The rule of the divisibility for 4 says that if a number is divisible by 4, the sum of place values of the last two digits of the number is divisible by 4.

For example, 1652, the place values of the last two digits are 50 and 2, which are summed up to 52. Since 52 is Divisible 4. Hence, 1652 is divisible by 4 without leaving remainder.

1652 \div 4=413

Then how to prove the rule in general form?

Let’s first evaluate 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 4

\dfrac{\overline{abcd}}{4} =\dfrac{1000a+100b}{4}+ \dfrac{10c+d}{4}

=250a+25b+ \dfrac{10c+d}{4}

Now it’s clear that if the quotient remains as integer, the sum of place values of the last two digits must be divisible by 4, that is, 10c+d is divisible by 4.

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

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

\dfrac{a}{4}=\dfrac{100\cdotp a_n 10^{n-2}+100\cdotp a_{n-1}10^{n-3}+\dots+100\cdotp a_2}{4} +\dfrac{10a_1+a_0}{4}

=25\cdotp a_n10^{n-2}+ 25\cdotp a_{n-1}10^{n-3}+\dots+25a_2+\dfrac{10a_1+a_0}{4}

Therefore, if a is divisible by 4 on the left hand side, the sum of place values of the last two digits of the number must be divisible by 4.