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2022133101433 is a prime number
BaseRepresentation
bin11101011011010000100…
…001100111001101111001
321011022110221011220021121
4131123100201213031321
5231112312033221213
64144542150105241
7266044225522414
oct35332041471571
97138427156247
102022133101433
1170a644467555
12287aa034b221
131188bcb496b3
146dc2c19ab7b
153790114d88d
hex1d6d0867379

2022133101433 has 2 divisors, whose sum is σ = 2022133101434. Its totient is φ = 2022133101432.

The previous prime is 2022133101419. The next prime is 2022133101437. The reversal of 2022133101433 is 3341013312202.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 1273169209104 + 748963892329 = 1128348^2 + 865427^2 .

It is a cyclic number.

It is not a de Polignac number, because 2022133101433 - 29 = 2022133100921 is a prime.

It is a super-2 number, since 2×20221331014332 (a number of 25 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 2022133101398 and 2022133101407.

It is not a weakly prime, because it can be changed into another prime (2022133101437) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1011066550716 + 1011066550717.

It is an arithmetic number, because the mean of its divisors is an integer number (1011066550717).

Almost surely, 22022133101433 is an apocalyptic number.

It is an amenable number.

2022133101433 is a deficient number, since it is larger than the sum of its proper divisors (1).

2022133101433 is an equidigital number, since it uses as much as digits as its factorization.

2022133101433 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 2592, while the sum is 25.

Adding to 2022133101433 its reverse (3341013312202), we get a palindrome (5363146413635).

The spelling of 2022133101433 in words is "two trillion, twenty-two billion, one hundred thirty-three million, one hundred one thousand, four hundred thirty-three".