Base | Representation |
---|---|
bin | 10000011111100110010… |
… | …010100101101010101111 |
3 | 11000100121101101101001111 |
4 | 100133212102211222233 |
5 | 122032241122001112 |
6 | 2224410053252451 |
7 | 144613462206325 |
oct | 20374622455257 |
9 | 4010541341044 |
10 | 1133440031407 |
11 | 3a7764392592 |
12 | 163802934127 |
13 | 82b62a710c8 |
14 | 3cc0473dd15 |
15 | 1e73b5dd5a7 |
hex | 107e64a5aaf |
1133440031407 has 2 divisors, whose sum is σ = 1133440031408. Its totient is φ = 1133440031406.
The previous prime is 1133440031393. The next prime is 1133440031413. The reversal of 1133440031407 is 7041300443311.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1133440031407 is a prime.
It is a super-2 number, since 2×11334400314072 (a number of 25 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (1133440031417) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 566720015703 + 566720015704.
It is an arithmetic number, because the mean of its divisors is an integer number (566720015704).
Almost surely, 21133440031407 is an apocalyptic number.
1133440031407 is a deficient number, since it is larger than the sum of its proper divisors (1).
1133440031407 is an equidigital number, since it uses as much as digits as its factorization.
1133440031407 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 12096, while the sum is 31.
Adding to 1133440031407 its reverse (7041300443311), we get a palindrome (8174740474718).
The spelling of 1133440031407 in words is "one trillion, one hundred thirty-three billion, four hundred forty million, thirty-one thousand, four hundred seven".
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