Base | Representation |
---|---|
bin | 11000111110001001… |
… | …10110100000101001 |
3 | 1021121022020221121110 |
4 | 30133010312200221 |
5 | 204424001024213 |
6 | 10054142500533 |
7 | 653135251536 |
oct | 143704664051 |
9 | 37538227543 |
10 | 13406267433 |
11 | 575a541169 |
12 | 2721894749 |
13 | 13585c8c30 |
14 | 9126c1d8d |
15 | 536e525c3 |
hex | 31f136829 |
13406267433 has 32 divisors (see below), whose sum is σ = 19489631616. Its totient is φ = 8147801088.
The previous prime is 13406267429. The next prime is 13406267449. The reversal of 13406267433 is 33476260431.
It is not a de Polignac number, because 13406267433 - 22 = 13406267429 is a prime.
It is a Harshad number since it is a multiple of its sum of digits (39).
It is a junction number, because it is equal to n+sod(n) for n = 13406267391 and 13406267400.
It is not an unprimeable number, because it can be changed into a prime (13406277433) by changing a digit.
It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 1401535 + ... + 1411067.
It is an arithmetic number, because the mean of its divisors is an integer number (609050988).
Almost surely, 213406267433 is an apocalyptic number.
13406267433 is a gapful number since it is divisible by the number (13) formed by its first and last digit.
It is an amenable number.
13406267433 is a deficient number, since it is larger than the sum of its proper divisors (6083364183).
13406267433 is a wasteful number, since it uses less digits than its factorization.
13406267433 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 9993.
The product of its (nonzero) digits is 217728, while the sum is 39.
The spelling of 13406267433 in words is "thirteen billion, four hundred six million, two hundred sixty-seven thousand, four hundred thirty-three".
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