Find all the integers equal to twice the sum of their digits

Problem statement

Find all the integer numbers equal to twice the sum of their digits.

Solution #1

Any integer number can be written as the sum of its digits times the appropriate power of \(10\).
Let \(a_n \in [0,9] \subset \mathbb{N}\) be the digits of a number, where subscript \(n\) represents the \(n\)-th digit. Then any \(N\)-digit number, \(A_N\), can be expressed as $$ A_N = \sum_{n=1}^N 10^{n-1} a_n = a_1 + 10a_2 + \ldots + 10^{N-1}a_N. $$

The condition that \(A_N\) is equal to twice the sum of its digits can be expressed as $$ \sum_{n=1}^N 10^{n-1} a_n = 2 \sum_{n=1}^N a_n $$ which is equivalent to $$ a_1 = \sum_{n=2}^N (10^{n-1}-2) a_n. $$ For any \(n>2\) we have \(10^{n-1}-2 > 9\) unless \(a_n = 0\), which implies that there is no \(N>2\)-digit number that can satisfy the requested conditions. In other words, only 2-digit numbers (\(N=2\)) can be equal to twice the sum of their digits.

With \(N=2\) the equality simplifies to $$ \begin{align} & a_1 + 10a_2 = 2a_1 + 2a_2 \newline \Rightarrow \; & a_1 = 8a_2 \end{align} $$ whose only acceptable solution is \(a_2 = 1\), \(a_1 = 8\), i.e., there is only a single integer number that is equal to twice the sum of its digits, and that number is \(18\).

A simple numerical check can confirm that, for any given number \(A\), the sum of its digits grows much slower than \(A\) itself, and there is only one solution to the problem.

    A = 1:100
    y = @. 2 * sum(digits(A))
    sol = findall(A .== y)

1-element Vector{Int64}:
 18