Base | Representation |
---|---|
bin | 111110111110010001… |
… | …0010000111111001011 |
3 | 110221001122101122012110 |
4 | 1331330202100333023 |
5 | 4203424213442220 |
6 | 142043023513403 |
7 | 12525131311251 |
oct | 1757442207713 |
9 | 427048348173 |
10 | 135233343435 |
11 | 52396773553 |
12 | 22261458863 |
13 | c9a2193458 |
14 | 678c54c5d1 |
15 | 37b750dbe0 |
hex | 1f7c890fcb |
135233343435 has 32 divisors (see below), whose sum is σ = 229298688000. Its totient is φ = 67823318528.
The previous prime is 135233343433. The next prime is 135233343457. The reversal of 135233343435 is 534343332531.
It is a cyclic number.
It is not a de Polignac number, because 135233343435 - 21 = 135233343433 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 135233343393 and 135233343402.
It is not an unprimeable number, because it can be changed into a prime (135233343433) 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, 68566 + ... + 524564.
It is an arithmetic number, because the mean of its divisors is an integer number (7165584000).
Almost surely, 2135233343435 is an apocalyptic number.
135233343435 is a gapful number since it is divisible by the number (15) formed by its first and last digit.
135233343435 is a deficient number, since it is larger than the sum of its proper divisors (94065344565).
135233343435 is a wasteful number, since it uses less digits than its factorization.
135233343435 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 457187.
The product of its digits is 583200, while the sum is 39.
Adding to 135233343435 its reverse (534343332531), we get a palindrome (669576675966).
The spelling of 135233343435 in words is "one hundred thirty-five billion, two hundred thirty-three million, three hundred forty-three thousand, four hundred thirty-five".
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