Assertion (A): $4^n$ ends with digit 0 for some natural number $n$.
Reason (R): For a number '$x$' having 2 and 5 as its prime factors, $x^n$ always ends with digit 0 for every natural number $n$.

Question

Accepted Answer

Step 1: $4^n$ cycles through 4, 16, 64, 256, ... (never ends with 0). So A is false. R is true since if $x$ has both 2 and 5 as prime factors, $x$ is divisible by 10.

Question

Solution