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130123211312131 is a prime number
BaseRepresentation
bin11101100101100010101011…
…110010011000110000000011
3122001201122202022211110100121
4131211202223302120300003
5114023414314043442011
61140425440020001111
736260043323450062
oct3545425362306003
9561648668743317
10130123211312131
1138508a42531714
1212716900230197
13577b748b6bc3b
14241bdd5bd61d9
151009c0b433e71
hex7658abc98c03

130123211312131 has 2 divisors, whose sum is σ = 130123211312132. Its totient is φ = 130123211312130.

The previous prime is 130123211312093. The next prime is 130123211312137. The reversal of 130123211312131 is 131213112321031.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 130123211312131 - 241 = 127924188056579 is a prime.

It is a super-2 number, since 2×1301232113121312 (a number of 29 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 130123211312096 and 130123211312105.

It is not a weakly prime, because it can be changed into another prime (130123211312137) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65061605656065 + 65061605656066.

It is an arithmetic number, because the mean of its divisors is an integer number (65061605656066).

Almost surely, 2130123211312131 is an apocalyptic number.

130123211312131 is a deficient number, since it is larger than the sum of its proper divisors (1).

130123211312131 is an equidigital number, since it uses as much as digits as its factorization.

130123211312131 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 648, while the sum is 25.

Adding to 130123211312131 its reverse (131213112321031), we get a palindrome (261336323633162).

The spelling of 130123211312131 in words is "one hundred thirty trillion, one hundred twenty-three billion, two hundred eleven million, three hundred twelve thousand, one hundred thirty-one".