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QUESTION

Without using your calculator, find

\begin{description}

\item[(i)]
the least positive residue of 24.17 mod 29.

\item[(ii)]
the least absolute residue of 19.14 mod 23.

\item[(iii)]
the remainder when $5^10$ is divided by 19.

\item[(iv)]
the final digit of $1!+2!+3!+4!+5!+6!+7!+8!+9!+10!$

\end{description}



ANSWER

We use least absolute residues, to keep the numbers in the
calculations small:

\begin{description}

\item[(i)]
$24.17\equiv(-5).(-12)\equiv60\equiv2$ mod 29.

\item[(ii)]
$19,14\equiv(-4)(-9)\equiv36\equiv13$ mod 23. Thus the least
positive residue is 13. The least absolute residue is $-10$.

\item[(iii)]
$5^2\equiv25\equiv6$ mod 19. Hence $5^4equiv36\equiv-2$ mod 9 and
$5^10\equiv(5^4)^2,5^2\equiv(-2)^2.6\equiv24\equiv5$ mod 19.

Thus the remainder when $5^10$ is divided by 19 is 5.

\item[(iv)]
The final digit of a number is given by its congruence class mod
10. (e.g. $1527=1.10^3+5.10^2+2.10+7$ so is congruent to 7 mod
10.)

Now $n=1.2.3.\ldots.n$, so if $n\geq5,\ n!$ is divisible by both 5
and 2, and hence by 10. Hence $n!\equiv0$ mod 10 for all $n\geq5$.

Thus
$1!+2!+\ldots+10!\equiv1!+2!+3!+4!\equiv1+2+6+24\equiv33\equiv3$
mod 10, and so the final digit of $1!+2!+\ldots+10!$ is 3.

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