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11240043331553 is a prime number
BaseRepresentation
bin1010001110010000011011…
…0010100011111111100001
31110210112111221112101000122
42203210012302203333201
52433124102043102203
635523334333130025
72240031515633231
oct243440662437741
943715457471018
1011240043331553
11364396a484853
12131648b929915
13636c1578784a
142ac0418a3cc1
151475a5324538
hexa3906ca3fe1

11240043331553 has 2 divisors, whose sum is σ = 11240043331554. Its totient is φ = 11240043331552.

The previous prime is 11240043331529. The next prime is 11240043331591. The reversal of 11240043331553 is 35513334004211.

11240043331553 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 11196465516544 + 43577815009 = 3346112^2 + 208753^2 .

It is a cyclic number.

It is not a de Polignac number, because 11240043331553 - 216 = 11240043266017 is a prime.

It is not a weakly prime, because it can be changed into another prime (11240043331513) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5620021665776 + 5620021665777.

It is an arithmetic number, because the mean of its divisors is an integer number (5620021665777).

Almost surely, 211240043331553 is an apocalyptic number.

It is an amenable number.

11240043331553 is a deficient number, since it is larger than the sum of its proper divisors (1).

11240043331553 is an equidigital number, since it uses as much as digits as its factorization.

11240043331553 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 64800, while the sum is 35.

Adding to 11240043331553 its reverse (35513334004211), we get a palindrome (46753377335764).

The spelling of 11240043331553 in words is "eleven trillion, two hundred forty billion, forty-three million, three hundred thirty-one thousand, five hundred fifty-three".