Use congruences to prove that the divisibility test 5 works.

Question

Divisibility rule for 5: Let a be any integer, 5 divides a if and only if the last digit of a is 0 or 5.

Answer 1

We write the integer number a as a=a_k 10^k + a_{k-1}10^{k-1}+...+a_1 10+a_0 where the numbers a_{k}, a_{k-1},...a_{0} are integers in [0,9]. Since 5 divides a and 5 divides all the previously mentioned powers of 10 we have that 5 must divide a_{0}. But a_{0} is either 0 or 1 or 2 or 3,..,or 9 and hence, the only options are a_{0}=0 or a_{0}=5.