Show the rule of divisibility for 10

The rule of the divisibility for 10 says that if a number is divisible by 10, the last digit of the number is 0.

For example, 16520, the last digit of the number is 0. Hence, 16520 is divisible by 10 without leaving remainder.

16520 \div 10=1652

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 10

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

=10a+b+\dfrac{c}{10}

The equation shows that if \overline{abc} is divisible by 10 if and only if \dfrac{c}{10} is zero, that is

( is 5.

In a more general form, any integers can be expressed as

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

=10\cdotp a_n10^{n-1}+10\cdotp a_{n-1}10^{n-2}+\dots+10a_1+a_0

Divide the equation by 10

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

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

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

Collected in the board: Number Theory

Steven Zheng Steven Zheng posted 5 days ago

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