Problem D and Solution

Choices

Matilda and Yolanda each chose an integer. When Matilda’s integer is multiplied by the sum of Matilda’s and Yolanda’s integers, the product is \(299\).

If Matilda’s integer is smaller than Yolanda’s integer, determine all possible pairs of integers that Matilda and Yolanda could have chosen.

Let \(x\) represent Matilda’s integer and \(y\) represent the Yolanda’s integer. We’re given \(x<y\).

The sum of the two integers is \(x+y\). We multiply this sum by Matilda’s integer, \(x\). The resulting expression is \(x(x+y)\). Thus, we want to find all pairs of integers satisfying \(x(x+y)=299\) with \(x<y\).

We want the product of two integers to be \(299\). The factors of \(299\) are \(\pm 1\), \(\pm 13\), \(\pm 23\), \(\pm 299\). Thus, the possible values for \(x\) are \(\pm 1\), \(\pm 13\), \(\pm 23\), \(\pm 299\).

In the following table, we list all the possible values for \(x\) and then determine the corresponding value for \(y\). If \(x<y\), then this is a valid possibility. For example, if \(x=1\), then \(x+y\) must be \(299\). Therefore, \(y\) must be \(298\). Since \(x<y\), one possibility is Matilda chooses \(1\) and Yolanda chooses \(298\).

\(x\) | \(x+y\) | \(y\) | \(x<y\)? |
---|---|---|---|

\(1\) | \(299\) | \(298\) | Yes |

\(13\) | \(23\) | \(10\) | No |

\(23\) | \(13\) | \(-10\) | No |

\(299\) | \(1\) | \(-298\) | No |

\(-1\) | \(-299\) | \(-298\) | No |

\(-13\) | \(-23\) | \(-10\) | Yes |

\(-23\) | \(-13\) | \(10\) | Yes |

\(-299\) | \(-1\) | \(298\) | Yes |

Therefore, there are four pairs of integers that Matilda and Yolanda could have chosen. Matilda could have chosen \(1\) and Yolanda chose \(298\), Matilda could have chosen \(-13\) and Yolanda chose \(-10\), Matilda could have chosen \(-23\) and Yolanda chose \(10\), or Matilda could have chosen \(-299\) and Yolanda chose \(298\).