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1311144204031 is a prime number
BaseRepresentation
bin10011000101000110010…
…010001101011011111111
311122100021220010200012121
4103011012102031123333
5132440210404012111
62442155322343411
7163504242564214
oct23050622153377
94570256120177
101311144204031
114660647a3672
1219213765ab67
1396842b717cc
1447661675b0b
152418c49b871
hex1314648d6ff

1311144204031 has 2 divisors, whose sum is σ = 1311144204032. Its totient is φ = 1311144204030.

The previous prime is 1311144203953. The next prime is 1311144204047. The reversal of 1311144204031 is 1304024411131.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 1311144204031 - 239 = 761388390143 is a prime.

It is a super-2 number, since 2×13111442040312 (a number of 25 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 1311144203987 and 1311144204005.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (1311144204631) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 655572102015 + 655572102016.

It is an arithmetic number, because the mean of its divisors is an integer number (655572102016).

Almost surely, 21311144204031 is an apocalyptic number.

1311144204031 is a deficient number, since it is larger than the sum of its proper divisors (1).

1311144204031 is an equidigital number, since it uses as much as digits as its factorization.

1311144204031 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 1152, while the sum is 25.

Adding to 1311144204031 its reverse (1304024411131), we get a palindrome (2615168615162).

The spelling of 1311144204031 in words is "one trillion, three hundred eleven billion, one hundred forty-four million, two hundred four thousand, thirty-one".