Base | Representation |
---|---|
bin | 11110011110101111011111… |
… | …101101100000111010110111 |
3 | 122120122102202011221102011111 |
4 | 132132233133231200322313 |
5 | 120032320021221312233 |
6 | 1153035314101415451 |
7 | 40144036350132106 |
oct | 3636573755407267 |
9 | 576572664842144 |
10 | 134053977525943 |
11 | 39793a77595a72 |
12 | 13050685a0a587 |
13 | 59a530b1a3787 |
14 | 25163661b203d |
15 | 10770b91ceacd |
hex | 79ebdfb60eb7 |
134053977525943 has 2 divisors, whose sum is σ = 134053977525944. Its totient is φ = 134053977525942.
The previous prime is 134053977525851. The next prime is 134053977526061. The reversal of 134053977525943 is 349525779350431.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 134053977525943 - 229 = 134053440655031 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (134053977523943) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 67026988762971 + 67026988762972.
It is an arithmetic number, because the mean of its divisors is an integer number (67026988762972).
Almost surely, 2134053977525943 is an apocalyptic number.
134053977525943 is a deficient number, since it is larger than the sum of its proper divisors (1).
134053977525943 is an equidigital number, since it uses as much as digits as its factorization.
134053977525943 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 428652000, while the sum is 67.
The spelling of 134053977525943 in words is "one hundred thirty-four trillion, fifty-three billion, nine hundred seventy-seven million, five hundred twenty-five thousand, nine hundred forty-three".
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