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215440404044513 is a prime number
BaseRepresentation
bin110000111111000100100000…
…110000001101101011100001
31001020210220100011212221011212
4300333010200300031223201
5211214233421413411023
62042111542202115505
763244025023234364
oct6077044060155341
91036726304787155
10215440404044513
116271184262194a
12201b59634abb95
139329c347c872b
143b2b52a7a40db
1519d91752e5e78
hexc3f120c0dae1

215440404044513 has 2 divisors, whose sum is σ = 215440404044514. Its totient is φ = 215440404044512.

The previous prime is 215440404044473. The next prime is 215440404044609. The reversal of 215440404044513 is 315440404044512.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 178825173737809 + 36615230306704 = 13372553^2 + 6051052^2 .

It is a cyclic number.

It is not a de Polignac number, because 215440404044513 - 224 = 215440387267297 is a prime.

It is a super-3 number, since 3×2154404040445133 (a number of 44 digits) contains 333 as substring.

It is not a weakly prime, because it can be changed into another prime (215440404040513) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 107720202022256 + 107720202022257.

It is an arithmetic number, because the mean of its divisors is an integer number (107720202022257).

Almost surely, 2215440404044513 is an apocalyptic number.

It is an amenable number.

215440404044513 is a deficient number, since it is larger than the sum of its proper divisors (1).

215440404044513 is an equidigital number, since it uses as much as digits as its factorization.

215440404044513 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 614400, while the sum is 41.

Subtracting 215440404044513 from its reverse (315440404044512), we obtain a palindrome (99999999999999).

The spelling of 215440404044513 in words is "two hundred fifteen trillion, four hundred forty billion, four hundred four million, forty-four thousand, five hundred thirteen".