Question 15
An $\ell$-code is an integer $n \ge 0$ of at most $\ell$ digits, if necessary supplemented by leading zeros, so that it consists of $\ell$ digits in total. Thus, you can make a 4-code out of 310 by writing it as 0310. An $\ell$-code is called *self-squared* if the last $\ell$ digits of the square of that code form exactly the original $\ell$-code. Thus, the 1-codes 5 and 6 are self-squared, but the 3-code 006 is not self-squared, because $6^2 = 36 = 036$ does not end in 006. The 2-code 76 is self-squared, because $76^2 = 5776$ ends in 76.
a. Prove that an $\ell$-code $n$ is self-squared if and only if $n(n-1)$ is divisible by $10^\ell$.
b. Prove that any self-squared $\ell$-code $n \ge 2$ ends in the digit 5 or the digit 6.
c. Prove that any self-squared $\ell$-code $n \ge 2$ is extendable to a self-squared $(\ell+1)$-code in exactly one way by placing a digit in front of it.
d. It follows from (b) and (c) that for every $\ell$, there exist exactly two self-squared $\ell$-codes $m, n \ge 2$. What is their sum $m+n$ (in terms of $\ell$)?
a. Prove that an $\ell$-code $n$ is self-squared if and only if $n(n-1)$ is divisible by $10^\ell$.
b. Prove that any self-squared $\ell$-code $n \ge 2$ ends in the digit 5 or the digit 6.
c. Prove that any self-squared $\ell$-code $n \ge 2$ is extendable to a self-squared $(\ell+1)$-code in exactly one way by placing a digit in front of it.
d. It follows from (b) and (c) that for every $\ell$, there exist exactly two self-squared $\ell$-codes $m, n \ge 2$. What is their sum $m+n$ (in terms of $\ell$)?