Base | Representation |
---|---|
bin | 11110000010110010000010… |
… | …110000100011011101110111 |
3 | 122022211202020211022001021022 |
4 | 132002302002300203131313 |
5 | 114304324444310113112 |
6 | 1145004510312011355 |
7 | 36555156126553346 |
oct | 3602620260433567 |
9 | 568752224261238 |
10 | 132132567660407 |
11 | 3911311a834436 |
12 | 129a0215690b5b |
13 | 5896080950162 |
14 | 248b37113bb5d |
15 | 1042110a2cd72 |
hex | 782c82c23777 |
132132567660407 has 2 divisors, whose sum is σ = 132132567660408. Its totient is φ = 132132567660406.
The previous prime is 132132567660361. The next prime is 132132567660451. The reversal of 132132567660407 is 704066765231231.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 132132567660407 - 218 = 132132567398263 is a prime.
It is a super-4 number, since 4×1321325676604074 (a number of 58 digits) contains 4444 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (132132567669407) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66066283830203 + 66066283830204.
It is an arithmetic number, because the mean of its divisors is an integer number (66066283830204).
Almost surely, 2132132567660407 is an apocalyptic number.
132132567660407 is a deficient number, since it is larger than the sum of its proper divisors (1).
132132567660407 is an equidigital number, since it uses as much as digits as its factorization.
132132567660407 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 7620480, while the sum is 53.
The spelling of 132132567660407 in words is "one hundred thirty-two trillion, one hundred thirty-two billion, five hundred sixty-seven million, six hundred sixty thousand, four hundred seven".
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