Today’s generation does not have a monopoly on good ideas. Here is some ‘new maths’ from the ancient Egyptians.

To multiply 26 by 33, successively ‘halve’ the 26 (dropping 1 where necessary), and double the 33 as follows (below right):

Now add the multiples of 33 corresponding to the odd numbers in the first column.

Thus, 66 + 264 + 528 = 858 – the right answer!

Hint 1

Try some simple examples to verify that the method works. What happens when the first number is, say, 32?

Hint 2

It is the dropping of the 1’ which is a worry here. Carefully work out the exact effect of repeatedly dividing 26 by 2. Now do the multiplication.

Solution

We work through the given example in a way which will hopefully make the method clear.

26 x 33 = (13 x 2) x 33 = 13 x (2 x 33) =

= 13 x 66

= ((6 x 2) + 1) x 66 = (6 x 2) x 66 + (1 x 66)

= (6 x 132) + 66

= (3 x 2) x 132 + 66

= 3 x 264 + 66

= ((1 x 2) + 1) x 264 + 66 = 1 x 528 + 1 x 264 + 66

= 528 + 264 + 66, as the Egyptians 'predicted'.

It is clear that whenever a factor of 26 is even, the process of dividing by 2 and multiplying by 2 causes no difficulty. However, when the factor (red) is odd the process ‘throws out’ a term which needs to be added to the answer total.

More briefly,

26 x 33 = (16 + 8 + 2) x 33 = 528 + 264 + 66.

Extensions

1. Try working through some other multiplications in detail like this to understand what is happening.

2. You may be interested in finding out what other mathematics was known by early civilizations. Use your library to look up An introduction to the history of mathematics – Howard Eves (Holt Rinehart and Winston)^.

3. There are fast ways to multiply by certain numbers. For example, to multiply by 10, just add a 0 on the end. Can you suggest fast ways of multiplying by 5? by 25?

Hint 1

Hint 2

Solution

Extensions

^{26 33}

^{13 66 <–}

^{6 132}

^{3 264 <–}

^{1 528 <–}