Base | Representation |
---|---|
bin | 111100100011010001… |
… | …0110101011000100011 |
3 | 110102122012012020110010 |
4 | 1321012202311120203 |
5 | 4112301313234210 |
6 | 135423000540003 |
7 | 12252220325340 |
oct | 1710642653043 |
9 | 412565166403 |
10 | 130032555555 |
11 | 50167a95651 |
12 | 2124b781603 |
13 | c353862a86 |
14 | 64179417c7 |
15 | 35b0b44920 |
hex | 1e468b5623 |
130032555555 has 32 divisors (see below), whose sum is σ = 238186013184. Its totient is φ = 59340404736.
The previous prime is 130032555553. The next prime is 130032555557. The reversal of 130032555555 is 555555230031.
It is an interprime number because it is at equal distance from previous prime (130032555553) and next prime (130032555557).
It is not a de Polignac number, because 130032555555 - 21 = 130032555553 is a prime.
It is a super-2 number, since 2×1300325555552 (a number of 23 digits) contains 22 as substring.
It is not an unprimeable number, because it can be changed into a prime (130032555551) 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, 1012557 + ... + 1133726.
It is an arithmetic number, because the mean of its divisors is an integer number (7443312912).
Almost surely, 2130032555555 is an apocalyptic number.
130032555555 is a gapful number since it is divisible by the number (15) formed by its first and last digit.
130032555555 is a deficient number, since it is larger than the sum of its proper divisors (108153457629).
130032555555 is a wasteful number, since it uses less digits than its factorization.
130032555555 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 2146875.
The product of its (nonzero) digits is 281250, while the sum is 39.
Adding to 130032555555 its reverse (555555230031), we get a palindrome (685587785586).
The spelling of 130032555555 in words is "one hundred thirty billion, thirty-two million, five hundred fifty-five thousand, five hundred fifty-five".
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