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 the units digit. If it is a multiple of 7, then number is divisible by 7.

We will prove the above theorem.
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 \( M_k + M_{k-1} + \ldots + M_0\) 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 \( \overline{abc} \) 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. \(_\square\)