Palindromic Decomposition

Number Theory Level pending

We can define the palindrome $P(n)$ of an integer $n$ as the integer with the same digits but in reverse :

$P(1) = 1 \\ P(124) = 421 \\ P(1300) = \space 31$

Some integers can be written as the sum of $n+P(n)$ (e.g. $44 = 13 + 31$ )

Which of these integers cannot be written as the sum of an integer and its palindrome ?

1555 1551 1553 1554

1 solution

Romain Bouchard
Dec 20, 2017

First of all here are the palindromic decompositions for $1551$ , $1554$ and $1555$ :

$1551 = 1050 + 501,\\ 1554=579+975 \\ 1555 = 629+926$

Now to show that there is no palindromic decomposition for $1553$ .

Let's assume there is an integer $n$ such that $n+P(n)=1553$ . There are two possibilities : either $n=\overline{abc}$ or $n=\overline{1ab0}$ .

We can exclude directly $n=\overline{1ab0}$ because the unit digit would be $1$ and not $3$ .
If $n=\overline{abc}$ then we have $\overline{abc}+\overline{cba}=1553$ which means that $a+c$ should be equal to either $15$ with $b<5$ or to $14$ with $b>5$ . In both cases the unit digit will never be $3$ so there is no solution either with $n=\overline{abc}$ .

You didn't show that $1553$ cannot be written as $n+P(n)$ , you just show that other number can.

Chan Tin Ping - 3 years, 5 months ago