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121221223021433 is a prime number
BaseRepresentation
bin11011100100000000000011…
…111011111111101101111001
3120220012122011110100100011202
4123210000003323333231321
5111342042031043141213
61105452132220303545
734350641624454233
oct3344000373775571
9526178143310152
10121221223021433
11356966a518a437
121171959a3a1bb5
13528415c8b4212
1421d11d2083253
15e03397643b58
hex6e4003effb79

121221223021433 has 2 divisors, whose sum is σ = 121221223021434. Its totient is φ = 121221223021432.

The previous prime is 121221223021399. The next prime is 121221223021439. The reversal of 121221223021433 is 334120322122121.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 63302836990864 + 57918386030569 = 7956308^2 + 7610413^2 .

It is a cyclic number.

It is not a de Polignac number, because 121221223021433 - 214 = 121221223005049 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 121221223021396 and 121221223021405.

It is not a weakly prime, because it can be changed into another prime (121221223021439) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 60610611510716 + 60610611510717.

It is an arithmetic number, because the mean of its divisors is an integer number (60610611510717).

Almost surely, 2121221223021433 is an apocalyptic number.

It is an amenable number.

121221223021433 is a deficient number, since it is larger than the sum of its proper divisors (1).

121221223021433 is an equidigital number, since it uses as much as digits as its factorization.

121221223021433 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 6912, while the sum is 29.

Adding to 121221223021433 its reverse (334120322122121), we get a palindrome (455341545143554).

The spelling of 121221223021433 in words is "one hundred twenty-one trillion, two hundred twenty-one billion, two hundred twenty-three million, twenty-one thousand, four hundred thirty-three".