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Friday, September 22, 2017

A Multiplication Shortcut

Suppose I want to multiply 8 and 256.  I note that 8 × 256 = 23 × 28 = 211 = 2048.   I have turned multiplication into addition. A voice in my head says, “Hey, that could be a useful trick.”  Another voice says “What if you want to multiply 7 and 652? Stupid!”   It is true that 7 = bx or  652 = by have no whole number solutions (except the obvious and useless b = b1.)  But we do not have to restrict ourselves to whole numbers.

Consider the equations √10 × 10 = 10 and 100.5 × 100.5 = 101 = 10.  In these equations 10 and 100.5 have exactly the same job.  We can consider them equivalent.   Hence the exponent ½ (or 0.5) represents a square root.  Similarly  the exponent represents a cube root. In general, the exponent n/m represents the mth root of a number to the power n. For example, 105/6 is the sixth root of 100 000.  Note that 100 000 = 105.

Lets look again at the product of 7 and 652.

P = 7 × 652

P = 10a × 10b = 10a+b.  

I want 7 = 10a and I want 652 = 10b. In other words I want the logarithms of 7 and of 652 in base 10.  By definition of a logarithm y = bx ó logb y = x.   Therefore 7 = 10a ó log10 7= a   and 652 = 10b ó log10 652= b.  The numbers a and b are 0.845 098 040 and  2.814 247 595 respectively.     

Therefore P = 10 0.845 098 040 +2.814 247 595  = 103.659345635 = 4563.999 = 4564  

Ok, that’s impressive. But where did you get log10 7= 0.845 098 040  and log10 652 = 2.814 247 595?  Agreed , that’s the hard part.  John Napier was the first person to calculate a table of logarithms. He published his table in the year 1614; after 20 years of work!  
Picture from
Napier's "invention was quickly and widely met with acclaim. The works of Bonaventura Cavalieri (Italy), Edmund Wingate (France), Xue Fengzuo (China), and Johannes Kepler's Chilias logarithmorum (Germany) helped spread the concept.”
Napier’s original table did not use base 10.  A few years later Napier worked with Henry Briggs  to produce a table of common (base 10) logarithms.  Briggs continued after Napier died and published their table in 1624.
Picture from

Tuesday, June 27, 2017

Arnold Sunrise Problem

In a interview in 1995 the notable Russian mathematician Vladimir Igorevich Arnold  recalled a problem set  by his schoolteacher, I. V. Morozkin, when he (Arnold) was 11 or 12 years old. (An Interview with Vladimir Arnold)

“Two old women started at sunrise and each walked at a constant velocity. One went from A to B and the other from B to A. They met at noon and, continuing with no stop, arrived respectively at B at 4 p.m. and at A at 9 p.m. At what time was the sunrise on this day?”

Arnold says “I spent a whole day thinking on this.”  I presume, from Arnold’s statement, the sunrise problem was an extension problem  set for high achievers after they completed  routine exercises.  That Arnold spent a day thinking about the problem implies Mr Morozkin did not teach the class a technique to solve it. Even as an extension exercise the sunrise problem seems far beyond the curriculum for 11-12 year olds in most countries now or then. The year was 1949.  

The sunrise problem seems to require fairly advanced abstract thinking at an age when Piaget believed that children were just making the transition from the concrete operational stage of development to the formal operation stage. Arnold said his solution was based on what  are “now called scaling arguments” and “came as a revelation.”  Did he draw something like this?

The women (lets us call them Ekaterina and Yelena) are walking at constant velocity.
Using Arnold's hint we can write VE = kVY
Rearranging, k= VE /VY .
Then from the diagram, or otherwise,  t/9 = 4/t.  Obviously, t = 6.
Sunrise occurs 6 hours before noon, that is  06.00 or 6 am.

Actually, I have omitted one step. (AP/t)/(AP/9) = (PB/4)/(PB/t ) But this seems trivial.

Saturday, May 13, 2017

Using Algebra and Geometry


In the regular hexagon below, compare the shaded area to the entire area. 

 There appear to be three tasks.

·         Find the grey area. 

·         Find the total area.

·         Compare the two results.

Where to start? It is often a good idea to make a rough sketch. 

Central congruent triangles

We don’t know if the hexagon has sides of 3 cm or 2 miles or ….  Actually it doesn’t matter.  Each of the sides is 1s.  Partition the hexagon into 6 identical equilateral triangles (One has been shaded pink).  The apothem (or “radius”) is the height of the pink triangle, h_p.

The area of the pink triangle can be found using A= 1/2  (side)(side)  sin (included angle).  Your calculation should show the area of a pink triangle is √3/4. Therefore the area of the entire hexagon is  6√3/4 .

The grey area

Now consider the grey triangle.  The sum of the interior angles is  180(n - 2)°=720°. Therefore one interior angle is 120°. The area of the grey triangle is 1/2  (side)(side)  sin (included angle) =                   1/2 (1)(1)  sin (120°) = √3/4.  

The ratio of  kX to X  is √3/4  :  6√3/4  so k is 1 6 =  1/6 .


Is the figure below a cube or a hexagon?   Actually we can consider the hexagon to be the ‘projection’  (or shadow) of a cube.  Immediately, we see the shaded area is one-sixth of the total area.

Input: A cube(3-D shape) casts a shadow on a wall. Output: A regular hexagon (2-D shape) is partitioned into 6 congruent triangles.  Thanks to students S.H. and A.L. for partitioning the hexagon into congruent triangles.


When you have the answer, don't stop. A hard-won solution may prompt additional insight.