Question

Show that one and only one out of $n,n+4,n+8,n\hspace{0.17em}+12\mathrm{and}n+16$ is divisible by 5 , where n is any positive integer .

Answer 1

Consider the numbers $n,\left(n+4\right),\left(n+8\right),\left(n+12\right)\mathrm{and}\left(n+16\right)$, where n is any positive integer.
$$\begin{gathered} \mathrm{Suppose}n=5q+r,\mathrm{where}0\le r<5 \\ n=5q,5q+1,5q+2,5q+3,5q+4 \end{gathered}$$
(By Euclid's division algorithm)

Case: 1
$$\begin{gathered} \mathrm{When}n=5q. \\ n=5q\mathrm{is}\mathrm{divisible}\mathrm{by}5. \\ n+4=5q+4\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5. \\ n+8=5q+5+5+3=5\left(q+1\right)+3\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5. \\ n+12=5q+10+2=5\left(q+2\right)+2\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5. \\ n+16=5q+15+1=5\left(q+3\right)+1\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5. \end{gathered}$$

Case: 2
$$\begin{gathered} \mathrm{When}n=5q+1. \\ n=5q+1\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5. \\ n+4=5q+1+4=5\left(q+1\right)\mathrm{is}\mathrm{divisible}\mathrm{by}5. \\ n+8=5q+1+5+3=5\left(q+1\right)+4\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5. \\ n+12=5q+1+12=5\left(q+2\right)+3\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5. \\ n+16=5q+1+16=5\left(q+3\right)+2\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5. \end{gathered}$$

Case: 3
$$\begin{gathered} \mathrm{When}n=5q+2. \\ n=5q+2\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5. \\ n+4=5q+2+4=5\left(q+1\right)+1\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5. \\ n+8=5q+2+8=5\left(q+2\right)\mathrm{is}\mathrm{divisible}\mathrm{by}5. \\ n+12=5q+2+12=5\left(q+2\right)+4\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5. \\ n+16=5q+2+16=5\left(q+3\right)+3\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5. \end{gathered}$$

Case: 4
$$\begin{gathered} \mathrm{When}n=5q+3. \\ n=5q+3\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5. \\ n+4=5q+3+4=5\left(q+1\right)+2\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5. \\ n+8=5q+3+8=5\left(q+2\right)+1\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5. \\ n+12=5q+3+12=5\left(q+3\right)\mathrm{is}\mathrm{divisible}\mathrm{by}5. \\ n+16=5q+3+16=5\left(q+3\right)+4\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5. \end{gathered}$$

Case: 5
$$\begin{gathered} \mathrm{When}n=5q+4. \\ n=5q+4\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5. \\ n+4=5q+4+4=5\left(q+1\right)+3\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5. \\ n+8=5q+4+8=5\left(q+2\right)+2\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5. \\ n+12=5q+4+12=5\left(q+3\right)+1\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}5. \\ n+16=5q+4+16=5\left(q+4\right)\mathrm{is}\mathrm{divisible}\mathrm{by}5. \end{gathered}$$

Hence, in each case, one and only one out of $n,n+4,n+8,n+12\mathrm{and}n+16$ is divisible by 5.