Base | Representation |
---|---|
bin | 11110100001000001111100… |
… | …110110001010110111101011 |
3 | 122121012110122120011221012021 |
4 | 132201001330312022313223 |
5 | 120042404101023443431 |
6 | 1153235442240325311 |
7 | 40161304313425606 |
oct | 3641017466126753 |
9 | 577173576157167 |
10 | 134211232640491 |
11 | 39844733a45056 |
12 | 13077052313837 |
13 | 59b70ac86c506 |
14 | 251dc0332b33d |
15 | 107b21ebbcb11 |
hex | 7a107cd8adeb |
134211232640491 has 2 divisors, whose sum is σ = 134211232640492. Its totient is φ = 134211232640490.
The previous prime is 134211232640477. The next prime is 134211232640501. The reversal of 134211232640491 is 194046232112431.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 134211232640491 - 27 = 134211232640363 is a prime.
It is a super-4 number, since 4×1342112326404914 (a number of 58 digits) contains 4444 as substring.
It is not a weakly prime, because it can be changed into another prime (134211232440491) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 67105616320245 + 67105616320246.
It is an arithmetic number, because the mean of its divisors is an integer number (67105616320246).
Almost surely, 2134211232640491 is an apocalyptic number.
134211232640491 is a deficient number, since it is larger than the sum of its proper divisors (1).
134211232640491 is an equidigital number, since it uses as much as digits as its factorization.
134211232640491 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 248832, while the sum is 43.
The spelling of 134211232640491 in words is "one hundred thirty-four trillion, two hundred eleven billion, two hundred thirty-two million, six hundred forty thousand, four hundred ninety-one".
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