Why does the number $2701$ always return to $703$?

Question

The number $2701$ has a sum of thousands equal to $703$, i.e. for any power $2701$ is raised to, the sum of the groups of three always reduces to $703$.

Some examples:

$$2\text{,}701 = 2+701 = 703;$$

$$2\text{,}701^2 = 7\text{,}295\text{,}401,\ 7+295+401=703;$$

$$2\text{,}701^3 = 19\text{,}704\text{,}878\text{,}101,\ 19+704+878+101=1\text{,}702,\ 1+702=703;$$

and so on.

What is the reason for this phenomenon?

Answer 1

This happens for the following reasons:

1. $999=27\cdot37$.
2. $2701\equiv1\pmod{27}$ and $2701\equiv0\pmod{37}$. Therefore the same holds for all the powers $2701^n$.
3. Adding the numbers formed by three digits computes the remainder modulo $999$.
4. $703$ is the remainder of $2701$ modulo $999$ (and the only remainder satisfying the two key congruences of item 2). In view of item 2 and the Chinese remainder theorem the same holds for all the powers $2701^n$.

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As some kind of a confirmation do the same exercise with $1296$. It is congruent to $0\pmod{27}$ and congruent to $1\pmod{37}$. The same process will thus leave $297$ as the answer for all the powers $1296^n$.

Answer 2

Suppose you have some number $n$ written in the following form (which is actually its digit representation base $1000$): $$n=a_0+a_1\cdot 1000+a_2\cdot 1000^2+...+a_k\cdot 1000^k$$ where each $a_i$ is between $0$ and $999$ inclusive. To use your example, we might consider $$2701^3=19704878101=(101)+(878\cdot 1000)+(704\cdot 1000^2) + (197\cdot 1000^3)$$ so that $a_0=101$, $a_1=878$, $a_2=704$, and $a_3=197$, and $k=3$. Now let’s consider this whole expression modulo $999$ (that is, take the remainder of $n$ modulo $999$). Since $1000\equiv 1 \bmod 999$, we have that $1000^m \equiv 1 \bmod 999$ for all positive integers $m$, which implies that $$n\equiv a_0+a_1+a_2+...+a_k\bmod 999$$ This means that when you repeatedly apply your “sum of thousands” process to any number, the result must always stay the same modulo $999$.

Now let’s consider the number $2701$ modulo $999$. It turns out that $2701\equiv 703\bmod 999$ (obviously, since $2+701=703$ and the “sum of thousands” preserves the residue modulo $999$). However, it also turns out that $703^2=494209\equiv 703\bmod 999$, meaning that all powers of $703$ are equivalent to $703$ mod $999$. This, in turn, implies that all powers of $2701$ are also congruent to $703$ modulo $999$.

This means that when you start with a large number like $2701^m$, repeatedly applying your “sum of thousands” calculation will steadily decrease the number but always return a number congruent to $703$ modulo $999$, and as this gets smaller and smaller, it eventually settles at the number $703$.

Please let me know if there’s anything I haven’t explained well enough, and I’ll be happy to clarify in the comments.