October
11, 2006
The
probability that the square of a positive integer (in decimal notation)
ends with the digit 1 is $2/10$ because out of every 10 numbers, those
and only those ending with the digits 1 or 9 have squares ending with
1. What is the probability that the cube of a positive integer
chosen at random ends with the digits 11?
Solution. If $a^3$ ends
in the digits 11, then $a^3\equiv 11 \pmod{100}$. This is equivalent to
the two conditions $a^3 \equiv 11 \equiv 3 \pmod{4}$ and $a^3 \equiv 11
\pmod{25}$. By an easy calculation, 3 is the only residue satisfying
$a^3 \equiv 11\bmod 4$. Now, $a^3\equiv 11 \pmod{25} \Longrightarrow
a^3 \equiv 1 \pmod{5}$. By another easy calculation, $a=1$ is the only
residue modulo 5 that satisfies this equation.
This means that we need only check 1, 6, 11, 16, and 21 modulo 25:
$\begin{eqnarray}
1^3&=&1, \quad 6^3=36\cdot 6\equiv 11\cdot 6 = 66 \equiv 16, \\
11^3&=&121\cdot 11 \equiv -4\cdot 11 = -44\equiv 19, \\
16^3&=&256\cdot 16 \equiv 6\cdot 16 = 96 \equiv 21, \\
21^3&\equiv& (-4)^3= -64 \equiv -14 \equiv 11.
\end{eqnarray}$
So $a=21$ is the only solution modulo 25.
By the Chinese Remainder Theorem,
there is only one number satisfying $a\equiv 3 \pmod{4}$ and $a\equiv
11\pmod{25}$ (that number is 71). It follows that the probability that
a random cube ends with the digits 11 is $1/100$.