Base | Representation |
---|---|
bin | 111100100011001110… |
… | …1110111101111000111 |
3 | 110102122002202010200211 |
4 | 1321012131313233013 |
5 | 4112301000120341 |
6 | 135422513020251 |
7 | 12252203266156 |
oct | 1710635675707 |
9 | 412562663624 |
10 | 130031254471 |
11 | 50167287078 |
12 | 2124b254687 |
13 | c3534c87c1 |
14 | 64176c359d |
15 | 35b0989181 |
hex | 1e46777bc7 |
130031254471 has 2 divisors, whose sum is σ = 130031254472. Its totient is φ = 130031254470.
The previous prime is 130031254421. The next prime is 130031254487. The reversal of 130031254471 is 174452130031.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 130031254471 - 211 = 130031252423 is a prime.
It is a super-3 number, since 3×1300312544713 (a number of 34 digits) contains 333 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (130031254421) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65015627235 + 65015627236.
It is an arithmetic number, because the mean of its divisors is an integer number (65015627236).
Almost surely, 2130031254471 is an apocalyptic number.
130031254471 is a deficient number, since it is larger than the sum of its proper divisors (1).
130031254471 is an equidigital number, since it uses as much as digits as its factorization.
130031254471 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 10080, while the sum is 31.
The spelling of 130031254471 in words is "one hundred thirty billion, thirty-one million, two hundred fifty-four thousand, four hundred seventy-one".
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