## Reverse Digits

#### Problem

If the number 41 is added to its reverse, 41 + 14 = 55.

The total, 55, is divisible by 5. But not many numbers have this property. For example, 27 + 72 = 99, which is not divisible by 5.

How many numbers under 100 have this property?

#### Solution

Let the original number, `n` = 10`a` + `b`, so its reverse, `m` = 10`b` + `a`.

Therefore, `n` + `m` = 11`a` + 11`b` = 11(`a` + `b`).

Clearly, `n` + `m` is divisible by 5 iff `a` + `b` 0 mod 5.

Listing combinations: | ||

a = 0, b = 5 | 05,50 | |

a = 1, b = 4 or 9 | 14, 41, 19, 91 | |

a = 2, b = 3 or 8 | 23, 32, 28, 82 | |

a = 3, b = 7 | 37, 73 | |

a = 4, b = 6 | 46, 64 | |

a = 5, b = 5 | 55 | |

a = 6, b = 9 | 69, 96 | |

a = 7, b = 8 | 78, 87 |

That is, 19 numbers in total.

What about under 1000?

Problem ID: 82 (Oct 2002) Difficulty: 2 Star