Base | Representation |
---|---|
bin | 10001011110101110111… |
… | …101000000010101110111 |
3 | 11020211120212020122021212 |
4 | 101132232331000111313 |
5 | 124140110130301401 |
6 | 2315500555232035 |
7 | 152533424623436 |
oct | 21365675002567 |
9 | 4224525218255 |
10 | 1201231103351 |
11 | 423491707284 |
12 | 174981a4861b |
13 | 8937767c426 |
14 | 421d5c3311d |
15 | 213a7d1bdbb |
hex | 117aef40577 |
1201231103351 has 2 divisors, whose sum is σ = 1201231103352. Its totient is φ = 1201231103350.
The previous prime is 1201231103269. The next prime is 1201231103353. The reversal of 1201231103351 is 1533011321021.
It is a strong prime.
It is an emirp because it is prime and its reverse (1533011321021) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 1201231103351 - 210 = 1201231102327 is a prime.
Together with 1201231103353, it forms a pair of twin primes.
It is a Chen prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (1201231103353) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 600615551675 + 600615551676.
It is an arithmetic number, because the mean of its divisors is an integer number (600615551676).
Almost surely, 21201231103351 is an apocalyptic number.
1201231103351 is a deficient number, since it is larger than the sum of its proper divisors (1).
1201231103351 is an equidigital number, since it uses as much as digits as its factorization.
1201231103351 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 540, while the sum is 23.
Adding to 1201231103351 its reverse (1533011321021), we get a palindrome (2734242424372).
The spelling of 1201231103351 in words is "one trillion, two hundred one billion, two hundred thirty-one million, one hundred three thousand, three hundred fifty-one".
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