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131011013212153 is a prime number
BaseRepresentation
bin11101110010011101100000…
…111001110111011111111001
3122011212112022202201212011021
4131302131200321313133321
5114132441023340242103
61142345343435502441
736411143016211654
oct3562354071673771
9564775282655137
10131011013212153
1138820505565448
121283a98b2b7a21
135814393422cb9
14244cd7715489b
151022d6cb0dbbd
hex772760e777f9

131011013212153 has 2 divisors, whose sum is σ = 131011013212154. Its totient is φ = 131011013212152.

The previous prime is 131011013212081. The next prime is 131011013212259. The reversal of 131011013212153 is 351212310110131.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 123130714359184 + 7880298852969 = 11096428^2 + 2807187^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-131011013212153 is a prime.

It is not a weakly prime, because it can be changed into another prime (131011013282153) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65505506606076 + 65505506606077.

It is an arithmetic number, because the mean of its divisors is an integer number (65505506606077).

Almost surely, 2131011013212153 is an apocalyptic number.

It is an amenable number.

131011013212153 is a deficient number, since it is larger than the sum of its proper divisors (1).

131011013212153 is an equidigital number, since it uses as much as digits as its factorization.

131011013212153 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 540, while the sum is 25.

Adding to 131011013212153 its reverse (351212310110131), we get a palindrome (482223323322284).

The spelling of 131011013212153 in words is "one hundred thirty-one trillion, eleven billion, thirteen million, two hundred twelve thousand, one hundred fifty-three".