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Interesting observations in number theory

September 17, 2011

Mohan, mdashf

(perhaps never known before at-least most of these)
I will post another article for my 50cent theorem and related facts

an interesting pattern for powers of 6: (6^2 .. 6^3.. 6^12), the last two digits you get by subtracting 20, for all powers from 2 to 12 and more

The last two digits I have separated by hyphen – ;
‎36, 2-16, 12-96, 77-76, 466-56, 2799-36, 16796-16, 100776-96, 604661-76, ***56, ***36, ***16, ***96 ….

Therefore you can discard all numbers whose last 2 digits are not 16, 36, 56, 76 or 96 : I call this my 50 cent axiom for power exponents of 6..

note that 20+16 =36, so start with 16 and go on adding 20..

new 50cent axiom for powers of 6, a number is not a power of 6 if it’s last two digits are not one of these: 16, 36, 56, 76 or 96, always…

for 6 one can also note it’s all odd numbers less than 10 in the 2nd digit from right .. and always 6 in the last

for powers of 4 last digit is always 4 or 6 (which add upto 10)

‎50cent axiom for powers of 7: last digit is always an odd number less than 10 but not it is never 5. (contrary to powers of 6, all odd numbers but in 2nd-last digit)

‎50cent axiom for powers of 3: all last-digits odd numbers except 5, which is same exactly as that of powers of 7. But 2nd-last digit for 7 and 3 differ.

given my 50cent axioms for testing a number is a possible power of (3,4,5,6,7) or not one can write one code which can take any combination of these base (3,4,5,6,7) and create a sample from list of any numbers

one can build such a routine to every good scientific software and after one has tested a list of numbers one can employ the “50cent theorem for powerfree numbers” in addition to this ..

look at this random number: 6776775546: this is not a power of 3,5,6,7 (just by looking at it, using my 50cent axioms above)

i.e. if you keep on dividing this number by any one of the 3,5,6,7 you will never get 3 for 3, 5 for 5, 6 for 6 and 7 for 7 .. you will be stopped somewhere that says noy further divisible by the given number among 3,5,6,7

useful if you want to buy pizza for 5 people size and you have “6776775546” number of people in the world, you will either go shortage of pizza or some pizza will be thrown away..

what a puzzle: if you have that many number of people for whom you want to buy pizza and you have got pizza coming only for 3,5,6 or 7 people size and assuming Stephen Hawking does not eat more than his expected apetite, assuming everyone eats equally, you will run into problem either having less pizza or pizza that are thrown …

Talking about powers of 4 again, alternative powers of 4 when added is divisible by 5. [4^k + 4^(k+1)] is always divsible by 5 and 10 and return integers. ie [4^k + 4^(k+1)] is not a prime number.

it’s divisibility by 5 is clear as you can take 4^k a common factor, rest adds to 5, but divisibility by 10 is not clear..

the divisibility of 10 follows from my 50cent axiom for powers of 4

powers of 6 and next power of 6 when added .. are divisible eg by 2, 4, 7, 8, 14, 16, 28, 56, 112 etc, if the power ends at 16. (check for other endings)

nature may speak mathematics but it’s often quiet..


Post a comment
  1. September 21, 2011

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  2. Mohan #
    September 28, 2011

    please note that above is a spam comment (nothing it mentions in special but a general comment, a sign of spamming)

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