Sunday, December 27, 2015


At the end of the 12/24/15 post entitled 'NUMBERS", I posed a problem:

What if I presented you with the sequence 0, 108, 27648 … and asked you to supply the next number in the sequence? The most likely response would be “How the (expletive deleted) should I know? But this is a legitimate, logical sequence with a real answer that turns out to have relevance in the TRUE analysis of the natural elements. I was going to give you the answer, i.e. the next number in the sequence 0, 108, 27648 … but I think I’ll wait and see if anyone can come up with it.

Thank you to all who responded, whether you gave a numerical answer or not. Most numerical responses were on the right track, even though none were correct. Responses, in the order received were:

D.O. 256, followed by 7077888

JM 86400000, followed by 4,031,078,400,000

D.C. 86400000

JA 86400000 added: “It doesn't fit what [you] asked though.
So if I knew for certain that [you] didn't accidentally leave 1^1 out of the sequence, I'd have to assume it was skipped on purpose and so my answer would be 1^1 * 2^2 *3^3 *4^4 * 5^5 * 6^6”
= 32400”

AJ “86400000 makes sense?”

All responses came after my 2nd hint was posted. The first hint posted at the bottom of the blog entry ‘NUMBERS’ on December 24 was “what are the prime factors of 108?” The second hint, posted December 26, about 1 hour before the first response, was “The answer is surprisingly simple, but it requires thinking in terms of powers of prime numbers.” Some of you didn’t see the first post on FB or on the Transcendental Physics blog.

I knew finding the fourth number in the sequence would be hard for anyone not used to thinking in terms of integer algebra, but I thought someone would get it. I guess it was a harder problem than I thought.

When confronted with the sequence 0, 108, 27648 … and asked to find the next number, most people, i.e, most non-mathematicians, will not know where to start, and will therefore pass or just guess. So I’m going to give you my approach for attacking this kind of problem. If you use this approach, it will probably raise your score on a standard IQ test by about 10 points! But beyond that, my reason for posting this was because putting consciousness into the equations, the subject of the current posts on this blog requires thinking in terms of integer mathematics, and 108 = 11 x 22 x 33 is a key number in the TRUE analysis of quantum physics.

Here’s my approach (designed to be understood by anyone) to finding the next number of any numerical sequence (even if it involves non-integers), given the first three:

First, number the integers of the sequence: #1, #2, #3 … #N. Then each member of the sequence is a function of N, call it f(N). Now don’t be put off by the technical terminology. ‘Function’ is just a word used by mathematicians to describe the fact that each member of the sequence will depend on N, the sequence number of the member. The f in f(N) could just as well stand for formula. What you are asked to find is the formula that will give you the value of each and every member of the sequence. In this way, the value of each member of the sequence will depend upon its ‘number’, or place in the sequence.

Next, the formula must describe the mathematical operations needed to produce the member of the sequence for any given N. There are only four fundamental mathematical operations: addition, subtraction multiplication and division, so that means the formula is limited to combinations of these four fundamental operations. So, even without any intuitive mathematical insight, we can find the correct answer with a finite number of trials. Try it on any sequence. Here it is applied to the problem at hand: 0, 108, 27648 …?

Since the first member of the sequence is zero, we start with a function of N that this equal to zero when N = 1. That function is N - 1.  Proof: if N = 1, N – 1 = 0. But when N =2, N – 1 = 1, So since the second member of the sequence is 108, not 1, the function is more complicated; but it must contain N – 1 as a factor because the first member is zero.

Next, factor the second member, 108, into its prime whole number factors: 108 = 1x2x2x3x3x3.
We see that for N = 2: f(N) = 11 x 22 x 33 = (N – 1)N-1x(N)Nx(N + 1)N +1 = 108

For N = 3, f(N) = 22 x 33 x 44 = 27648. Also, for N = 1, f(N) = 0 x 11x22=0. So the function (formula) f(N) = (N – 1)N-1x(N)Nx(N + 1)N +1 is the function that works for all three members of the given sequence 0, 108, 27648. And thus the next number is given by f(N) when N = 4:

So, f(4) = 33 x 44 x 55 = 256 x 3125 x 46656 = 21600000 the correct answer..

I could see the reasoning behind all of the incorrect answers; none of them were completely off the beam.  It is interesting that four respondents gave 86400000 as the answer, although JA noted that it did not fit the set up. The big problem that seemed to throw everyone was having zero as the first number in the sequence. Sequences with zero or negative integers as the first member of the sequence will have to have subtraction in the formula. This sequence has addition, subtraction and multiplication in the formula; of the four fundamental operations of mathematics, only division is missing.

1 comment:

  1. CBSE 8th Class Exemplar Maths Solutions
    CBSE 8th Class Maths Exemplar Problems with Solutions 2023 has been Divided into 3 parts, CBSE 8th Exemplar 2023 Subjects like Maths Pdf free Download was Designed by Expert Teachers from Latest Edition of CBSE 8th Maths Exemplar Books to get good marks in board exams. Here we have given CBSE 8th Maths Exemplar Problems with Solutions 2023. CBSE 8th Class Exemplar Maths Solutions CBSE Maths Exemplar Problems for 8 Class are Available in this web page. You will get the Download links of Exemplar Problems Students which will help to Prepare Final Examination Prepare of CBSE Board, CBSE 8 Class Students your Exemplar Problems with Solutions Online Provide our website.