Base | Representation |
---|---|
bin | 1110010001111101000… |
… | …11100111100010010001 |
3 | 1201220112001002002002211 |
4 | 13020332203213202101 |
5 | 31014401000433423 |
6 | 1013225141112121 |
7 | 50310255310426 |
oct | 7107643474221 |
9 | 1656461062084 |
10 | 490675796113 |
11 | 17a104771078 |
12 | 7b11a3b8641 |
13 | 37369476c79 |
14 | 19a6ab37d4d |
15 | cb6c26610d |
hex | 723e8e7891 |
490675796113 has 2 divisors, whose sum is σ = 490675796114. Its totient is φ = 490675796112.
The previous prime is 490675796101. The next prime is 490675796123. The reversal of 490675796113 is 311697576094.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 310312501249 + 180363294864 = 557057^2 + 424692^2 .
It is a cyclic number.
It is not a de Polignac number, because 490675796113 - 221 = 490673698961 is a prime.
It is a super-2 number, since 2×4906757961132 (a number of 24 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (490675796123) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 245337898056 + 245337898057.
It is an arithmetic number, because the mean of its divisors is an integer number (245337898057).
Almost surely, 2490675796113 is an apocalyptic number.
It is an amenable number.
490675796113 is a deficient number, since it is larger than the sum of its proper divisors (1).
490675796113 is an equidigital number, since it uses as much as digits as its factorization.
490675796113 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 8573040, while the sum is 58.
The spelling of 490675796113 in words is "four hundred ninety billion, six hundred seventy-five million, seven hundred ninety-six thousand, one hundred thirteen".
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