Base | Representation |
---|---|
bin | 11110100010110011011001… |
… | …111110100011000111000011 |
3 | 122121122010002111221110202011 |
4 | 132202303121332203013003 |
5 | 120101403043333433212 |
6 | 1153411430403233351 |
7 | 40203146211551062 |
oct | 3642633176430703 |
9 | 577563074843664 |
10 | 134333054202307 |
11 | 39891368050564 |
12 | 13096790078857 |
13 | 59c5725255bbc |
14 | 2525a7c5601d9 |
15 | 107e49ea3a4a7 |
hex | 7a2cd9fa31c3 |
134333054202307 has 2 divisors, whose sum is σ = 134333054202308. Its totient is φ = 134333054202306.
The previous prime is 134333054202301. The next prime is 134333054202343. The reversal of 134333054202307 is 703202450333431.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 134333054202307 - 223 = 134333045813699 is a prime.
It is a self number, because there is not a number n which added to its sum of digits gives 134333054202307.
It is not a weakly prime, because it can be changed into another prime (134333054202301) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 67166527101153 + 67166527101154.
It is an arithmetic number, because the mean of its divisors is an integer number (67166527101154).
Almost surely, 2134333054202307 is an apocalyptic number.
134333054202307 is a deficient number, since it is larger than the sum of its proper divisors (1).
134333054202307 is an equidigital number, since it uses as much as digits as its factorization.
134333054202307 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 544320, while the sum is 40.
The spelling of 134333054202307 in words is "one hundred thirty-four trillion, three hundred thirty-three billion, fifty-four million, two hundred two thousand, three hundred seven".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.072 sec. • engine limits •