Base | Representation |
---|---|
bin | 1000101110111110111… |
… | …1100100100000110011 |
3 | 112100022022122022202001 |
4 | 2023233233210200303 |
5 | 4424301041104021 |
6 | 152533232034431 |
7 | 13561256262532 |
oct | 2135757444063 |
9 | 470268568661 |
10 | 150051113011 |
11 | 586aaa3a108 |
12 | 250b7992a17 |
13 | 111c4037347 |
14 | 7396498319 |
15 | 3d83329091 |
hex | 22efbe4833 |
150051113011 has 2 divisors, whose sum is σ = 150051113012. Its totient is φ = 150051113010.
The previous prime is 150051112999. The next prime is 150051113039. The reversal of 150051113011 is 110311150051.
It is a weak prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-150051113011 is a prime.
It is a super-2 number, since 2×1500511130112 (a number of 23 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 150051113011.
It is not a weakly prime, because it can be changed into another prime (150051113041) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 75025556505 + 75025556506.
It is an arithmetic number, because the mean of its divisors is an integer number (75025556506).
Almost surely, 2150051113011 is an apocalyptic number.
150051113011 is a deficient number, since it is larger than the sum of its proper divisors (1).
150051113011 is an equidigital number, since it uses as much as digits as its factorization.
150051113011 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 75, while the sum is 19.
Adding to 150051113011 its reverse (110311150051), we get a palindrome (260362263062).
The spelling of 150051113011 in words is "one hundred fifty billion, fifty-one million, one hundred thirteen thousand, eleven".
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