LCM problem with remainder

Question

Let us find the least number divisible by $13$ such that when that number divided by $8,12,16,20$ it leaves remainder $1$ in all cases

$help !$ $Answer$ $4812$ $how ?$

Answer 1

Do you know the Chinese Remainder Theorem?. Note that $1$ satisfies the last half and solutions to the last half recur at intervals of $\operatorname{LCM}(8,12,16,20)=240$ so you want a number of the form $240k+1$ that is a multiple of $13$. As the remainder on dividing $240$ by $13$ is $6$, you need two of them because $1+6+6=13$. The answer is $481$.

Answer 2

These are the given conditions.

\begin{align} N \equiv 0 \pmod{13} \\ N \equiv 1 \pmod{ 8} \\ N \equiv 1 \pmod{12} \\ N \equiv 1 \pmod{16} \\ N \equiv 1 \pmod{20} \\ \end{align}

The last four conditions can be simplified to one condition.

\begin{array}{r|ccc|c} & 2 & 3 & 5 \\ \hline 8 & 2^3 & 1 & 1 \\ 12 & 2^2 & 3^1 & 1 \\ 16 & 2^4 & 1 & 1 \\ 20 & 2^2 & 1 & 5^1 \\ \hline &2^4 & 3^1 & 5^1 & 240 \end{array}

$$\operatorname{lcm}(8,12,16,20) = 240$$

\begin{align} N \equiv 0 \pmod{13} \\ N \equiv 1 \pmod{240} \\ \end{align}

From the first congruence, we know that $N = 13n$. From the second we know that $13n \equiv 1 \pmod{240}$.

\begin{align} 13n &\equiv 1 \pmod{240} \\ 13n &= 240m + 1 \\ 13n &= 13\cdot 18m + 6m + 1 \\ 6m+1 &\equiv 0 \pmod{13} \\ 6m &\equiv 12 \pmod{13} \\ m &= 2 \\ 13n &= 240 \cdot 2 + 1 \\ 13n &= 481 \\ N &= 481 \\ \end{align}

So $\color{\red}{N = 481}$.