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3545145055433 is a prime number
BaseRepresentation
bin110011100101101011000…
…111001001100011001001
3110112220122020212102101202
4303211223013021203021
5431040424113233213
611312341145553545
7514061666526032
oct63455307114311
913486566772352
103545145055433
11114753aa64543
124930a5aa58b5
131c93c8949217
14c382bb47a89
156233d8b9b58
hex3396b1c98c9

3545145055433 has 2 divisors, whose sum is σ = 3545145055434. Its totient is φ = 3545145055432.

The previous prime is 3545145055379. The next prime is 3545145055457. The reversal of 3545145055433 is 3345505415453.

3545145055433 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 2425628698249 + 1119516357184 = 1557443^2 + 1058072^2 .

It is a cyclic number.

It is not a de Polignac number, because 3545145055433 - 214 = 3545145039049 is a prime.

It is a super-2 number, since 2×35451450554332 (a number of 26 digits) contains 22 as substring.

It is a Sophie Germain prime.

It is a Curzon number.

It is not a weakly prime, because it can be changed into another prime (3545141055433) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1772572527716 + 1772572527717.

It is an arithmetic number, because the mean of its divisors is an integer number (1772572527717).

Almost surely, 23545145055433 is an apocalyptic number.

It is an amenable number.

3545145055433 is a deficient number, since it is larger than the sum of its proper divisors (1).

3545145055433 is an equidigital number, since it uses as much as digits as its factorization.

3545145055433 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 5400000, while the sum is 47.

The spelling of 3545145055433 in words is "three trillion, five hundred forty-five billion, one hundred forty-five million, fifty-five thousand, four hundred thirty-three".