Base | Representation |
---|---|
bin | 10111010000000111001000… |
… | …10000001110011000010011 |
3 | 20201001002100200021110020001 |
4 | 23220003210100032120103 |
5 | 23200213013224300120 |
6 | 300425151243430431 |
7 | 13525046552153062 |
oct | 1350034420163023 |
9 | 221032320243201 |
10 | 51131120150035 |
11 | 1532364a380442 |
12 | 5899678825417 |
13 | 226b84a97a7c3 |
14 | c8aa8bbc21d9 |
15 | 5da087750e0a |
hex | 2e80e440e613 |
51131120150035 has 8 divisors (see below), whose sum is σ = 62662819588416. Its totient is φ = 40034579181120.
The previous prime is 51131120150023. The next prime is 51131120150083. The reversal of 51131120150035 is 53005102113115.
It is a sphenic number, since it is the product of 3 distinct primes.
It is not a de Polignac number, because 51131120150035 - 239 = 50581364336147 is a prime.
It is a super-2 number, since 2×511311201500352 (a number of 28 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 = 51131120149985 and 51131120150012.
It is an unprimeable number.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 108789617106 + ... + 108789617575.
It is an arithmetic number, because the mean of its divisors is an integer number (7832852448552).
Almost surely, 251131120150035 is an apocalyptic number.
51131120150035 is a deficient number, since it is larger than the sum of its proper divisors (11531699438381).
51131120150035 is a wasteful number, since it uses less digits than its factorization.
51131120150035 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 217579234733.
The product of its (nonzero) digits is 2250, while the sum is 28.
The spelling of 51131120150035 in words is "fifty-one trillion, one hundred thirty-one billion, one hundred twenty million, one hundred fifty thousand, thirty-five".
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