Base | Representation |
---|---|
bin | 11110011110000111110111… |
… | …101100000000110010010111 |
3 | 122120111100220222110212020112 |
4 | 132132013313230000302113 |
5 | 120031120402102044112 |
6 | 1153004000122553235 |
7 | 40141002111106511 |
oct | 3636076754006227 |
9 | 576440828425215 |
10 | 134011430112407 |
11 | 39777a2369a491 |
12 | 1304439096881b |
13 | 59a12ba430a33 |
14 | 251428b570bb1 |
15 | 1075e28c53a22 |
hex | 79e1f7b00c97 |
134011430112407 has 2 divisors, whose sum is σ = 134011430112408. Its totient is φ = 134011430112406.
The previous prime is 134011430112401. The next prime is 134011430112421. The reversal of 134011430112407 is 704211034110431.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 134011430112407 - 214 = 134011430096023 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (134011430112401) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 67005715056203 + 67005715056204.
It is an arithmetic number, because the mean of its divisors is an integer number (67005715056204).
Almost surely, 2134011430112407 is an apocalyptic number.
134011430112407 is a deficient number, since it is larger than the sum of its proper divisors (1).
134011430112407 is an equidigital number, since it uses as much as digits as its factorization.
134011430112407 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 8064, while the sum is 32.
Adding to 134011430112407 its reverse (704211034110431), we get a palindrome (838222464222838).
The spelling of 134011430112407 in words is "one hundred thirty-four trillion, eleven billion, four hundred thirty million, one hundred twelve thousand, four hundred seven".
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