Base | Representation |
---|---|
bin | 1000010001011101010… |
… | …1010111111111111001 |
3 | 111120211222000101111211 |
4 | 2010113111113333321 |
5 | 4312033100300212 |
6 | 145142544323121 |
7 | 13160666652634 |
oct | 2042725277771 |
9 | 446758011454 |
10 | 142125400057 |
11 | 553030a0713 |
12 | 236656134a1 |
13 | 10530004174 |
14 | 6c43a0ac1b |
15 | 3a6c6049a7 |
hex | 2117557ff9 |
142125400057 has 2 divisors, whose sum is σ = 142125400058. Its totient is φ = 142125400056.
The previous prime is 142125400049. The next prime is 142125400181. The reversal of 142125400057 is 750004521241.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 138893363856 + 3232036201 = 372684^2 + 56851^2 .
It is a cyclic number.
It is not a de Polignac number, because 142125400057 - 23 = 142125400049 is a prime.
It is not a weakly prime, because it can be changed into another prime (142125400457) 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, 71062700028 + 71062700029.
It is an arithmetic number, because the mean of its divisors is an integer number (71062700029).
Almost surely, 2142125400057 is an apocalyptic number.
It is an amenable number.
142125400057 is a deficient number, since it is larger than the sum of its proper divisors (1).
142125400057 is an equidigital number, since it uses as much as digits as its factorization.
142125400057 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 11200, while the sum is 31.
Adding to 142125400057 its reverse (750004521241), we get a palindrome (892129921298).
The spelling of 142125400057 in words is "one hundred forty-two billion, one hundred twenty-five million, four hundred thousand, fifty-seven".
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