Track / Stage 6 / 215 of 400 #1215 of 2000

Problem 1215

National olympiad, first round

Number theory Difficulty 6.4 Prove it

## Task 4 - 290814

For every six-digit natural number $n$ , whose unit digit is different from zero, one can form the number $n^{\prime}$ obtained by writing the digits of $n$ in reverse order. Subsequently, the number $n+n^{\prime}$ can be calculated.

a) Form some examples! Determine if there is a prime number by which the number $n+n^{\prime}$ is divisible in your examples! State a conjecture!

b) Try to prove your conjecture!

c) Now let $k$ be any even natural number greater than zero. For every $k$ -digit natural number $n$ , whose unit digit is different from zero, one can form the number obtained by writing the digits of $n$ in reverse order.

Does a corresponding statement like in a), b) also hold for $n+n^{\prime}$ ?

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

a) From examples such as

$\begin{aligned} & 110222+222071=332233=11 \cdot 30203 \\ & 130333+333031=463364=11 \cdot 42124=2^{2} \cdot 11 \cdot 10531 \\ & 118862+268811=387673=11 \cdot 35243=11 \cdot 13 \cdot 2711 \end{aligned}$

one can arrive at the conjecture:

For any six-digit natural number $n$ , whose unit digit is not zero, $n+n^{\prime}$ is divisible by 11.

b) Proof of this conjecture:

If $a, b, c, d, e, f$ are the digits of $n$ in this order, then

$\begin{aligned} n & =100000 a+10000 b+1000 c+100 d+10 e+f \\ n^{\prime} & =100000 f+10000 e+1000 d+100 c+10 b+a \quad \text { so } \\ n+n^{\prime} & =100001 a+10010 b+1100 c+1100 d+10010 e+100001 f \\ & =11(9091(a+f)+910(b+e)+100(c+d)) \end{aligned}$

is divisible by 11.

c) Let $k=2 m$ with a natural number $m \geq 1$ , and let $n$ be any $k$ -digit natural number whose unit digit is not zero. If $a_{0}, a_{1}, \ldots, a_{2 m-1}$ are the digits of $n$ in this order, then

$\begin{aligned} n & =a_{0} \cdot 10^{2 m-1}+a_{1} \cdot 10^{2 m-2}+\ldots+a_{2 m-2} \cdot 10+a_{2 m-1} \\ n^{\prime} & =a_{0}+a_{1} \cdot 10+\ldots+a_{2 m-2} \cdot 10^{2 m-2}+a_{2 m-1} \cdot 10^{2 m-1} \end{aligned}$

After addition and factoring out $a_{0}, a_{1}, \ldots, a_{2 m-1}$ , in $n+n^{\prime}$

the factor $10^{2 m-1}+1$ appears for $a_{0}$ and $a_{2 m-1}$ ,

the factor $10^{2 m-2}+10=10\left(10^{2 m-3}+1\right)$ appears for $a_{1}$ and $a_{2 m-2}$ , ...

the factor $10^{m}+10^{m-1}=10^{m-1}(10+1)$ appears for $a_{m-1}$ and $a_{m}$ .

Now it can be proven that the numbers $10+1, 10^{3}+1, \ldots, 10^{2 m-3}+1, 10^{2 m-1}+1$ that appear here are divisible by 11:

For $10+1$ this is clear, and the other numbers have 1 as the first and last digit, and an even number of zeros in between. Subtracting 11 results in a number with 0 as the last digit and an even number of 9s before it.

Each such number is divisible by 11; this is also proven for $n+n^{\prime}$ .

Solutions of the 1st Round 1989 taken from [5]

### 5.31.2 2nd Round 1989, Class 8

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.