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13405504222541 is a prime number
BaseRepresentation
bin1100001100010011011001…
…0100010011010101001101
31202110112221122020101010102
43003010312110103111031
53224113433040110131
644302223103220445
72552341614541424
oct303046624232515
952415848211112
1013405504222541
1142a927371aa42
1216060b02a7725
13763197758372
14344b892202bb
15183a943d61cb
hexc313651354d

13405504222541 has 2 divisors, whose sum is σ = 13405504222542. Its totient is φ = 13405504222540.

The previous prime is 13405504222489. The next prime is 13405504222619. The reversal of 13405504222541 is 14522240550431.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 13130687129641 + 274817092900 = 3623629^2 + 524230^2 .

It is a cyclic number.

It is not a de Polignac number, because 13405504222541 - 210 = 13405504221517 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 13405504222495 and 13405504222504.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (13405504222741) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6702752111270 + 6702752111271.

It is an arithmetic number, because the mean of its divisors is an integer number (6702752111271).

Almost surely, 213405504222541 is an apocalyptic number.

It is an amenable number.

13405504222541 is a deficient number, since it is larger than the sum of its proper divisors (1).

13405504222541 is an equidigital number, since it uses as much as digits as its factorization.

13405504222541 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 38.

Adding to 13405504222541 its reverse (14522240550431), we get a palindrome (27927744772972).

The spelling of 13405504222541 in words is "thirteen trillion, four hundred five billion, five hundred four million, two hundred twenty-two thousand, five hundred forty-one".