Base | Representation |
---|---|
bin | 11110010101010000100110… |
… | …101011000011011010100011 |
3 | 122111100010201221212100101212 |
4 | 132111100212223003122203 |
5 | 114441130434224043042 |
6 | 1151420104042420335 |
7 | 40046000113264442 |
oct | 3625204653033243 |
9 | 574303657770355 |
10 | 133402333034147 |
11 | 3956268022a3a1 |
12 | 12b663235016ab |
13 | 5958a2ab935a8 |
14 | 24d29c9044d59 |
15 | 106517a430482 |
hex | 795426ac36a3 |
133402333034147 has 2 divisors, whose sum is σ = 133402333034148. Its totient is φ = 133402333034146.
The previous prime is 133402333034053. The next prime is 133402333034153. The reversal of 133402333034147 is 741430333204331.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 133402333034147 - 212 = 133402333030051 is a prime.
It is not a weakly prime, because it can be changed into another prime (133402333034167) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66701166517073 + 66701166517074.
It is an arithmetic number, because the mean of its divisors is an integer number (66701166517074).
Almost surely, 2133402333034147 is an apocalyptic number.
133402333034147 is a deficient number, since it is larger than the sum of its proper divisors (1).
133402333034147 is an equidigital number, since it uses as much as digits as its factorization.
133402333034147 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 653184, while the sum is 41.
Adding to 133402333034147 its reverse (741430333204331), we get a palindrome (874832666238478).
The spelling of 133402333034147 in words is "one hundred thirty-three trillion, four hundred two billion, three hundred thirty-three million, thirty-four thousand, one hundred forty-seven".
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