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150441951311 is a prime number
BaseRepresentation
bin1000110000011100001…
…0100000000001001111
3112101022120001222002122
42030013002200001033
54431101114420221
6153040113043155
713604043326021
oct2140702400117
9471276058078
10150441951311
1158890617773
12251a68564bb
1311256cbb802
1473d2356011
153da77cceab
hex23070a004f

150441951311 has 2 divisors, whose sum is σ = 150441951312. Its totient is φ = 150441951310.

The previous prime is 150441951277. The next prime is 150441951313. The reversal of 150441951311 is 113159144051.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 150441951311 - 210 = 150441950287 is a prime.

It is a super-2 number, since 2×1504419513112 (a number of 23 digits) contains 22 as substring.

Together with 150441951313, it forms a pair of twin primes.

It is a Chen prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (150441951313) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (13) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 75220975655 + 75220975656.

It is an arithmetic number, because the mean of its divisors is an integer number (75220975656).

Almost surely, 2150441951311 is an apocalyptic number.

150441951311 is a deficient number, since it is larger than the sum of its proper divisors (1).

150441951311 is an equidigital number, since it uses as much as digits as its factorization.

150441951311 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 10800, while the sum is 35.

The spelling of 150441951311 in words is "one hundred fifty billion, four hundred forty-one million, nine hundred fifty-one thousand, three hundred eleven".