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361315560131 is a prime number
BaseRepresentation
bin1010100001000000001…
…01011110101011000011
31021112121200222012201122
411100200011132223003
521404433240411011
6433552554222455
735050510405331
oct5204005365303
91245550865648
10361315560131
1112a26223a139
125a037a24a2b
13280c1cb1336
14136b8606b51
1595ea6773db
hex542015eac3

361315560131 has 2 divisors, whose sum is σ = 361315560132. Its totient is φ = 361315560130.

The previous prime is 361315560043. The next prime is 361315560139. The reversal of 361315560131 is 131065513163.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 361315560131 - 214 = 361315543747 is a prime.

It is a super-3 number, since 3×3613155601313 (a number of 36 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a junction number, because it is equal to n+sod(n) for n = 361315560091 and 361315560100.

It is not a weakly prime, because it can be changed into another prime (361315560139) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 180657780065 + 180657780066.

It is an arithmetic number, because the mean of its divisors is an integer number (180657780066).

Almost surely, 2361315560131 is an apocalyptic number.

361315560131 is a deficient number, since it is larger than the sum of its proper divisors (1).

361315560131 is an equidigital number, since it uses as much as digits as its factorization.

361315560131 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 24300, while the sum is 35.

The spelling of 361315560131 in words is "three hundred sixty-one billion, three hundred fifteen million, five hundred sixty thousand, one hundred thirty-one".