Base | Representation |
---|---|
bin | 11110010011000100100111… |
… | …011101100111010101110011 |
3 | 122110210202202112221200001021 |
4 | 132103010213131213111303 |
5 | 114431200120220330024 |
6 | 1151223044540520311 |
7 | 40032066223252051 |
oct | 3623044735472563 |
9 | 573722675850037 |
10 | 133252022433139 |
11 | 395049578770a5 |
12 | 12b41174850097 |
13 | 59477c517406c |
14 | 24c960a3211d1 |
15 | 10612ce44ade4 |
hex | 793127767573 |
133252022433139 has 2 divisors, whose sum is σ = 133252022433140. Its totient is φ = 133252022433138.
The previous prime is 133252022433053. The next prime is 133252022433151. The reversal of 133252022433139 is 931334220252331.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 133252022433139 - 211 = 133252022431091 is a prime.
It is a super-2 number, since 2×1332520224331392 (a number of 29 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 133252022433095 and 133252022433104.
It is not a weakly prime, because it can be changed into another prime (133252022433539) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66626011216569 + 66626011216570.
It is an arithmetic number, because the mean of its divisors is an integer number (66626011216570).
Almost surely, 2133252022433139 is an apocalyptic number.
133252022433139 is a deficient number, since it is larger than the sum of its proper divisors (1).
133252022433139 is an equidigital number, since it uses as much as digits as its factorization.
133252022433139 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 699840, while the sum is 43.
The spelling of 133252022433139 in words is "one hundred thirty-three trillion, two hundred fifty-two billion, twenty-two million, four hundred thirty-three thousand, one hundred thirty-nine".
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