Base | Representation |
---|---|
bin | 10000011101100010100… |
… | …111111000100110001011 |
3 | 11000010220102202001001221 |
4 | 100131202213320212023 |
5 | 122013230120141301 |
6 | 2223402542422511 |
7 | 144504650013136 |
oct | 20354247704613 |
9 | 4003812661057 |
10 | 1131231021451 |
11 | 3a6830469723 |
12 | 1632a6b91a37 |
13 | 828a020671b |
14 | 3ca751d921d |
15 | 1e65c6e1da1 |
hex | 107629f898b |
1131231021451 has 2 divisors, whose sum is σ = 1131231021452. Its totient is φ = 1131231021450.
The previous prime is 1131231021413. The next prime is 1131231021497. The reversal of 1131231021451 is 1541201321311.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 1131231021451 - 211 = 1131231019403 is a prime.
It is a super-3 number, since 3×11312310214513 (a number of 37 digits) contains 333 as substring.
It is not a weakly prime, because it can be changed into another prime (1131231021551) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 565615510725 + 565615510726.
It is an arithmetic number, because the mean of its divisors is an integer number (565615510726).
Almost surely, 21131231021451 is an apocalyptic number.
1131231021451 is a deficient number, since it is larger than the sum of its proper divisors (1).
1131231021451 is an equidigital number, since it uses as much as digits as its factorization.
1131231021451 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 720, while the sum is 25.
Adding to 1131231021451 its reverse (1541201321311), we get a palindrome (2672432342762).
The spelling of 1131231021451 in words is "one trillion, one hundred thirty-one billion, two hundred thirty-one million, twenty-one thousand, four hundred fifty-one".
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