A problem with numbers

Question

Aliens have been found in Mars who have six fingers in each of their hands, total 12 fingers in their two hands. We use 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 to do all the calculation, and they use 0, 1, 2, 3, 4, a, 5, 6, 7, 8, 9, b. So, 10, 11, 12 in earth is 9, b, 10 in Mars. Find the product of 1b and 16 in aliens’ system.

Answer 1

1b = 23 base 10 16 = 19 base 10

this product is equal to 437 base 10

reverting to base 12 gives 437 = 3(12)^2 + 0(12) + 5

This corresponds to the base 12 number 30a

Answer 2

The aliens use a slightly modified duodecimal number system with the following digits (and their corresponding decimal numbers in parenthesis). $$0(0), 1(1), 2(2), 3(3), 4(4), a(5), 5(6), 6(7), 7(8), 8(9), 9(10), b(11)$$

Converting the numbers to decimal, $$1b_{12} = b_{12} \times 12^0 + 1_{12} \times 12^1 = 11 + 12 = 23$$ and $$16_{12} = 6_{12} \times 12^0 + 1_{12} \times 12^1 = 7 + 12 = 19$$ So the product is $23 \times 19 = 437$ (in decimal). The number in the aliens' system is obtained by factorising 437 by 12 and writing the remainder in reverse order. $$437mod12=5$$ $$36mod12=0$$ $$3mod12=3$$ Thus the number is $30a_{12}$