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35325151131997 is a prime number
BaseRepresentation
bin10000000100000110001110…
…00101011011100101011101
311122002001101112000210120201
420002003013011123211131
514112231402142210442
6203044052252543501
710304105114231326
oct1002030705334535
9148061345023521
1035325151131997
111028a348986611
123b66301781b91
1316931bbc7282c
148a1a6431d44d
15413d4cc817b7
hex2020c715b95d

35325151131997 has 2 divisors, whose sum is σ = 35325151131998. Its totient is φ = 35325151131996.

The previous prime is 35325151131983. The next prime is 35325151132009. The reversal of 35325151131997 is 79913115152353.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 34075016939161 + 1250134192836 = 5837381^2 + 1118094^2 .

It is a cyclic number.

It is not a de Polignac number, because 35325151131997 - 215 = 35325151099229 is a prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (35325151131397) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 17662575565998 + 17662575565999.

It is an arithmetic number, because the mean of its divisors is an integer number (17662575565999).

Almost surely, 235325151131997 is an apocalyptic number.

It is an amenable number.

35325151131997 is a deficient number, since it is larger than the sum of its proper divisors (1).

35325151131997 is an equidigital number, since it uses as much as digits as its factorization.

35325151131997 is an evil number, because the sum of its binary digits is even.

The product of its digits is 3827250, while the sum is 55.

The spelling of 35325151131997 in words is "thirty-five trillion, three hundred twenty-five billion, one hundred fifty-one million, one hundred thirty-one thousand, nine hundred ninety-seven".