3)Find the last non-zero digit of $30^{2345}$
- $3^1=3$
- $3^2=9$
- $3^3=27$
- $3^4=81$
- $3^5=243$ ... as last digit is following a cycle of $4$ so $2345/4$ gives remainder of 1, and $3^1=3$, so the last non-zero digit is 3.
This solution confuses me. Why can they use the powers of $3$ to determine the last digit of the $2345$th entry in the series of $30$?