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12333131214149 is a prime number
BaseRepresentation
bin1011001101111000011111…
…1001111010110101000101
31121200000222010020012020002
42303132013321322311011
53104031223042323044
642121432352051045
72412016326425054
oct263360771726505
947600863205202
1012333131214149
113a254a7056719
1214722b0833a85
136b601751a105
14308cd8d6a39b
15165c2e06a74e
hexb3787e7ad45

12333131214149 has 2 divisors, whose sum is σ = 12333131214150. Its totient is φ = 12333131214148.

The previous prime is 12333131214037. The next prime is 12333131214163. The reversal of 12333131214149 is 94141213133321.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 6717131778049 + 5615999436100 = 2591743^2 + 2369810^2 .

It is an emirp because it is prime and its reverse (94141213133321) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 12333131214149 - 212 = 12333131210053 is a prime.

It is a super-2 number, since 2×123331312141492 (a number of 27 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (12333131214179) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6166565607074 + 6166565607075.

It is an arithmetic number, because the mean of its divisors is an integer number (6166565607075).

Almost surely, 212333131214149 is an apocalyptic number.

It is an amenable number.

12333131214149 is a deficient number, since it is larger than the sum of its proper divisors (1).

12333131214149 is an equidigital number, since it uses as much as digits as its factorization.

12333131214149 is an evil number, because the sum of its binary digits is even.

The product of its digits is 46656, while the sum is 38.

The spelling of 12333131214149 in words is "twelve trillion, three hundred thirty-three billion, one hundred thirty-one million, two hundred fourteen thousand, one hundred forty-nine".