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131063000489 is a prime number
BaseRepresentation
bin111101000001111110…
…1101010110110101001
3110112022000011021011022
41322003331222312221
54121404112003424
6140113130542225
712316602060101
oct1720375526651
9415260137138
10131063000489
1150646713133
1221498897975
13c48919c5c5
1464b4736201
15362133b95e
hex1e83f6ada9

131063000489 has 2 divisors, whose sum is σ = 131063000490. Its totient is φ = 131063000488.

The previous prime is 131063000473. The next prime is 131063000491. The reversal of 131063000489 is 984000360131.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 130057603225 + 1005397264 = 360635^2 + 31708^2 .

It is a cyclic number.

It is not a de Polignac number, because 131063000489 - 24 = 131063000473 is a prime.

It is a super-2 number, since 2×1310630004892 (a number of 23 digits) contains 22 as substring.

Together with 131063000491, 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 (131063000459) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65531500244 + 65531500245.

It is an arithmetic number, because the mean of its divisors is an integer number (65531500245).

Almost surely, 2131063000489 is an apocalyptic number.

It is an amenable number.

131063000489 is a deficient number, since it is larger than the sum of its proper divisors (1).

131063000489 is an equidigital number, since it uses as much as digits as its factorization.

131063000489 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 15552, while the sum is 35.

The spelling of 131063000489 in words is "one hundred thirty-one billion, sixty-three million, four hundred eighty-nine", and thus it is an aban number.