Application of Divisibility Rules

Prove the divisibility rule of 7: For a number $N,$ double the units digit and take its difference with the remaining number except for the units digit. If it is a multiple of 7, then the number is divisible by 7.

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This follows because $10^6 -1 = 999999 = 142857 \times 7$, so $10^{6k}M_k + 10^{6(k-1)}M_{k-1} + \cdots + 10^6 M_1 + M_0$ is a multiple of 7 if and only if $M_k + M_{k-1} + \cdots + M_0$ is a multiple of 7. If we use $0 \leq M_i \leq 999999$, we get the result as stated. $_\square$

Show that if the last 3 digits of a number $N$ are $\overline{abc}$, then $N$ is a multiple of 8 if and only if $4a + 2b + c$ is a multiple of 8.

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This follows because $100a + 10 b + c = 8 (12a + b) + 4a + 2b + c$.

Hence, by the divisibility rule of 8, $N$ is a multiple of 8 if and only if $4a+2b+c$ is a multiple of 8. $_\square$

Show that the 6-digit number $\overline{abcdef}$ is a multiple of 7 if and only if $5a + 4b + 6c + 2d + 3e + f$ is a multiple of 7.

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This follows because

$$\begin{aligned} &1 & = &0 \times 7 & + 1\\ &10 & = &1 \times 7 & + 3\\ &100 & =& 14 \times 7 & + 2\\ &1000 & = &142 \times 7 &+ 6\\ &10000 & = &1428 \times 7 &+ 4 \\ &100000& = &14285 \times 7 &+ 5 &.\\ \end{aligned}$$

Hence $\overline{abcdef}$ is a multiple of 7 if and only if $5a + 4b + 6c + 2d + 3e + f$ is a multiple of 7. $_\square$