order to solve for y, move every term with a "y" to one side. |

\(8y-xy=3-x\) | Let's factor out a "y" since that is the GCF of the left hand side of the equation. |

\(y(8-x)=3-x\) | Divide by 8-x to isolate y completely. |

\(y=\frac{3-x}{8-x}, x\neq 8\) | We can now begin to guess x-values and hope they output integers in between 1 to 10 for y. |

Now, I will not guess that is 4 or below because that would result my friend's corresponding x-value to be negative, and that can never be one more than my product. Let's start guessing at 5 to 10.

\(\frac{3-5}{8-5}=\frac{-2}{3}\\ \frac{3-6}{8-6}=\frac{-3}{2}\\ \frac{3-7}{8-7}=\frac{-4}{1}=-4\\ \frac{3-9}{8-9}=\frac{-6}{-1}=6\\ \frac{3-10}{8-10}=\frac{-7}{-2}\)

Look at that! A match has appeaed! x=9 and y=6. That's the only solution, too.

When x=9 and y=6, the product of my numbers is 54.

TheXSquaredFactor Apr 22, 2018

#1**+3 **

Best Answer

I think I have a way of figuring out this problem. It may not be the most efficient, but it gets the problem done.

Both ranging from 1 to 10, "x" and "y" are two numbers, and their product plus one equals the the number your friend got. I can create an equation with this.

\(xy+1=(x-4)(2y-1)\) | I am going to solve for y and see if I can make any more observations. First, let's expand. |

\(xy+1=2xy-x-8y+4\) | In order to solve for y, move every term with a "y" to one side. |

\(8y-xy=3-x\) | Let's factor out a "y" since that is the GCF of the left hand side of the equation. |

\(y(8-x)=3-x\) | Divide by 8-x to isolate y completely. |

\(y=\frac{3-x}{8-x}, x\neq 8\) | We can now begin to guess x-values and hope they output integers in between 1 to 10 for y. |

Now, I will not guess that is 4 or below because that would result my friend's corresponding x-value to be negative, and that can never be one more than my product. Let's start guessing at 5 to 10.

\(\frac{3-5}{8-5}=\frac{-2}{3}\\ \frac{3-6}{8-6}=\frac{-3}{2}\\ \frac{3-7}{8-7}=\frac{-4}{1}=-4\\ \frac{3-9}{8-9}=\frac{-6}{-1}=6\\ \frac{3-10}{8-10}=\frac{-7}{-2}\)

Look at that! A match has appeaed! x=9 and y=6. That's the only solution, too.

When x=9 and y=6, the product of my numbers is 54.

TheXSquaredFactor Apr 22, 2018