Base | Representation |
---|---|
bin | 11110000010110000000011… |
… | …101110011111101110001111 |
3 | 122022211112202021222112022012 |
4 | 132002300003232133232033 |
5 | 114304311113210224401 |
6 | 1145003515020011435 |
7 | 36555052250351114 |
oct | 3602600356375617 |
9 | 568745667875265 |
10 | 132130436414351 |
11 | 391122267a4761 |
12 | 1299b91b990b7b |
13 | 5895b0124210b |
14 | 248b20c07a90b |
15 | 10420388928bb |
hex | 782c03b9fb8f |
132130436414351 has 2 divisors, whose sum is σ = 132130436414352. Its totient is φ = 132130436414350.
The previous prime is 132130436414293. The next prime is 132130436414363. The reversal of 132130436414351 is 153414634031231.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 132130436414351 - 222 = 132130432220047 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 132130436414299 and 132130436414308.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (132130436414371) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66065218207175 + 66065218207176.
It is an arithmetic number, because the mean of its divisors is an integer number (66065218207176).
Almost surely, 2132130436414351 is an apocalyptic number.
132130436414351 is a deficient number, since it is larger than the sum of its proper divisors (1).
132130436414351 is an equidigital number, since it uses as much as digits as its factorization.
132130436414351 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 311040, while the sum is 41.
The spelling of 132130436414351 in words is "one hundred thirty-two trillion, one hundred thirty billion, four hundred thirty-six million, four hundred fourteen thousand, three hundred fifty-one".
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