Solve it Sunday: Tribal Mathematics

Hello everybody, I hope you have your thinking caps on! I’m back for another Solve it Sunday post, and this week is a tricky one.

As always, answer is in the comments, and you think you have the right answer, make a comment of your own.

During the 19th century, a European colonel in Ethiopia recorded a report of an encounter with local tribesmen, from whom he was purchasing cattle.

He wanted seven beasts, at a cost of 22 birr each. Not being numerate, the herder called a local priest to verify the total price.

When he arrived, the priest dug two parallel columns of holes. The right-hand column represented the purchase price, so in the first hole he placed 22 stones, and then halved the number of stones for each subsequent hole, rounding down. This gave him 22, 11, 5, 2 and 1 stone.

The left-hand column then represented the cattle, and in the first hole he placed seven small stones. He then doubled the number of stones for each subsequent hole in the column, so that the holes contained 7, 14, 28, and 112 stones.

Declaring even values to be evil, the priest then went down to the right-hand column and whenever he encountered an even number of stones – the 22 and 2 holes, in this instance – he removed the stones from that hole and its neighbor in the left-hand column – 14, 28, and 112 respectively – into one pile, which he counted out one by one.

They came to 154 birr, which was indeed 22×7.

Indeed, this technique of multiplication will always work for whole numbers, but why?

Think you know the answer: let me know in the comments or on social media!
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One thought on “Solve it Sunday: Tribal Mathematics

    Despite the spiritual trappings, this is sophisticated physical implementation of binary multiplication. The right-hand column, where the number value is halved repeatedly (rounding down), and then even numbers discarded, becomes the binary equivalent of the number in base ten. So here, 22 becomes represented by 10110.

    Then the left-hand column serves as a way of multiplying the powers of two through for the designated volume number. By starting from he unit value, 7, and then doubling, each column becomes 7x that binary digit. 14 is 7×2, 28 is 7×4, and so on. Because 22 is 10110, 7×22 is (0x1) + (7×2) + (7×4) + (0x8) + (7×16), and adding those stones gives your final answer.

    It is just as effective the other way around, of course: 7, halved and rounded, gives 3, which then gives 1. No “evil” evens, so just three holes for 22, 44, and 88 – which, added together, give you 154.


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