Competition problem — Number theory

Library / /58 of 196

Number theory Difficulty 4.8 AIME Prove it Soviet Union

Problem:

If the numbers from $11111$ to $99999$ are arranged in an arbitrary order show that the resulting $444445$ digit number is not a power of $2$ .

Solution

Solution:

Let the set of numbers be $S$ . Define the function $f$ on $S$ as follows. Replace each digit $i$ in $n$ by $9 - i$ for $0 < i < 9$ . This gives $f(n)$ . Then $f(f(n)) = n$ , so $f$ is a bijection. The fixed points have only the digits $0$ and $9$ and so are all divisible by $9$ . The other points divide into pairs $(n, f(n))$ and the sum of each pair is divisible by $9$ . Hence the sum of all the numbers in $S$ is divisible by $9$ .

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 reproduced verbatim; metadata (topic, difficulty) added by this project.