Question

If four whole numbers taken at random are multiplied together, the chance that the last digit in the product is 1, 3, 7 or 9 is $a/b$, then $a+b=$

Answer 1

The last digit of any number is $0,1,2,3,\mathrm{4...9}$
The last digit of any of the four whole numbers is $0,2,4,5,6$ and $8$
then the product of the four whole number cannot be $1,3,7$ or $9$
Hence if the last digit of product of four whole number is $1,3,7$ or $9$, the last digit be $1,3,7$ or $9$
Therefore probability that the last digit in any number is $1,3,7$ or $9$ $=\frac{4}{10}=\frac{2}{5}$
Therefore probability that the product have last digit $1,3,7$ or $9$ $={\left(P\right)}^4={\left(\frac{2}{5}\right)}^2=\frac{16}{625}$
$\therefore a+b=16+625=641$