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13254332540377 is a prime number
BaseRepresentation
bin1100000011100000001111…
…0010000110010111011001
31201221002202020001020111011
43000320003302012113121
53214124333112243002
644104542321333521
72535410500463245
oct300700362062731
951832666036434
1013254332540377
1142501491a7725
1215a09411972a1
13751b5551ab66
1433b727dd1425
1517eb97a118d7
hexc0e03c865d9

13254332540377 has 2 divisors, whose sum is σ = 13254332540378. Its totient is φ = 13254332540376.

The previous prime is 13254332540311. The next prime is 13254332540381. The reversal of 13254332540377 is 77304523345231.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 11556573054016 + 1697759486361 = 3399496^2 + 1302981^2 .

It is a cyclic number.

It is not a de Polignac number, because 13254332540377 - 223 = 13254324151769 is a prime.

It is a self number, because there is not a number n which added to its sum of digits gives 13254332540377.

It is not a weakly prime, because it can be changed into another prime (13254332540077) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6627166270188 + 6627166270189.

It is an arithmetic number, because the mean of its divisors is an integer number (6627166270189).

Almost surely, 213254332540377 is an apocalyptic number.

It is an amenable number.

13254332540377 is a deficient number, since it is larger than the sum of its proper divisors (1).

13254332540377 is an equidigital number, since it uses as much as digits as its factorization.

13254332540377 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 6350400, while the sum is 49.

The spelling of 13254332540377 in words is "thirteen trillion, two hundred fifty-four billion, three hundred thirty-two million, five hundred forty thousand, three hundred seventy-seven".