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1102203312433 is a prime number
BaseRepresentation
bin10000000010100000011…
…011111101100100110001
310220100222102222122011221
4100002200123331210301
5121024303021444213
62202202322105041
7142426424400646
oct20024033754461
93810872878157
101102203312433
11395495043263
121597457ab181
137cc253c9542
143b4bdd67dcd
151da0e12438d
hex100a06fd931

1102203312433 has 2 divisors, whose sum is σ = 1102203312434. Its totient is φ = 1102203312432.

The previous prime is 1102203312361. The next prime is 1102203312443. The reversal of 1102203312433 is 3342133022011.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 867155788944 + 235047523489 = 931212^2 + 484817^2 .

It is a cyclic number.

It is not a de Polignac number, because 1102203312433 - 217 = 1102203181361 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 1102203312398 and 1102203312407.

It is not a weakly prime, because it can be changed into another prime (1102203312443) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 551101656216 + 551101656217.

It is an arithmetic number, because the mean of its divisors is an integer number (551101656217).

Almost surely, 21102203312433 is an apocalyptic number.

It is an amenable number.

1102203312433 is a deficient number, since it is larger than the sum of its proper divisors (1).

1102203312433 is an equidigital number, since it uses as much as digits as its factorization.

1102203312433 is an odious number, because the sum of its binary digits is odd.

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

Adding to 1102203312433 its reverse (3342133022011), we get a palindrome (4444336334444).

The spelling of 1102203312433 in words is "one trillion, one hundred two billion, two hundred three million, three hundred twelve thousand, four hundred thirty-three".