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1332022420433 is a prime number
BaseRepresentation
bin10011011000100010101…
…110010010001111010001
311201100011221011221212202
4103120202232102033101
5133310440214423213
62455531143132545
7165143520521615
oct23304256221721
94640157157782
101332022420433
114739a8a26671
121961a3744155
13987bc486903
1448682447345
15249b038eb58
hex13622b923d1

1332022420433 has 2 divisors, whose sum is σ = 1332022420434. Its totient is φ = 1332022420432.

The previous prime is 1332022420321. The next prime is 1332022420441. The reversal of 1332022420433 is 3340242202331.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 1312200033169 + 19822387264 = 1145513^2 + 140792^2 .

It is a cyclic number.

It is not a de Polignac number, because 1332022420433 - 212 = 1332022416337 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 1332022420396 and 1332022420405.

It is not a weakly prime, because it can be changed into another prime (1332022420453) 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, 666011210216 + 666011210217.

It is an arithmetic number, because the mean of its divisors is an integer number (666011210217).

Almost surely, 21332022420433 is an apocalyptic number.

It is an amenable number.

1332022420433 is a deficient number, since it is larger than the sum of its proper divisors (1).

1332022420433 is an equidigital number, since it uses as much as digits as its factorization.

1332022420433 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 20736, while the sum is 29.

Adding to 1332022420433 its reverse (3340242202331), we get a palindrome (4672264622764).

The spelling of 1332022420433 in words is "one trillion, three hundred thirty-two billion, twenty-two million, four hundred twenty thousand, four hundred thirty-three".