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, 2014

Combinatorics Difficulty 5.1 AIME, harder Prove it Ireland

The 11-digit number $52014641025$ has two interesting properties: it contains the string of digits $2014$ , and it is unchanged if we reverse the digits. How many 11-digit numbers have this property?
(A number cannot begin with the digit $0$ .)

Solution

An integer $N$ that satisfies these conditions is determined by its first (leftmost) six digits. The first instance of either the string $2014$ or its reverse, $4102$ , must start in position $1$ , $2$ or $3$ . In each case, there are two other digits to be chosen. These digits can be chosen arbitrarily if $2014$ or its reverse start in position $1$ , but the first digit must be non-zero in the other cases. This gives a total of $2(100 + 90 + 90) = 560$ numbers having the required form.

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