Base | Representation |
---|---|
bin | 100000011011100101111010… |
… | …111011001001110110010011 |
3 | 200201000122100000002110202121 |
4 | 200123211322323021312103 |
5 | 122143402134100310201 |
6 | 1223204552520555111 |
7 | 42020640230450041 |
oct | 4033457273116623 |
9 | 621018300073677 |
10 | 142633631260051 |
11 | 414a1646439018 |
12 | 13bb742a622a97 |
13 | 61783a83a0b39 |
14 | 273170c835391 |
15 | 1175364b499a1 |
hex | 81b97aec9d93 |
142633631260051 has 2 divisors, whose sum is σ = 142633631260052. Its totient is φ = 142633631260050.
The previous prime is 142633631260013. The next prime is 142633631260097. The reversal of 142633631260051 is 150062136336241.
It is a weak prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-142633631260051 is a prime.
It is a super-3 number, since 3×1426336312600513 (a number of 43 digits) contains 333 as substring.
It is not a weakly prime, because it can be changed into another prime (142633631260151) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 71316815630025 + 71316815630026.
It is an arithmetic number, because the mean of its divisors is an integer number (71316815630026).
Almost surely, 2142633631260051 is an apocalyptic number.
142633631260051 is a deficient number, since it is larger than the sum of its proper divisors (1).
142633631260051 is an equidigital number, since it uses as much as digits as its factorization.
142633631260051 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 466560, while the sum is 43.
Adding to 142633631260051 its reverse (150062136336241), we get a palindrome (292695767596292).
The spelling of 142633631260051 in words is "one hundred forty-two trillion, six hundred thirty-three billion, six hundred thirty-one million, two hundred sixty thousand, fifty-one".
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