Base | Representation |
---|---|
bin | 101111110000011001101100… |
… | …110000101011111001101010 |
3 | 1000112200002021120100210211210 |
4 | 233300121230300223321222 |
5 | 210012200234221123200 |
6 | 2022412233224533550 |
7 | 62145323064111234 |
oct | 5760315460537152 |
9 | 1015602246323753 |
10 | 210034315411050 |
11 | 60a18064514609 |
12 | 1b6820833212b6 |
13 | 90271c6866450 |
14 | 39c1a02377254 |
15 | 194371b8d6650 |
hex | bf066cc2be6a |
210034315411050 has 192 divisors, whose sum is σ = 567627291778176. Its totient is φ = 51088394649600.
The previous prime is 210034315410919. The next prime is 210034315411091. The reversal of 210034315411050 is 50114513430012.
It is a super-2 number, since 2×2100343154110502 (a number of 29 digits) contains 22 as substring.
It is a Harshad number since it is a multiple of its sum of digits (30).
It is a junction number, because it is equal to n+sod(n) for n = 210034315410999 and 210034315411017.
It is an unprimeable number.
It is a polite number, since it can be written in 95 ways as a sum of consecutive naturals, for example, 71495994 + ... + 74375706.
It is an arithmetic number, because the mean of its divisors is an integer number (2956392144678).
Almost surely, 2210034315411050 is an apocalyptic number.
It is a practical number, because each smaller number is the sum of distinct divisors of 210034315411050, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (283813645889088).
210034315411050 is an abundant number, since it is smaller than the sum of its proper divisors (357592976367126).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
210034315411050 is a wasteful number, since it uses less digits than its factorization.
210034315411050 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 2880185 (or 2880180 counting only the distinct ones).
The product of its (nonzero) digits is 7200, while the sum is 30.
Adding to 210034315411050 its reverse (50114513430012), we get a palindrome (260148828841062).
The spelling of 210034315411050 in words is "two hundred ten trillion, thirty-four billion, three hundred fifteen million, four hundred eleven thousand, fifty".
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