Base | Representation |
---|---|
bin | 110100111000010110… |
… | …1100100010100011011 |
3 | 101212010012102222122002 |
4 | 1221300231210110123 |
5 | 3330032344040011 |
6 | 124100243551215 |
7 | 11130061201331 |
oct | 1516055442433 |
9 | 355105388562 |
10 | 113560143131 |
11 | 44184816366 |
12 | 1a01308b50b |
13 | a929c63686 |
14 | 56d3d59b51 |
15 | 2e49934c3b |
hex | 1a70b6451b |
113560143131 has 2 divisors, whose sum is σ = 113560143132. Its totient is φ = 113560143130.
The previous prime is 113560143071. The next prime is 113560143139. The reversal of 113560143131 is 131341065311.
It is a happy number.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 113560143131 - 214 = 113560126747 is a prime.
It is a super-2 number, since 2×1135601431312 (a number of 23 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 113560143094 and 113560143103.
It is not a weakly prime, because it can be changed into another prime (113560143139) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 56780071565 + 56780071566.
It is an arithmetic number, because the mean of its divisors is an integer number (56780071566).
Almost surely, 2113560143131 is an apocalyptic number.
113560143131 is a deficient number, since it is larger than the sum of its proper divisors (1).
113560143131 is an equidigital number, since it uses as much as digits as its factorization.
113560143131 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 3240, while the sum is 29.
The spelling of 113560143131 in words is "one hundred thirteen billion, five hundred sixty million, one hundred forty-three thousand, one hundred thirty-one".
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