Base | Representation |
---|---|
bin | 11011000100010010… |
… | …10100010000001011 |
3 | 1101111201121021000111 |
4 | 31202021110100023 |
5 | 214230031330212 |
6 | 10401540415151 |
7 | 1023050544115 |
oct | 154211242013 |
9 | 41451537014 |
10 | 14531511307 |
11 | 6187728585 |
12 | 29966a74b7 |
13 | 14a7777b01 |
14 | 9bbd123b5 |
15 | 5a0b22ea7 |
hex | 36225440b |
14531511307 has 2 divisors, whose sum is σ = 14531511308. Its totient is φ = 14531511306.
The previous prime is 14531511271. The next prime is 14531511331. The reversal of 14531511307 is 70311513541.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 14531511307 - 211 = 14531509259 is a prime.
It is a super-2 number, since 2×145315113072 (a number of 21 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (14531511347) 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, 7265755653 + 7265755654.
It is an arithmetic number, because the mean of its divisors is an integer number (7265755654).
Almost surely, 214531511307 is an apocalyptic number.
14531511307 is a deficient number, since it is larger than the sum of its proper divisors (1).
14531511307 is an equidigital number, since it uses as much as digits as its factorization.
14531511307 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 6300, while the sum is 31.
The spelling of 14531511307 in words is "fourteen billion, five hundred thirty-one million, five hundred eleven thousand, three hundred seven".
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