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13141455010403 is a prime number
BaseRepresentation
bin1011111100111011101111…
…0000011111101001100011
31201112022101121001121001212
42333032323300133221203
53210302144440313103
643541033500220335
72524303335346466
oct277167360375143
951468347047055
1013141455010403
114207294835775
121582a9a83a6ab
13744306b199c5
1433609a9a52dd
1517bc8ceb77d8
hexbf3bbc1fa63

13141455010403 has 2 divisors, whose sum is σ = 13141455010404. Its totient is φ = 13141455010402.

The previous prime is 13141455010399. The next prime is 13141455010433. The reversal of 13141455010403 is 30401055414131.

It is a weak prime.

It is an emirp because it is prime and its reverse (30401055414131) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 13141455010403 - 22 = 13141455010399 is a prime.

It is a super-2 number, since 2×131414550104032 (a number of 27 digits) contains 22 as substring.

It is a self number, because there is not a number n which added to its sum of digits gives 13141455010403.

It is not a weakly prime, because it can be changed into another prime (13141455010433) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6570727505201 + 6570727505202.

It is an arithmetic number, because the mean of its divisors is an integer number (6570727505202).

Almost surely, 213141455010403 is an apocalyptic number.

13141455010403 is a deficient number, since it is larger than the sum of its proper divisors (1).

13141455010403 is an equidigital number, since it uses as much as digits as its factorization.

13141455010403 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 14400, while the sum is 32.

The spelling of 13141455010403 in words is "thirteen trillion, one hundred forty-one billion, four hundred fifty-five million, ten thousand, four hundred three".