Base | Representation |
---|---|
bin | 110111001001010101111001… |
… | …101100110000010111011011 |
3 | 1011210202002001210120011110112 |
4 | 313021111321230300113123 |
5 | 223242141224300412103 |
6 | 2215454500100434535 |
7 | 102041354150130104 |
oct | 6711257154602733 |
9 | 1153662053504415 |
10 | 242534550013403 |
11 | 7030836369008a |
12 | 232509a20a8a4b |
13 | a543baa16888b |
14 | 43c6a406253ab |
15 | 1d08d30bb2cd8 |
hex | dc9579b305db |
242534550013403 has 2 divisors, whose sum is σ = 242534550013404. Its totient is φ = 242534550013402.
The previous prime is 242534550013333. The next prime is 242534550013429. The reversal of 242534550013403 is 304310055435242.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 242534550013403 - 216 = 242534549947867 is a prime.
It is a super-2 number, since 2×2425345500134032 (a number of 30 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (242534550513403) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 121267275006701 + 121267275006702.
It is an arithmetic number, because the mean of its divisors is an integer number (121267275006702).
Almost surely, 2242534550013403 is an apocalyptic number.
242534550013403 is a deficient number, since it is larger than the sum of its proper divisors (1).
242534550013403 is an equidigital number, since it uses as much as digits as its factorization.
242534550013403 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 864000, while the sum is 41.
The spelling of 242534550013403 in words is "two hundred forty-two trillion, five hundred thirty-four billion, five hundred fifty million, thirteen thousand, four hundred three".
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