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131404102109149 is a prime number
BaseRepresentation
bin11101111000001011100110…
…110101000000011111011101
3122020021002221210211211010011
4131320023212311000133131
5114210411100114443044
61143250113335043221
736451430101543432
oct3570134665003735
9566232853754104
10131404102109149
1138962191a40871
12128a2bb3b10511
135842479cb227c
142463dc74dbb89
15102d1c78ed134
hex7782e6d407dd

131404102109149 has 2 divisors, whose sum is σ = 131404102109150. Its totient is φ = 131404102109148.

The previous prime is 131404102109117. The next prime is 131404102109153. The reversal of 131404102109149 is 941901201404131.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 130995052324249 + 409049784900 = 11445307^2 + 639570^2 .

It is an emirp because it is prime and its reverse (941901201404131) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 131404102109149 - 25 = 131404102109117 is a prime.

It is a super-2 number, since 2×1314041021091492 (a number of 29 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (131404102409149) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65702051054574 + 65702051054575.

It is an arithmetic number, because the mean of its divisors is an integer number (65702051054575).

Almost surely, 2131404102109149 is an apocalyptic number.

It is an amenable number.

131404102109149 is a deficient number, since it is larger than the sum of its proper divisors (1).

131404102109149 is an equidigital number, since it uses as much as digits as its factorization.

131404102109149 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 31104, while the sum is 40.

The spelling of 131404102109149 in words is "one hundred thirty-one trillion, four hundred four billion, one hundred two million, one hundred nine thousand, one hundred forty-nine".