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53345354025013 is a prime number
BaseRepresentation
bin11000010000100011011101…
…11000001000100000110101
320222212202200102221202102021
430020101232320020200311
523443002241112300023
6305242303424113141
714144034413436001
oct1410215670104065
9228782612852367
1053345354025013
1115aa7701700366
125b9682b1657b1
13239c5a3427b6b
14d25d00d09701
1562797d5d235d
hex30846ee08835

53345354025013 has 2 divisors, whose sum is σ = 53345354025014. Its totient is φ = 53345354025012.

The previous prime is 53345354024983. The next prime is 53345354025031. The reversal of 53345354025013 is 31052045354335.

Together with next prime (53345354025031) it forms an Ormiston pair, because they use the same digits, order apart.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 51477424999729 + 1867929025284 = 7174777^2 + 1366722^2 .

It is a cyclic number.

It is not a de Polignac number, because 53345354025013 - 241 = 51146330769461 is a prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (53345354025043) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 26672677012506 + 26672677012507.

It is an arithmetic number, because the mean of its divisors is an integer number (26672677012507).

Almost surely, 253345354025013 is an apocalyptic number.

It is an amenable number.

53345354025013 is a deficient number, since it is larger than the sum of its proper divisors (1).

53345354025013 is an equidigital number, since it uses as much as digits as its factorization.

53345354025013 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 1620000, while the sum is 43.

Adding to 53345354025013 its reverse (31052045354335), we get a palindrome (84397399379348).

The spelling of 53345354025013 in words is "fifty-three trillion, three hundred forty-five billion, three hundred fifty-four million, twenty-five thousand, thirteen".