Question

If $a$is a non-negative integer, prove that$a$ is congruent to its last digit mod 10 [for example, $27\equiv 7\left(mod10\right)$].

Short answer

It is proved that$a$ is congruent to its last digit mod 10.

Step-by-step solution

Consider the given situation

Let us consider that$a=a_n{a}_{n-1}.......a_0$ , where$a_0,a_1,......,a_n$ are digits of $a$. This implies that$a=a_n{10}^n+a_{n-1}{10}^{n-1}+.....+10a_1+a_0$ .

Prove that a is congruent to its last digit mod 10

Rewrite the given equation asfollows:

$\begin{array}{rcl}a-a_0& =& a_n{10}^n+a_{n-1}{10}^{n-1}+.....+10a_1\\ a-a_0& =& 10\left(a_n{10}^{n-1}+a_{n-1}{10}^{n-2}+....a_1\right)\end{array}$

It is observed that 10 is a divisor of$a-a_0$ , that is,$10a-a_0$ . Then by definition of congruency, we have $a\equiv a_0\left(mod10\right)$.

Hence,it is proved that$a$ is congruent to its last digit mod 10.