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131960504057 is a prime number
BaseRepresentation
bin111101011100101110…
…1010111111011111001
3110121121112222011121012
41322321131113323321
54130223342112212
6140342151323305
712351050530424
oct1727135277371
9417545864535
10131960504057
1150a6729a918
12216a9380535
13c5a01008c5
14655ba029bb
153675023922
hex1eb9757ef9

131960504057 has 2 divisors, whose sum is σ = 131960504058. Its totient is φ = 131960504056.

The previous prime is 131960504041. The next prime is 131960504059. The reversal of 131960504057 is 750405069131.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 125493771001 + 6466733056 = 354251^2 + 80416^2 .

It is a cyclic number.

It is not a de Polignac number, because 131960504057 - 24 = 131960504041 is a prime.

Together with 131960504059, it forms a pair of twin primes.

It is a Chen prime.

It is not a weakly prime, because it can be changed into another prime (131960504059) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65980252028 + 65980252029.

It is an arithmetic number, because the mean of its divisors is an integer number (65980252029).

Almost surely, 2131960504057 is an apocalyptic number.

It is an amenable number.

131960504057 is a deficient number, since it is larger than the sum of its proper divisors (1).

131960504057 is an equidigital number, since it uses as much as digits as its factorization.

131960504057 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 113400, while the sum is 41.

The spelling of 131960504057 in words is "one hundred thirty-one billion, nine hundred sixty million, five hundred four thousand, fifty-seven".