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14132211241433 is a prime number
BaseRepresentation
bin1100110110100110100101…
…1011100101000111011001
31212001000201002200220101012
43031221221123211013121
53323020232034211213
650020125340503305
72656006240353665
oct315515133450731
955030632626335
1014132211241433
11455948990217a
121702b00b4ab35
137b687ac94a5c
1436c00749d8a5
15197927e4e4a8
hexcda696e51d9

14132211241433 has 2 divisors, whose sum is σ = 14132211241434. Its totient is φ = 14132211241432.

The previous prime is 14132211241429. The next prime is 14132211241451. The reversal of 14132211241433 is 33414211223141.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 10281052262464 + 3851158978969 = 3206408^2 + 1962437^2 .

It is a cyclic number.

It is not a de Polignac number, because 14132211241433 - 22 = 14132211241429 is a prime.

It is a Sophie Germain prime.

It is a Curzon number.

It is a self number, because there is not a number n which added to its sum of digits gives 14132211241433.

It is not a weakly prime, because it can be changed into another prime (14132211241453) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7066105620716 + 7066105620717.

It is an arithmetic number, because the mean of its divisors is an integer number (7066105620717).

Almost surely, 214132211241433 is an apocalyptic number.

It is an amenable number.

14132211241433 is a deficient number, since it is larger than the sum of its proper divisors (1).

14132211241433 is an equidigital number, since it uses as much as digits as its factorization.

14132211241433 is an evil number, because the sum of its binary digits is even.

The product of its digits is 13824, while the sum is 32.

Adding to 14132211241433 its reverse (33414211223141), we get a palindrome (47546422464574).

The spelling of 14132211241433 in words is "fourteen trillion, one hundred thirty-two billion, two hundred eleven million, two hundred forty-one thousand, four hundred thirty-three".