Question

The digit in the unit place of $2009! + 3^{7886}$ is

• A. 9
• B. 7
• C. 3
• D. 1

Answer 1

Answer: (A)

Unit's digit of $2009! = 0$
Now, $3^1 = 3 \Rightarrow $ unit's digit = 3
$3^2 = 9 \Rightarrow $ unit's digit = 9
$3^3 = 27 \Rightarrow $ unit's digit = 7
$3^4 = 81 \Rightarrow $ unit's digit = 1
$3^5 = 243\Rightarrow $ unit's digit = 3
Continuing the process, we get
$3^{7866} = 3^{7860 +6} = 3^{4 \times 1965} \times 3^6$
$\therefore $ Unit's digit of $3^{7866}$
= unit's digit of $(3^{4 \times 1965} \times 3^6) $
$ = (1)^{1965} \times 9 = 9$
So, unit's digit of $2009! + 3^{7866} = 0 + 9 = 9$