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.
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Have you missed out a between 9 and b? – badjohn Apr 30 '17 at 07:31
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you mean 9, a, b – Saketh Malyala Apr 30 '17 at 07:34
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All you have to do is multiply in base 12 and convert the base 12 result to "alien" characters. – Millie Smith Apr 30 '17 at 07:35
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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
Saketh Malyala
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2I think there's a point in the a being in the "5" slot for the "alien" system. I think the student is supposed to notice this and convert to base 10 / 12 and back to the alien system with this in mind. So I think 16 in alien is actually 19 in base 10. – Millie Smith Apr 30 '17 at 07:44
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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 $30a_{12}$
