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The partial sums of the series 1 + 2 + 3 + 4 + 5 + 6 + ⋯ are 1, 3, 6, 10, 15, etc.The nth partial sum is given by a simple formula: = = (+). This equation was known ...
For any integer n, n ≡ 1 (mod 2) if and only if 3n + 1 ≡ 4 (mod 6). Equivalently, n − 1 / 3 ≡ 1 (mod 2) if and only if n ≡ 4 (mod 6). Conjecturally, this inverse relation forms a tree except for the 1–2–4 loop (the inverse of the 4–2–1 loop of the unaltered function f defined in the Statement of the problem section of ...
The idea becomes clearer by considering the general series 1 − 2x + 3x 2 − 4x 3 + 5x 4 − 6x 5 + &c. that arises while expanding the expression 1 ⁄ (1+x) 2, which this series is indeed equal to after we set x = 1.
The Basel problem is a problem in mathematical analysis with relevance to number theory, concerning an infinite sum of inverse squares. It was first posed by Pietro Mengoli in 1650 and solved by Leonhard Euler in 1734, [1] and read on 5 December 1735 in The Saint Petersburg Academy of Sciences. [2] Since the problem had withstood the attacks of ...
Perfect number. In number theory, a perfect number is a positive integer that is equal to the sum of its positive proper divisors, that is, divisors excluding the number itself. For instance, 6 has proper divisors 1, 2 and 3, and 1 + 2 + 3 = 6, so 6 is a perfect number. The next perfect number is 28, since 1 + 2 + 4 + 7 + 14 = 28.
This sum can also be found in the four outer numbers clockwise from the corners (3+8+14+9) and likewise the four counter-clockwise (the locations of four queens in the two solutions of the 4 queens puzzle [50]), the two sets of four symmetrical numbers (2+8+9+15 and 3+5+12+14), the sum of the middle two entries of the two outer columns and rows ...
Every prime number p divides a Fibonacci number that can be determined by the value of p modulo 5. If p is congruent to 1 or 4 modulo 5, then p divides F p−1, and if p is congruent to 2 or 3 modulo 5, then, p divides F p+1. The remaining case is that p = 5, and in this case p divides F p.
Answer: 7 × 1 + 6 × 10 + 5 × 9 + 4 × 12 + 3 × 3 + 2 × 4 + 1 × 1 = 178 mod 13 = 9 Remainder = 9 A recursive method can be derived using the fact that = and that =. This implies that a number is divisible by 13 iff removing the first digit and subtracting 3 times that digit from the new first digit yields a number divisible by 13.