Current Research in Traffic Transportation Engineering(CRTTE)
ISSN: 3069-5538 | DOI: 10.33140/CRTTE
Mathematical Note - (2026) Volume 4, Issue 1
FLT. Formulas for Numbers A, B, C
Received Date: Feb 09, 2026 / Accepted Date: Feb 26, 2026 / Published Date: Mar 10, 2026
Copyright: ©2026 Victor Sorokine. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.
Citation: Sorokine, V. (2026). FLT. Formulas for Numbers A, B, C. Curr Res Traffic Transport Eng, 4(1), 01.
Abstract
Throughout the text, coprime natural numbers A, B, C with the last positive digits a, b, c are written in the number system with a prime base n > 2,
Theorems
Throughout the text, coprime natural numbers A, B, C with the last positive digits a, b, c are written in the number system with a prime base n > 2
In the equality An + Bn - Cn = 0, starting from k=1, the k-digit endings of the numbers A, B, C are equal to the numbers an^(k-1), bn^(k-1), cn^(k-1) mod nk , where each k generates the next value k, i.e. k+1. Without end.
Designations
A1 , A2 , … Ak – k-digit endings of the number A in the number system with base n.
k-end of the number – k-digit ending of a number Ð = an^(k-1).
The factors A + B and R in the expansion of the sum of powers An + Bn = (A+B)R are relatively prime, because the polynomial R can be represented as (A + B)D ± nAB. Therefore, factors in numbers
3. Cn = An +Bn = (A+B)R, An = Cn - Bn = (C - B)P, Bn = Cn - An = (C - A)Q are powers:
4. A+B = dn , R = rn . C-B = en , P = pn , C-A = f n , Q = qn . And, according to the little theorem, P1 = Q1 = R1 = 1.
5. According to Newton's binomial (A°n+a)n , the penultimate digit of a two-digit number to the power of n (that is, the number an) does not depend on the penultimate digit of the base (that is, the last digit of A°)
Consequence
The k-th digit from the end in the number An^k does not depend on the k-th digit of the number
A. Numbers outside the (k+1)-ending of the form an^k are not used in the proof.
***
Proof of the Theorem
For k=1 we have three identities: A1 = an^(k-1) = a, B1 = bn^(k-1) = b, C1 = cn^(k-1) = c, (mod n) Let's write the numbers A, B, C in the form:
8. A = A°n + an , where A° = (A - an )/n; B = B°n + bn , where B° = (B - bn )/n; C = C°n + cn , where C° = (C - cn )/n, then we will substitute these values into equalities 2, in which the power numbers P, Q, R end in 01, and expand Newton's binomials in which the penultimate terms end in 2 zeros. Therefore, we obtain a system of three equalities mod n2 :
9. A + B = cn^1; C - B = an^1; C - A = bn^1, solving which for A, B, C, we find:
10. A2 = an^1, B2 = bn^1, C2 = cn^1, (mod n2 ) with value k=2.
And then we return to step 6, but with k=2, and carry out similar operations to obtain k-endings with the value k=3. And so on - WITHOUT END!
==========================
https://docs.google.com/document/d/1IKEN4aunpT_eDA46Lwo7tnEOQNapTUxejBf6a4vm68/edit?tab=t.0