Archive for December, 2008

The number 5040 – according to Plato

December 30, 2008

5 040 – Greek philosopher Plato gave this number as the population of an ideal city.

This number has several factors, which implies an efficient division of population for the sake of land distribution, taxes, war etc.
In fact,
5, 040 = 7! = (1)(2)(3)(4)(5)(6)(7)
5, 040 = (24)(32)(51)(71)
Thus, 5 040 has (4+1)(2+1)(1+1)(1+1) = 60 factors

A Simple Technique #1

December 29, 2008

Compute in (less than) 30 seconds without using calculator.
My technique:
Since 2007 is the arithmetic mean of the numbers 2004, 2006, 2008 and 2010, then: Let x = 2007
So, x-3 = 2004, x-1 = 2006, x+1 = 2008, x+3 = 2010
Hence, the expression is equal to
Simplifying,
= x2 – 5 = (2007)2 – [...]

Value of a in ax+b

December 27, 2008

When the polynomial x2007 + 1 is divided by x2 – 3x +2, the remainder can be expressed in the form ax+b. Find the value of a.
My Solution:
Let k be the quotient, then:
x2007 + 1 = k(x2 – 3x +2) + (ax+b)
Since x2 – 3x +2 = (x-2)(x-1),
x2007 + 1 = k(x-2)(x-1) + (ax+b)
To omit [...]

Number of Perfect Squares

December 26, 2008

Compute the number of squares between 49 and 94
My short solution: The technique is to express these two numbers into perfect squares so that it’s easy to compute the number of squares between these two numbers.
49 = (22)9 = (29)2 = 5122 ;
94 = (32)4 = (34)2 = 812
Therefore, the number of squares between 49 [...]

25th number

December 26, 2008

Five digit numbers containing all digits from 1-5 are arranged from highest to lowest. What is the 25th number?
(Examples of these numbers are 54 321; 23 451; 41 532)
My Solution:
First, it’s impractical to list these (5)(4)(3)(2)(1)=120 numbers from highest to lowest and then determine the 25th number in the list. So, we can instead do [...]

A Simple Proof # 1

December 23, 2008

Prove that:    , for any
Solution: Since     , then in order to produce a short
proof, the expression   should be express in the form  
Doing so,

Hence,

Angle of Projection

December 23, 2008

At what angle of projection is the range maximum? Prove it mathematically.

To solve for the required angle of projection ( ), we need first to find a general expression for the range.
To find range, set y = 0 first and use:
,

Implies that,

(non zero value)
Solving for the general expression for range,

Regardless of the [...]