﻿ Show the rule of divisibility for 8

#### Question

Show the rule of divisibility for 8

Collected in the board: Number Theory

Steven Zheng posted 5 months ago

The rule of the divisibility for 8 says that if a number is divisible by 8, the sum of place values of the last three digits of the number is divisible by 8.

For example, 21496, the place values of the last three digits are 400, 90 and 6, which are summed up to 496. Since 496 is Divisible 8. Hence, 21496 is divisible by 8 without leaving remainder.

8| 21496

Then how to prove the rule in general form?

Let’s first evaluate a 4-digit number that is denoted as \overline{abcd}, in which a, b, c, d are four digits on thousands, hundreds, tens and ones places respectively. Since a number can be expressed as the sum of its place values, \overline{abcd} is rewritten as

\overline{abcd}=(1000a+100b)+(10c+d)

Divide both sides by 8

\dfrac{\overline{abcd}}{8} =\dfrac{1000a}{8}+ \dfrac{100b+10c+d}{8}

=125a+ \dfrac{100b+10c+d}{8}

which shows that if the quotient remains as integer, the sum of place values of the last three digits must be divisible by 8, that is, 100b+10c+d is divisible by 8.

In a general form, any integers can be expressed as

a=a_n10^n+a_{n-1}10^{n-1}+\dots+a_210^2+a_110+a_0

=(100\cdotp a_n 10^{n-2}+100\cdotp a_{n-1}10^{n-3}+\dots+(100\cdotp a_2)+(10a_1+a_0)

Divide it by 8,

\dfrac{a}{8} =\dfrac{1000\cdotp a_n 10^{n-3}+1000\cdotp a_{n-1}10^{n-4}+\dots+1000\cdotp a_3}{8} +\dfrac{100\cdotp a_2+10a_1+a_0}{8}

=125\cdotp a_n10^{n-3}+ 125\cdotp a_{n-1}10^{n-4}+\dots+125a_3+\dfrac{100a_2+10a_1+a_0}{8}

Therefore, if a is divisible by 8 on the left hand side, the sum of place values of the last three digits of the number must be divisible by 8.

Steven Zheng posted 5 months ago

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