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1411300344473 is a prime number
BaseRepresentation
bin10100100010011000000…
…011100100001010011001
311222220210221120202212122
4110202120003210022121
5141110320342010343
63000201544151025
7203651223665525
oct24423003441231
94886727522778
101411300344473
114a4590285358
121a9629768475
13a3114b20036
144c44333cd85
1526aa027a068
hex148980e4299

1411300344473 has 2 divisors, whose sum is σ = 1411300344474. Its totient is φ = 1411300344472.

The previous prime is 1411300344413. The next prime is 1411300344479. The reversal of 1411300344473 is 3744430031141.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 1245038650969 + 166261693504 = 1115813^2 + 407752^2 .

It is an emirp because it is prime and its reverse (3744430031141) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 1411300344473 - 210 = 1411300343449 is a prime.

It is not a weakly prime, because it can be changed into another prime (1411300344479) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 705650172236 + 705650172237.

It is an arithmetic number, because the mean of its divisors is an integer number (705650172237).

Almost surely, 21411300344473 is an apocalyptic number.

It is an amenable number.

1411300344473 is a deficient number, since it is larger than the sum of its proper divisors (1).

1411300344473 is an equidigital number, since it uses as much as digits as its factorization.

1411300344473 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 48384, while the sum is 35.

The spelling of 1411300344473 in words is "one trillion, four hundred eleven billion, three hundred million, three hundred forty-four thousand, four hundred seventy-three".