Explain why in order to determine whether an integer is a multiple of 3, it is sufficient to show that the sum of the digits of the integer is a multiple of 3.
@Nicolas
Let’s consider the decimal representation of an integer n :
n=a_k10^k+a_{k-1}10^{k-1}+\cdots+a_110+a_0,
where a_i is the i th digit from the right (the units digit being a_0 ) and k is the number of digits in n minus 1.
We want to show that n is divisible by 3 if and only if the sum of its digits is divisible by 3.
Note that 10\equiv 1\pmod 3 . This means that 10^k\equiv 1^k=1\pmod 3 for any nonnegative integer k . Therefore, we have
\begin{align*}
n&\equiv a_k10^k+a_{k-1}10^{k-1}+\cdots+a_110+a_0 \\
&\equiv a_k\cdot 1+a_{k-1}\cdot 1+\cdots+a_1\cdot 1+a_0\cdot 1 \\
&\equiv a_k+a_{k-1}+\cdots+a_1+a_0 \pmod 3.
\end{align*}
This means that the remainder when n is divided by 3 is the same as the remainder when the sum of its digits is divided by 3.
Since a number is divisible by 3 if and only if its remainder when divided by 3 is 0, we see that n is divisible by 3 if and only if the sum of its digits is divisible by 3.