﻿ Show the rule of divisibility for 5

# Show the rule of divisibility for 5

The rule of the divisibility for 5 says that if a number is divisible by 5, the last digit of the number is 5 or 0.

For example, 525, the last digit of the number is 5. Hence, 525 is divisible by 5 without leaving remainder.

525 \div 5=105

Then how to prove the rule in general form?

Let’s first evaluate 3-digit number that is denoted as \overline{abc}, in which a, b, c are three digits at hundreds, tens and ones places respectively. Since a number can be expressed as the sum of its place values, \overline{abc} is rewritten as

\overline{abc}=100a+10b+c

Divide both sides by 5

\dfrac{\overline{abc}}{5} =\dfrac{100a+10b}{5}+ \dfrac{c}{5}

=55a+5b+\dfrac{c}{5}

The equation shows that if \overline{abc} is divisible by 5 if and only if \dfrac{c}{5} is an integer or zero. Then, there are only 2 cases for c to be divisible by 5. One is c is 5; the other, c is 0.

In a more 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

=10\cdotp a_n 10^{n-1}+10\cdotp a_{n-1} 10^{n-2}+\dots+10a_1+a_0

Divide the equation by 5

\dfrac{a}{5}=\dfrac{10\cdotp a_n 10^{n-1}+10\cdotp a_{n-1}10^{n-2}+\dots+10a_1}{5}+\dfrac{a_0}{5}

=2\cdotp a_n 10^{n-1}+2\cdotp a_{n-1}10^{n-2}+\dots+2a_1+\dfrac{a_0}{5}

which shows, a is divisible by 5 if and only if a_0 is equal to 5 or 0 (a\geq 10).

Collected in the board: Number Theory

Steven Zheng posted 4 months ago

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