edited by
636 views

1 Answer

1 votes
1 votes
Here right most non-zero digit depends on the unit digit of $3^{2720}$

 $3^{2720}$ = $[3^{4}]^{680}$ =$81^{680}$

The unit digit of 81 is 1 so any power of 81 will always give its unit digit as 1.

Thus, required unit digit is 1.

Hence (1).1 is the Answer.

Related questions

0 votes
0 votes
0 answers
2
go_editor asked Dec 29, 2015
286 views
Let $\text{S}$ be a positive integer such that every element $n$ of $\text{S}$ satisfies the conditions$1000 \leq n \leq 1200$every digit in $n$ is oddThen how many elem...
0 votes
0 votes
1 answer
3
go_editor asked Dec 29, 2015
633 views
For a positive integer $n$, let $\text{P}_n$ denote product of the digits of $n$ and $\text{S}_n$ denote the sum of the digits of $n$ The number of integers between $10$ ...
0 votes
0 votes
0 answers
5
go_editor asked Dec 29, 2015
345 views
Let $n! = 1 \times 2 \times 3 \times \dots \times n$ for integer $n \geq 1$. If $p = 1! (2 \times 2!) + (3 \times 3!) + \dots + (10 \times 10!)$, then $p+2$ when divided ...