Base | Representation |
---|---|
bin | 1011111100101000010110… |
… | …1101010101001001110011 |
3 | 1201111210222002000111001122 |
4 | 2333022011231111021303 |
5 | 3210211021312302034 |
6 | 43534413400252455 |
7 | 2524030413150023 |
oct | 277120555251163 |
9 | 51453862014048 |
10 | 13136253244019 |
11 | 4205065549694 |
12 | 1581a8877112b |
13 | 743988235b29 |
14 | 335b25bc1a83 |
15 | 17ba864a862e |
hex | bf285b55273 |
13136253244019 has 2 divisors, whose sum is σ = 13136253244020. Its totient is φ = 13136253244018.
The previous prime is 13136253243989. The next prime is 13136253244021. The reversal of 13136253244019 is 91044235263131.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 13136253244019 - 216 = 13136253178483 is a prime.
It is a super-2 number, since 2×131362532440192 (a number of 27 digits) contains 22 as substring.
Together with 13136253244021, it forms a pair of twin primes.
It is a Chen prime.
It is not a weakly prime, because it can be changed into another prime (13136253244069) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6568126622009 + 6568126622010.
It is an arithmetic number, because the mean of its divisors is an integer number (6568126622010).
Almost surely, 213136253244019 is an apocalyptic number.
13136253244019 is a deficient number, since it is larger than the sum of its proper divisors (1).
13136253244019 is an equidigital number, since it uses as much as digits as its factorization.
13136253244019 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 466560, while the sum is 44.
The spelling of 13136253244019 in words is "thirteen trillion, one hundred thirty-six billion, two hundred fifty-three million, two hundred forty-four thousand, nineteen".
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