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311420152151 is a prime number
BaseRepresentation
bin1001000100000100001…
…01100110100101010111
31002202211101221212101022
410202002011212211113
520100242024332101
6355021530351355
731333165421513
oct4420205464527
91082741855338
10311420152151
11110088614679
1250431b8b55b
132349cb022a4
1411103b41143
15817a092a1b
hex4882166957

311420152151 has 2 divisors, whose sum is σ = 311420152152. Its totient is φ = 311420152150.

The previous prime is 311420152133. The next prime is 311420152181. The reversal of 311420152151 is 151251024113.

It is a weak prime.

It is a cyclic number.

It is not a de Polignac number, because 311420152151 - 218 = 311419890007 is a prime.

It is a super-2 number, since 2×3114201521512 (a number of 24 digits) contains 22 as substring.

It is a self number, because there is not a number n which added to its sum of digits gives 311420152151.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (311420152181) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 155710076075 + 155710076076.

It is an arithmetic number, because the mean of its divisors is an integer number (155710076076).

Almost surely, 2311420152151 is an apocalyptic number.

311420152151 is a deficient number, since it is larger than the sum of its proper divisors (1).

311420152151 is an equidigital number, since it uses as much as digits as its factorization.

311420152151 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 1200, while the sum is 26.

Adding to 311420152151 its reverse (151251024113), we get a palindrome (462671176264).

The spelling of 311420152151 in words is "three hundred eleven billion, four hundred twenty million, one hundred fifty-two thousand, one hundred fifty-one".