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11520504125483 is a prime number
BaseRepresentation
bin1010011110100101001110…
…0011100101100000101011
31111210100102201222212221212
42213221103203211200223
53002222443024003413
640300240241223335
72266220555225333
oct247512343454053
944710381885855
1011520504125483
113741900270361
121360901631b4b
136574c0517552
142bb8499cc0c3
1514ea1c438ca8
hexa7a538e582b

11520504125483 has 2 divisors, whose sum is σ = 11520504125484. Its totient is φ = 11520504125482.

The previous prime is 11520504125461. The next prime is 11520504125513. The reversal of 11520504125483 is 38452140502511.

11520504125483 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a weak prime.

It is a cyclic number.

It is not a de Polignac number, because 11520504125483 - 26 = 11520504125419 is a prime.

It is a self number, because there is not a number n which added to its sum of digits gives 11520504125483.

It is not a weakly prime, because it can be changed into another prime (11520504124483) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5760252062741 + 5760252062742.

It is an arithmetic number, because the mean of its divisors is an integer number (5760252062742).

Almost surely, 211520504125483 is an apocalyptic number.

11520504125483 is a deficient number, since it is larger than the sum of its proper divisors (1).

11520504125483 is an equidigital number, since it uses as much as digits as its factorization.

11520504125483 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 192000, while the sum is 41.

Adding to 11520504125483 its reverse (38452140502511), we get a palindrome (49972644627994).

The spelling of 11520504125483 in words is "eleven trillion, five hundred twenty billion, five hundred four million, one hundred twenty-five thousand, four hundred eighty-three".