Base | Representation |
---|---|
bin | 11110111011101100001110… |
… | …011110101001010000101011 |
3 | 122211200121121201101021122221 |
4 | 132323230032132221100223 |
5 | 120312413220443441042 |
6 | 1201201235520254511 |
7 | 40440543406236424 |
oct | 3673541636512053 |
9 | 584617551337587 |
10 | 136043332015147 |
11 | 3a390720309a47 |
12 | 13312138771437 |
13 | 5abaaa60a8b1b |
14 | 2584764caa64b |
15 | 10adbec7e1d67 |
hex | 7bbb0e7a942b |
136043332015147 has 2 divisors, whose sum is σ = 136043332015148. Its totient is φ = 136043332015146.
The previous prime is 136043332015081. The next prime is 136043332015169. The reversal of 136043332015147 is 741510233340631.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 136043332015147 - 239 = 135493576201259 is a prime.
It is not a weakly prime, because it can be changed into another prime (136043332025147) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 68021666007573 + 68021666007574.
It is an arithmetic number, because the mean of its divisors is an integer number (68021666007574).
Almost surely, 2136043332015147 is an apocalyptic number.
136043332015147 is a deficient number, since it is larger than the sum of its proper divisors (1).
136043332015147 is an equidigital number, since it uses as much as digits as its factorization.
136043332015147 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 544320, while the sum is 43.
Adding to 136043332015147 its reverse (741510233340631), we get a palindrome (877553565355778).
The spelling of 136043332015147 in words is "one hundred thirty-six trillion, forty-three billion, three hundred thirty-two million, fifteen thousand, one hundred forty-seven".
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