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1235040214981 is a prime number
BaseRepresentation
bin10001111110001110001…
…000100001111111000101
311101001212002001110111111
4101332032020201333011
5130213330243334411
62343211502005021
7155141306421136
oct21761610417705
94331762043444
101235040214981
11436861005251
1217b438576771
138c604c4c021
1443ac2101c8d
15221d6061721
hex11f8e221fc5

1235040214981 has 2 divisors, whose sum is σ = 1235040214982. Its totient is φ = 1235040214980.

The previous prime is 1235040214847. The next prime is 1235040215063. The reversal of 1235040214981 is 1894120405321.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 953327668225 + 281712546756 = 976385^2 + 530766^2 .

It is a cyclic number.

It is not a de Polignac number, because 1235040214981 - 217 = 1235040083909 is a prime.

It is a super-2 number, since 2×12350402149812 (a number of 25 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (1235040214381) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 617520107490 + 617520107491.

It is an arithmetic number, because the mean of its divisors is an integer number (617520107491).

Almost surely, 21235040214981 is an apocalyptic number.

It is an amenable number.

1235040214981 is a deficient number, since it is larger than the sum of its proper divisors (1).

1235040214981 is an equidigital number, since it uses as much as digits as its factorization.

1235040214981 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 69120, while the sum is 40.

The spelling of 1235040214981 in words is "one trillion, two hundred thirty-five billion, forty million, two hundred fourteen thousand, nine hundred eighty-one".