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114511315405433 is a prime number
BaseRepresentation
bin11010000010010110111110…
…010001110000101001111001
3120000110011201110101100102212
4122002112332101300221321
5110002123133220433213
61043313434433304505
733056112100666502
oct3202267621605171
9500404643340385
10114511315405433
1133539a812738a1
1210a15087a05135
134bb84a85a511b
14203c53d1d7ba9
15d38a7e8e39a8
hex6825be470a79

114511315405433 has 2 divisors, whose sum is σ = 114511315405434. Its totient is φ = 114511315405432.

The previous prime is 114511315405429. The next prime is 114511315405453. The reversal of 114511315405433 is 334504513115411.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 63029499348544 + 51481816056889 = 7939112^2 + 7175083^2 .

It is a cyclic number.

It is not a de Polignac number, because 114511315405433 - 22 = 114511315405429 is a prime.

It is a super-2 number, since 2×1145113154054332 (a number of 29 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (114511315405423) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 57255657702716 + 57255657702717.

It is an arithmetic number, because the mean of its divisors is an integer number (57255657702717).

Almost surely, 2114511315405433 is an apocalyptic number.

It is an amenable number.

114511315405433 is a deficient number, since it is larger than the sum of its proper divisors (1).

114511315405433 is an equidigital number, since it uses as much as digits as its factorization.

114511315405433 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 216000, while the sum is 41.

The spelling of 114511315405433 in words is "one hundred fourteen trillion, five hundred eleven billion, three hundred fifteen million, four hundred five thousand, four hundred thirty-three".