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11822605433 is a prime number
BaseRepresentation
bin10110000001010111…
…01010010001111001
31010111221022120202022
423000223222101321
5143203041333213
65233051244225
7565655326262
oct130053522171
933457276668
1011822605433
1150176094a1
12235b449675
131165493822
14802241969
15492dce308
hex2c0aea479

11822605433 has 2 divisors, whose sum is σ = 11822605434. Its totient is φ = 11822605432.

The previous prime is 11822605411. The next prime is 11822605441. The reversal of 11822605433 is 33450622811.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 9065515369 + 2757090064 = 95213^2 + 52508^2 .

It is an emirp because it is prime and its reverse (33450622811) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 11822605433 - 212 = 11822601337 is a prime.

It is a super-2 number, since 2×118226054332 (a number of 21 digits) contains 22 as substring.

It is a Sophie Germain prime.

It is a Curzon number.

It is a junction number, because it is equal to n+sod(n) for n = 11822605393 and 11822605402.

It is not a weakly prime, because it can be changed into another prime (11822605463) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5911302716 + 5911302717.

It is an arithmetic number, because the mean of its divisors is an integer number (5911302717).

Almost surely, 211822605433 is an apocalyptic number.

It is an amenable number.

11822605433 is a deficient number, since it is larger than the sum of its proper divisors (1).

11822605433 is an equidigital number, since it uses as much as digits as its factorization.

11822605433 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 34560, while the sum is 35.

The spelling of 11822605433 in words is "eleven billion, eight hundred twenty-two million, six hundred five thousand, four hundred thirty-three".