Expert: Scott A Wilson - 8/2/2011

Question
Dear Prof Scott

Writing New Mathematical Equations and Theorems
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Can Strong Fundamentals Numbers Theory with Basic Mathematical computations (Addition, Subtraction, Multiplication, Division, Exponentiation etc),Calculus, Algebra and Geometry is the Base for Creating New Mathematical Equations and Theorems ?.

For example we learn Multiplication Tables viz one digit number, Two Digit Number, Three Digit Number, should we start
practicing a four digit, five digit .... n digit multiplication table ?

Multiplication Table i mean is for example say 7 a single digit number.
7*1=7
7*2= 14
7*3 =21
7*4=28
.
.
.


We learn so many theorems and Formula/Equations in Various Branches of Mathematics viz Statistics, Trigonometry, Differential Equations,
Profit and Loss, Matrices, Complex Numbers, Simulate nous Equations, Quadratic Equations, Pythagoras Theorem, Fourier series, Laplace transforms,Logarithms,
Vector Algebra, Calculus and the List is Endless.

What would be your Guidelines to Budding Mathematicians who would like to write their own Theorems and Mathematical Equations in those Branches ?

For example similar to Pythagoras Theorem 3,4,5 -> 3 square + 4 square = 5 square.

Should we make our Base Fundamentals clear and explore in detail advanced Maths to achieve this Goals ?

Awaiting your views, inputs and opinion ?

Thanks & Regards,
Prashant S Akerkar

Answer
The best way I've found to find something new is to keep reviewing old problems.

For example, it is known that when a number is multiplied by 3, the digits add up to a number divisible by 3.  This also applies to 9.  What I'm not sure I've seen anywhere is that when a number is multiplied by 7, divide the result by separating off the last digit.  Multiply the first part by 3, then add in that last digit.  If the result is divisible by 7, so is the number that was started with.

For example, take 175; divide 175 up by taking off the last digit ( 5 ), and leaving 17.  Multiply 17 by 3 and get 51, then add 5, givin 56.  Since 56 is divisible by 7, 175 is also divisible by 7.

For a longer problem, take 32,475.

Numer   start   last   3*start + last
227325  22732    5          68201
68201   6820    1          20461
20461   2046    1          6139
 6139    613    9          1848
 1848    184    8          560
  560     56    0          168
  168     16    8          56
   56      5    6          21
   21      2    1          7
Since it was recognized as being divisible by 7, I could have stopped at 560.

To check divisibitity is far easier to do with calculators, but its kind of interesting.
This same pattern can be applied to many other digits as well by different use of factors.
It always involves taking off the last ditit, many times uses different multipliers, sometimes uses a multiplier on the last digit, and sometimes changes the plus to a minus.


Another thing I've found is the repeat of the last digits in powers.
For example, take 7.  It is known that the powers of 7 are 7, 49, 343, 2401, 16807, 117649, etc.
Looking at the last two digits gives us 07, 49, 43, 01, 07, 49, etc.
As can be seen, the pattern for 7 is 01, 07, 49, 43.
Another interesting not here is the 1st and 3rd numbers add to 50 and the 2nd and 4th numbers add to 50.  This is most likely since 7^0 = 1, 7^2 =  49, and 1+49 = 50,
so for any power n of 7, if the last two digits are taken from both numbers, the sum is 50.

The only way I've found to come up with something new is to keep reviewing what I know already.