Competition problem — Number theory

Library / /14 of 39

Number theory Difficulty 5.2 AIME, harder Prove it Ukraine

Let's designate through $P(n)$ the product of digits of the integer non-negative number $n$ . Prove that sets $A$ and $B$ are unbounded, where:

a. $A = \left\{ \frac{P(n)}{P(n^2)} \right\}$ , where $n$ belongs to the set of such whole non-negative numbers that the number $n^2$ does not contain zero in the decimal record;

b. $B = \left\{ \frac{P(n^2)}{P(n)} \right\}$ , where $n$ belongs to the set of such whole non-negative numbers that the number $n$ does not contain zero in the decimal record.

Solution

Both points are proved with the help of corresponding examples which are in turn proved by a method of mathematical induction.

a.
Let's consider the equality:
$(2\underbrace{66\dots68}_{n-1})^2 = 7\underbrace{11\dots18}_{n-1}\underbrace{22\dots24}_{n-1},$

Further, it is enough to calculate the corresponding ratio as $n \to \infty$ :
$\frac{P(n)}{P(n^2)} = \frac{16 \cdot 6^{n-1}}{7 \cdot 8 \cdot 4 \cdot 2^{n-1}} \to +\infty.$

b.
Let's consider the equality:
$(\underbrace{66\dots67}_{n-1})^2 = \underbrace{44\dots48}_{n-1}\underbrace{\dots89}_{n-1}.$

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.