This is a proof that for any positive integer, when summing its digits and multiplying by 3 repeatedly, will eventually equal 27. For example,

Step Value
To start with 97
Sum of digits 16
Multiplied by 3 48
Sum of digits 12
Multiplied by 3 36
Sum of digits 9
Multiplied by 3 27

Now let us prove this mathematically step by step.

(a) First let us prove:

Let \(n\) be a positive integer, and let \(s(n)\) denote the sum of the digits of \(n\). Then \(n\) and \(s(n)\) are congruent modulo 9, i.e., \(n \equiv s(n) \pmod{9}\).

Consider a positive integer \(n\) with decimal representation \(a_k\ a_{k-1} \ldots a_1\ a_0\), where \(a_i\) are the digits of \(n\). We can write \(n\) as:

\[\begin{align*} n &= 10^k \cdot a_k + 10^{k-1} \cdot a_{k-1} + \cdots + 10 \cdot a_1 + a_0 \\ &= (10^k - 1) \cdot a_k + a_k + (10^{k-1} - 1) \cdot a_{k-1} + a_{k-1} + \cdots + 9 \cdot a_1 + a_1 + a_0 \\ &= (10^k - 1) \cdot a_k + (10^{k-1} - 1) \cdot a_{k-1} + \cdots + 9 \cdot a_1 + s(n) \end{align*}\]

Since \((10^k - 1), 10^{k-1} - 1), \cdots, 9\) are all dividable by 9, \(n\) and \(s(n)\) are congruent modulo 9.

(b) Since \(n\) and \(s(n)\) are congruent modulo 9, \(3n\) and \(3s(n)\) are congruent modulo 9.

(c) By applying theorem in (a), \(3s(n)\) and \(s(3s(n))\) are congruent modulo 9.

(d) By combining (b) and (c), \(3n\) and \(s(3s(n))\) are congruent modulo 9.

(e) Since \(3n\) and \(s(3s(n))\) are congruent modulo 9, \(9n\) and \(3s(3s(n))\) are congruent modulo 9.

(f) Since \(9n\) dividable by 9, \(3s(3s(n))\) is a multiple of 9, which means that after two repetations of summing digits of an integer and multiplying by 3, the result becomes a multiple of 9.

(g) Let us continue with third repetition, by applying theorem in (a) again, \(s(3s(3s(n)))\) and \(3s(3s(n))\) are congruent modulo 9, i.e. \(s(3s(3s(n)))\) is multiple of 9. Therefore, \(3s(3s(3s(n)))\) is multiple of 27. That is, after third repetition, the result is a multiple of 27. Since this conclusion is based on the number before the repetition is a multiple of 9, the result will always be a multiple of 27 in any further repetitions.

(h) Now let us prove that the operation of summing its digits and multiplying by 3 will result in smaller number for a number greater or equal 30, i.e. \(3s(n) < n\) for \(n \geq 30\).

Given:

\[n = 10^k \cdot a_k + 10^{k-1} \cdot a_{k-1} + \cdots + 10 \cdot a_1 + a_0\]

We have:

\[\begin{align*} n - 3s(n) &= (10^k - 3) \cdot a_k + (10^{k-1} -3) \cdot a_{k-1} + \cdots + (10-3) \cdot a_1 + (1 - 3)\cdot a_1 \\ &> (10^k - 3) \cdot a_k + (10^{k-1} -3) \cdot a_{k-1} + \cdots + 7 \times 3 - 2 \times 9 \\ n - 3s(n) &> (10^k - 3) \cdot a_k + (10^{k-1} -3) \cdot a_{k-1} + \cdots +3 \\ &> 0 \end{align*}\]

(i) Based on (g), after third round, if the result is not 27, it is \(2\times27\) or greater. Based on (h), each further repetation will reduce the result until it is below 30, and (h) tells us the result is a multiple of 27, A number that is below 30 and a multiple of 27 is 27. \(_\blacksquare\)

P.S. Finding the smallest number that requires 4 repetitions of summing its digits and multiplying by 3 to reach 27 by brute force is challenging due to the size of the number. Instead, we can first find \(X\), the smallest multiple of 3 that requires 3 repetitions, and then work backward to find the smallest number leading to \(X\). The results of this discovery are tabulated below.

Step Value Note
To start with \(2 \times 10^{259} + (10^{259} - 1)\)     It is 2 followed by 259 digits of 9
Sum of digits 2333     Computed as \(2 + 259\times9\)
Multiplied by 3 6999     First round
Sum of digits 33    
Multiplied by 3 99     Second round, multiple of 9
Sum of digits 18    
Multiplied by 3 54     Third round, multiple of 27
Sum of digits 9    
Multiplied by 3 27     Fourth round