Base | Representation |
---|---|
bin | 101111101101000011… |
… | …0001110010110010111 |
3 | 100210102110011110221120 |
4 | 1133122012032112113 |
5 | 3134300442000120 |
6 | 115021203415023 |
7 | 10254432534444 |
oct | 1373206162627 |
9 | 323373143846 |
10 | 102443312535 |
11 | 3a49a640a63 |
12 | 17a30084473 |
13 | 9877a818c5 |
14 | 4d5b7569cb |
15 | 29e89b9140 |
hex | 17da18e597 |
102443312535 has 32 divisors (see below), whose sum is σ = 169582448640. Its totient is φ = 52753420800.
The previous prime is 102443312497. The next prime is 102443312537. The reversal of 102443312535 is 535213344201.
It is a de Polignac number, because none of the positive numbers 2k-102443312535 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 102443312496 and 102443312505.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (102443312537) 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, 46786 + ... + 455055.
It is an arithmetic number, because the mean of its divisors is an integer number (5299451520).
Almost surely, 2102443312535 is an apocalyptic number.
102443312535 is a gapful number since it is divisible by the number (15) formed by its first and last digit.
102443312535 is a deficient number, since it is larger than the sum of its proper divisors (67139136105).
102443312535 is a wasteful number, since it uses less digits than its factorization.
102443312535 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 502319.
The product of its (nonzero) digits is 43200, while the sum is 33.
Adding to 102443312535 its reverse (535213344201), we get a palindrome (637656656736).
The spelling of 102443312535 in words is "one hundred two billion, four hundred forty-three million, three hundred twelve thousand, five hundred thirty-five".
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