# proof that number of sum-product numbers in any base is finite

Let $b$ be the base of numeration.

Suppose that an integer $n$ has $m$ digits when expressed in base $b$ (not counting leading zeros, of course). Then $n\ge {b}^{m-1}$.

Since each digit is at most $b-1$, we have that the sum of the digits is at most $m(b-1)$ and the product is at most ${(b-1)}^{m}$, hence the sum of the digits of $n$ times the product of the digits of $n$ is at most $m{(b-1)}^{m+1}$.

If $n$ is a sum-product number, then $n$ equals the sum of its digits
times the product of its digits. In light of the inequalities^{} of the
last two paragraphs, this implies that $m{(b-1)}^{m+1}\ge n\ge {b}^{m-1}$, so $m{(b-1)}^{m+1}\ge {b}^{m-1}$. Dividing both sides, we
obtain ${(b-1)}^{2}m\ge {(b/(b-1))}^{m-1}$. By the growth of exponential
function, there can only be a finite number of values of $m$ for which
this is true. Hence, there is a finite limit to the number of digits
of $n$, so there can only be a finite number of sum-product numbers to
any given base $b$.

Title | proof that number of sum-product numbers in any base is finite |
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Canonical name | ProofThatNumberOfSumproductNumbersInAnyBaseIsFinite |

Date of creation | 2013-03-22 15:47:06 |

Last modified on | 2013-03-22 15:47:06 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 7 |

Author | rspuzio (6075) |

Entry type | Proof |

Classification | msc 11A63 |