Base | Representation |
---|---|
bin | 11110111011100010000110… |
… | …001101101111001111000111 |
3 | 122211122120112202010210020211 |
4 | 132323202012031233033013 |
5 | 120312223432210302312 |
6 | 1201152240400340251 |
7 | 40440010033212511 |
oct | 3673420615571707 |
9 | 584576482123224 |
10 | 136032455947207 |
11 | 3a38704a029146 |
12 | 13310002314687 |
13 | 5ab9a61a54071 |
14 | 258401265abb1 |
15 | 10ad7b2a833a7 |
hex | 7bb88636f3c7 |
136032455947207 has 2 divisors, whose sum is σ = 136032455947208. Its totient is φ = 136032455947206.
The previous prime is 136032455947147. The next prime is 136032455947211. The reversal of 136032455947207 is 702749554230631.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 136032455947207 - 211 = 136032455945159 is a prime.
It is a super-2 number, since 2×1360324559472072 (a number of 29 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (136032455947277) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 68016227973603 + 68016227973604.
It is an arithmetic number, because the mean of its divisors is an integer number (68016227973604).
Almost surely, 2136032455947207 is an apocalyptic number.
136032455947207 is a deficient number, since it is larger than the sum of its proper divisors (1).
136032455947207 is an equidigital number, since it uses as much as digits as its factorization.
136032455947207 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 38102400, while the sum is 58.
The spelling of 136032455947207 in words is "one hundred thirty-six trillion, thirty-two billion, four hundred fifty-five million, nine hundred forty-seven thousand, two hundred seven".
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