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1332055344353 is a prime number
BaseRepresentation
bin10011011000100100101…
…011111000010011100001
311201100021020010200000012
4103120210223320103201
5133311022132004403
62455534320534305
7165144400415666
oct23304453702341
94640236120005
101332055344353
11473a15573918
121961b2781395
13987c6243716
1448686977a6d
15249b31e4ed8
hex13624af84e1

1332055344353 has 2 divisors, whose sum is σ = 1332055344354. Its totient is φ = 1332055344352.

The previous prime is 1332055344349. The next prime is 1332055344379. The reversal of 1332055344353 is 3534435502331.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 1316967398464 + 15087945889 = 1147592^2 + 122833^2 .

It is a cyclic number.

It is not a de Polignac number, because 1332055344353 - 22 = 1332055344349 is a prime.

It is a self number, because there is not a number n which added to its sum of digits gives 1332055344353.

It is not a weakly prime, because it can be changed into another prime (1332055344313) 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, 666027672176 + 666027672177.

It is an arithmetic number, because the mean of its divisors is an integer number (666027672177).

Almost surely, 21332055344353 is an apocalyptic number.

It is an amenable number.

1332055344353 is a deficient number, since it is larger than the sum of its proper divisors (1).

1332055344353 is an equidigital number, since it uses as much as digits as its factorization.

1332055344353 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 972000, while the sum is 41.

The spelling of 1332055344353 in words is "one trillion, three hundred thirty-two billion, fifty-five million, three hundred forty-four thousand, three hundred fifty-three".