“Probably no symbol in mathematics has evoked as much mystery, misconception and human interest as the number pi”, according to well-known professor of mathematics, William L. Schaaf.
Pi has interested people around the world for over 4,000 years. Many mathematicians – from famous ones such as Fibonacci, Newton, Leibniz, and Gauss, to lesser well-known mathematical minds – have toiled over pi, calculated its digits, and applied it in numerous areas of mathematics. Some spent the better parts of their lives calculating just a few digits.
Mr. Subramani K, one of the staff members of IIIT Bangalore, who is always intrigued by various components of mathematics has solved mathematical problems on pi.
The following technique highlights an interesting aspect of (pi) when operated upon by a cyclic prime number. The procedure is as follows:
A). Consider the value of (pi), to 15 decimal places:
3.141592653589793
B). Now consider the recurring number 1/7
1/7 = 0.142857142857…, recurring after every 6 digits.
There being 6 digits, we are going to find what we call 6 “Special Multiplicands” for (pi) that gives some interesting results.
The 1st step is to write out (pi) (without the decimal point) & divide by 7:
7) 314159265358979306 ( 06 suffixed for perfect divisibility )
044879895051282758
A reminder of 4 occurs. In order for this not to happen & instead be perfectly divisible, we suffix any 2 digits to make the remainder go to zero. Appending 06 & continuing with the division by 7 yields the “Special Quotient” to be
0448798950551282758
C). We now write down the recurring number 1/7, again omitting the decimal point, and subtract this “Special Quotient” from it. We start off, for our FIRST case, by writing from the 1st digit of 1/7=0.142857.. Namely i.e .
142857142857142857
– 044879895051282758
——————————–
097977247805860099 = S1
——————————–
097977247805860099 is the number “S1” that we get.
Now write down that string of numbers that will always give 9 when added to S1
(let us call this T1):
S1: 097977247805860099
T1:902022752194139900
Let us now multiply T1 by 7
T1*7 =06314159265358979300
Omit the first two leading digits & the last two trailing digits and voila! We are again left with (pi) : 3141592653589793
D). We now move on to the SECOND digit of the recurring number 1/7 = 0.142857, namely 4 and repeat the process as in C- first writing 428571428571… & then subtracting the Special Quotient from it to get S2.
428571428571428571
044879895051282758
——————————–
383691533520145813 = S2
———————————-
The corresponding number T2 that on adding with S2 gives a string of 9’s is:
T2 =616308466479854186
Multiply T2 by 7, again omitting the leading digit & the last 2 trailing digits in the product:
T2 * 7 =04314159265358979302
We are again left with the digits in (pi) (again ignoring the decimal point)! This exact procedure can be repeated for each of the other 4 digits in the recurring number 0.142857 to yield “Special multiplicands” T3, T4, T5 & T6 ;
T3 =759165609336997043,
T4 =187737037908425615,
T5 =473451323622711329 &
T6=330594180765568472.
all of which have the interesting property that on multiplying any Ti by 7, and
omitting the last two trailing digits and also omitting the leading digits (including 0)
We are left with,
3141592653589793 !
It should be noted that all the Special Multiplicands are implicitly considered to have
a decimal point after the first digit. Hence in reality :
T1 is = 9.02022752194139900
T2 is = 6.16308466479854186, etc , etc ,——
Now we could extend this procedure to other Cyclic Primes,
( 17,19,23,29,47,59,61,97,109, —–503 )
Subramani.K
Email : Subramani.k@iiitb.ac.in
