Base | Representation |
---|---|
bin | 11101000111111100110… |
… | …110111001111101011011 |
3 | 21002022222011001002002221 |
4 | 131013330312321331123 |
5 | 230242332410404001 |
6 | 4131233103404511 |
7 | 264411415453441 |
oct | 35077466717533 |
9 | 7068864032087 |
10 | 2001402044251 |
11 | 701876306445 |
12 | 283a755b3737 |
13 | 116967b7a45c |
14 | 6cc22b74b91 |
15 | 370db1281a1 |
hex | 1d1fcdb9f5b |
2001402044251 has 2 divisors, whose sum is σ = 2001402044252. Its totient is φ = 2001402044250.
The previous prime is 2001402044249. The next prime is 2001402044261. The reversal of 2001402044251 is 1524402041002.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 2001402044251 - 21 = 2001402044249 is a prime.
It is a super-2 number, since 2×20014020442512 (a number of 25 digits) contains 22 as substring.
Together with 2001402044249, it forms a pair of twin primes.
It is not a weakly prime, because it can be changed into another prime (2001402044231) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1000701022125 + 1000701022126.
It is an arithmetic number, because the mean of its divisors is an integer number (1000701022126).
Almost surely, 22001402044251 is an apocalyptic number.
2001402044251 is a deficient number, since it is larger than the sum of its proper divisors (1).
2001402044251 is an equidigital number, since it uses as much as digits as its factorization.
2001402044251 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 2560, while the sum is 25.
Adding to 2001402044251 its reverse (1524402041002), we get a palindrome (3525804085253).
The spelling of 2001402044251 in words is "two trillion, one billion, four hundred two million, forty-four thousand, two hundred fifty-one".
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