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d() is the number of positive divisors of n, including 1 and n itself; σ() is the sum of the positive divisors of n, including 1 and n itselfs() is the sum of the proper divisors of n, including 1 but not n itself; that is, s(n) = σ(n) − n
Subtracting 2 times the last digit from the rest gives a multiple of 3. (Works because 21 is divisible by 3) 405: 40 − 5 × 2 = 40 − 10 = 30 = 3 × 10. 4: The last two digits form a number that is divisible by 4. [2] [3] 40,832: 32 is divisible by 4. If the tens digit is even, the ones digit must be 0, 4, or 8.
2012 – The number 8 × 10 2012 − 1 is a prime number [10] 2013 – number of widely totally strongly normal compositions of 17; 2014 – 5 × 2 2014 - 1 is prime [11] 2015 – Lucas–Carmichael number [12] 2016 – triangular number, number of 5-cubes in a 9-cube, Erdős–Nicolas number, [13] 2 11-2 5; 2017 – Mertens function zero, sexy ...
The first: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23 (sequence A005408 in the OEIS). All integers are either even or odd. All integers are either even or odd. A square has even multiplicity for all prime factors (it is of the form a 2 for some a ).
The rule is that if the year is divisible by 100 and not divisible by 400, the leap year is skipped. The year 2000 was a leap year, for example, but the years 1700, 1800, and 1900 were not.
Thus, 1/54, in sexagesimal, is 1/60 + 6/60 2 + 40/60 3, also denoted 1:6:40 as Babylonian notational conventions did not specify the power of the starting digit. Conversely 1/4000 = 54/60 3, so division by 1:6:40 = 4000 can be accomplished by instead multiplying by 54 and shifting three sexagesimal places.
the k given prime numbers p i must be precisely the first k prime numbers (2, 3, 5, ...); if not, we could replace one of the given primes by a smaller prime, and thus obtain a smaller number than n with the same number of divisors (for instance 10 = 2 × 5 may be replaced with 6 = 2 × 3; both have four divisors);
0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9, 24, 8, 25, 43, 62, ... "subtract if possible, otherwise add" : a (0) = 0; for n > 0, a ( n ) = a ( n − 1) − n if that number is positive and not already in the sequence, otherwise a ( n ) = a ( n − 1) + n , whether or not that number is already in the sequence.