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315551310356081 is a prime number
BaseRepresentation
bin100011110111111100000001…
…1110000110001111001110001
31112101021102002100222211120202
41013233320003300301321301
5312324443040422343311
63035042134252044545
7123315554430454112
oct10757700360617161
91471242070884522
10315551310356081
11915a9617321458
122b483b5b463755
131070c48cb349b7
1457cca9ccba209
152673331534e3b
hex11efe03c31e71

315551310356081 has 2 divisors, whose sum is σ = 315551310356082. Its totient is φ = 315551310356080.

The previous prime is 315551310355979. The next prime is 315551310356123. The reversal of 315551310356081 is 180653013155513.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 315516744109681 + 34566246400 = 17762791^2 + 185920^2 .

It is a cyclic number.

It is not a de Polignac number, because 315551310356081 - 218 = 315551310093937 is a prime.

It is a super-2 number, since 2×3155513103560812 (a number of 30 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (315551300356081) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 157775655178040 + 157775655178041.

It is an arithmetic number, because the mean of its divisors is an integer number (157775655178041).

Almost surely, 2315551310356081 is an apocalyptic number.

It is an amenable number.

315551310356081 is a deficient number, since it is larger than the sum of its proper divisors (1).

315551310356081 is an equidigital number, since it uses as much as digits as its factorization.

315551310356081 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 810000, while the sum is 47.

The spelling of 315551310356081 in words is "three hundred fifteen trillion, five hundred fifty-one billion, three hundred ten million, three hundred fifty-six thousand, eighty-one".