Base | Representation |
---|---|
bin | 1000101110111011001… |
… | …1000100110000111011 |
3 | 112100021012210121201112 |
4 | 2023232303010300323 |
5 | 4424233011114311 |
6 | 152531504110535 |
7 | 13561004422364 |
oct | 2135663046073 |
9 | 470235717645 |
10 | 150035254331 |
11 | 586a1a98266 |
12 | 250b260544b |
13 | 111c0973c01 |
14 | 739432ab6b |
15 | 3d81c4528b |
hex | 22eecc4c3b |
150035254331 has 4 divisors (see below), whose sum is σ = 150036067512. Its totient is φ = 150034441152.
The previous prime is 150035254273. The next prime is 150035254333. The reversal of 150035254331 is 133452530051.
It is a semiprime because it is the product of two primes, and also a brilliant number, because the two primes have the same length.
It is a cyclic number.
It is not a de Polignac number, because 150035254331 - 230 = 148961512507 is a prime.
It is a super-2 number, since 2×1500352543312 (a number of 23 digits) contains 22 as substring.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (150035254333) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 17876 + ... + 548078.
It is an arithmetic number, because the mean of its divisors is an integer number (37509016878).
Almost surely, 2150035254331 is an apocalyptic number.
150035254331 is a deficient number, since it is larger than the sum of its proper divisors (813181).
150035254331 is an equidigital number, since it uses as much as digits as its factorization.
150035254331 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 813180.
The product of its (nonzero) digits is 27000, while the sum is 32.
Adding to 150035254331 its reverse (133452530051), we get a palindrome (283487784382).
The spelling of 150035254331 in words is "one hundred fifty billion, thirty-five million, two hundred fifty-four thousand, three hundred thirty-one".
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