Base | Representation |
---|---|
bin | 11110000001100001000000… |
… | …010010010110111001100001 |
3 | 122022112102122122001111200212 |
4 | 132001201000102112321201 |
5 | 114301413243043234213 |
6 | 1144500520100023505 |
7 | 36545655621200405 |
oct | 3601410022267141 |
9 | 568472578044625 |
10 | 132045553102433 |
11 | 3908a228340103 |
12 | 12987390691b95 |
13 | 588aac44b6029 |
14 | 2487078937105 |
15 | 103ec1b81b9a8 |
hex | 781840496e61 |
132045553102433 has 2 divisors, whose sum is σ = 132045553102434. Its totient is φ = 132045553102432.
The previous prime is 132045553102427. The next prime is 132045553102451. The reversal of 132045553102433 is 334201355540231.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 72524436112384 + 59521116990049 = 8516128^2 + 7714993^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-132045553102433 is a prime.
It is not a weakly prime, because it can be changed into another prime (132045553102453) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66022776551216 + 66022776551217.
It is an arithmetic number, because the mean of its divisors is an integer number (66022776551217).
Almost surely, 2132045553102433 is an apocalyptic number.
It is an amenable number.
132045553102433 is a deficient number, since it is larger than the sum of its proper divisors (1).
132045553102433 is an equidigital number, since it uses as much as digits as its factorization.
132045553102433 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 648000, while the sum is 41.
The spelling of 132045553102433 in words is "one hundred thirty-two trillion, forty-five billion, five hundred fifty-three million, one hundred two thousand, four hundred thirty-three".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.081 sec. • engine limits •