Base | Representation |
---|---|
bin | 1110010010111011010… |
… | …10111111101111111001 |
3 | 1201221212102100211201202 |
4 | 13021131222333233321 |
5 | 31021433120331433 |
6 | 1013353030023545 |
7 | 50326224420542 |
oct | 7113552775771 |
9 | 1657772324652 |
10 | 491197823993 |
11 | 17a352403378 |
12 | 7b2451a7bb5 |
13 | 374206724c6 |
14 | 19aba1c54c9 |
15 | cb9ced0ee8 |
hex | 725dabfbf9 |
491197823993 has 2 divisors, whose sum is σ = 491197823994. Its totient is φ = 491197823992.
The previous prime is 491197823989. The next prime is 491197824043. The reversal of 491197823993 is 399328791194.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 486440687209 + 4757136784 = 697453^2 + 68972^2 .
It is a cyclic number.
It is not a de Polignac number, because 491197823993 - 22 = 491197823989 is a prime.
It is not a weakly prime, because it can be changed into another prime (491197823953) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 245598911996 + 245598911997.
It is an arithmetic number, because the mean of its divisors is an integer number (245598911997).
Almost surely, 2491197823993 is an apocalyptic number.
It is an amenable number.
491197823993 is a deficient number, since it is larger than the sum of its proper divisors (1).
491197823993 is an equidigital number, since it uses as much as digits as its factorization.
491197823993 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 26453952, while the sum is 65.
The spelling of 491197823993 in words is "four hundred ninety-one billion, one hundred ninety-seven million, eight hundred twenty-three thousand, nine hundred ninety-three".
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