Base | Representation |
---|---|
bin | 11110100100100000010110… |
… | …111100001100111010111111 |
3 | 122122001022001112110100112121 |
4 | 132210200112330030322333 |
5 | 120110312141011124203 |
6 | 1153541255102543411 |
7 | 40214460233110255 |
oct | 3644402674147277 |
9 | 578038045410477 |
10 | 134450041114303 |
11 | 39926a40116002 |
12 | 130b539a897b67 |
13 | 5a0377a125755 |
14 | 252b5b96545d5 |
15 | 108254a2140bd |
hex | 7a4816f0cebf |
134450041114303 has 2 divisors, whose sum is σ = 134450041114304. Its totient is φ = 134450041114302.
The previous prime is 134450041114289. The next prime is 134450041114307. The reversal of 134450041114303 is 303411140054431.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 134450041114303 - 217 = 134450040983231 is a prime.
It is a super-4 number, since 4×1344500411143034 (a number of 58 digits) contains 4444 as substring. Note that it is a super-d number also for d = 2.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (134450041114307) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 67225020557151 + 67225020557152.
It is an arithmetic number, because the mean of its divisors is an integer number (67225020557152).
Almost surely, 2134450041114303 is an apocalyptic number.
134450041114303 is a deficient number, since it is larger than the sum of its proper divisors (1).
134450041114303 is an equidigital number, since it uses as much as digits as its factorization.
134450041114303 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 34560, while the sum is 34.
Adding to 134450041114303 its reverse (303411140054431), we get a palindrome (437861181168734).
The spelling of 134450041114303 in words is "one hundred thirty-four trillion, four hundred fifty billion, forty-one million, one hundred fourteen thousand, three hundred three".
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