Question

The last non zero digit in 12! is

• A. 6
• B. 5
• C. 4
• D. 3

Answer 1

The correct option is A 6

$12!$
$=\mathrm{1.2.3....11.12}$
$=\left(\mathrm{2.4.6.8.10.12}\right)\left(3.9\right)\left(5\right)\left(7\right)\left(11\right)$
$=2^{10}{.3}^5{.5}^2.7.11$
Hence for ending digits we omit the multiples of 5, as 5 and 2 together will give 0's at the end of the number.
Hence
$2^8{.3}^5.7.11$
$=8\left(2^5{.3}^5.7.11\right)$
$=8\left(598752\right)$
Hence ending digit will be $8\times 2$
$=16$
Hence ending digit will be 6.