Base | Representation |
---|---|
bin | 10010110001001101… |
… | …110011100110010011 |
3 | 1221000111112022002222 |
4 | 102301031303212103 |
5 | 312233140240020 |
6 | 13131433531255 |
7 | 1312261101146 |
oct | 226115634623 |
9 | 57014468088 |
10 | 20153055635 |
11 | 8601968a89 |
12 | 3aa5273b2b |
13 | 1b923065bb |
14 | d9276805d |
15 | 7ce400b25 |
hex | 4b1373993 |
20153055635 has 4 divisors (see below), whose sum is σ = 24183666768. Its totient is φ = 16122444504.
The previous prime is 20153055557. The next prime is 20153055667. The reversal of 20153055635 is 53655035102.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is not a de Polignac number, because 20153055635 - 218 = 20152793491 is a prime.
It is a super-2 number, since 2×201530556352 (a number of 21 digits) contains 22 as substring.
It is a Duffinian number.
It is a junction number, because it is equal to n+sod(n) for n = 20153055595 and 20153055604.
It is an unprimeable number.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 2015305559 + ... + 2015305568.
It is an arithmetic number, because the mean of its divisors is an integer number (6045916692).
Almost surely, 220153055635 is an apocalyptic number.
20153055635 is a deficient number, since it is larger than the sum of its proper divisors (4030611133).
20153055635 is an equidigital number, since it uses as much as digits as its factorization.
20153055635 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 4030611132.
The product of its (nonzero) digits is 67500, while the sum is 35.
The spelling of 20153055635 in words is "twenty billion, one hundred fifty-three million, fifty-five thousand, six hundred thirty-five".
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