Competition problem — Number theory

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Number theory Difficulty 6.0 National Olympiad Prove it Romania

Let $M$ be the set of palindromic numbers of the form $5n + 4$ , where $n \ge 0$ is an integer. (A positive integer is a palindromic number if it remains the same when its digits are reversed. For instance, the numbers 7, 191, 23532, 3770773 are palindromic numbers.)
a) If we write the elements of $M$ in increasing order, which is the 50th number?
b) Among all numbers in $M$ , written with nonzero digits who sum up to 2014, which is the greatest one and which the smallest one?

Solution

a) The last (and hence, the first) digit of a number from $M$ equals 4 or 9. A direct count shows that $M$ contains 2 one digit numbers, 2 two digit numbers, 20 three digit and 20 four digit numbers, hence the 50th number is the 6th five digit number, that is, 40504.

b) The greatest number in $M$ has the maximum number of digits. Therefore, we put 4 as the first and last digit and complete the decimal representation

with 2006 digits 1, obtaining thus $\underbrace{411\ldots14}_{2006\ \text{digits}\ 1}$ . Similarly, the smallest number in $M$ has the least number of digits. The answer is $\underbrace{9899\ldots989}_{220\ \text{digits}\ 9}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.