Base | Representation |
---|---|
bin | 1011111010110101100111… |
… | …0000101011100011110101 |
3 | 1201101212110102110011112221 |
4 | 2332231121300223203311 |
5 | 3204204430443404103 |
6 | 43512321112442341 |
7 | 2521560203540215 |
oct | 276553160534365 |
9 | 51355412404487 |
10 | 13105451153653 |
11 | 41a2a99577261 |
12 | 1577b110a93b1 |
13 | 740ab9938561 |
14 | 334443026245 |
15 | 17ad822431bd |
hex | beb59c2b8f5 |
13105451153653 has 2 divisors, whose sum is σ = 13105451153654. Its totient is φ = 13105451153652.
The previous prime is 13105451153629. The next prime is 13105451153659. The reversal of 13105451153653 is 35635115450131.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 10101253993009 + 3004197160644 = 3178247^2 + 1733262^2 .
It is a cyclic number.
It is not a de Polignac number, because 13105451153653 - 217 = 13105451022581 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (13105451153659) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6552725576826 + 6552725576827.
It is an arithmetic number, because the mean of its divisors is an integer number (6552725576827).
Almost surely, 213105451153653 is an apocalyptic number.
It is an amenable number.
13105451153653 is a deficient number, since it is larger than the sum of its proper divisors (1).
13105451153653 is an equidigital number, since it uses as much as digits as its factorization.
13105451153653 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 405000, while the sum is 43.
The spelling of 13105451153653 in words is "thirteen trillion, one hundred five billion, four hundred fifty-one million, one hundred fifty-three thousand, six hundred fifty-three".
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