Base | Representation |
---|---|
bin | 11111000000001101111000… |
… | …100110001110110001010011 |
3 | 122212210100102002122122212111 |
4 | 133000031320212032301103 |
5 | 120333012202102042003 |
6 | 1202000153524010151 |
7 | 40502163622623043 |
oct | 3700157046166123 |
9 | 585710362578774 |
10 | 136354350034003 |
11 | 3a4a0611930952 |
12 | 13362477b30957 |
13 | 5b112107a07ca |
14 | 259582b44d723 |
15 | 10b6d523ede6d |
hex | 7c037898ec53 |
136354350034003 has 2 divisors, whose sum is σ = 136354350034004. Its totient is φ = 136354350034002.
The previous prime is 136354350033883. The next prime is 136354350034061. The reversal of 136354350034003 is 300430053453631.
It is a strong prime.
It is an emirp because it is prime and its reverse (300430053453631) is a distict prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-136354350034003 is a prime.
It is not a weakly prime, because it can be changed into another prime (136354350034403) 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, 68177175017001 + 68177175017002.
It is an arithmetic number, because the mean of its divisors is an integer number (68177175017002).
Almost surely, 2136354350034003 is an apocalyptic number.
136354350034003 is a deficient number, since it is larger than the sum of its proper divisors (1).
136354350034003 is an equidigital number, since it uses as much as digits as its factorization.
136354350034003 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 583200, while the sum is 40.
The spelling of 136354350034003 in words is "one hundred thirty-six trillion, three hundred fifty-four billion, three hundred fifty million, thirty-four thousand, three".
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