Base | Representation |
---|---|
bin | 10010101101010110001011… |
… | …01000010100100010110111 |
3 | 12101122222212001022002120110 |
4 | 21112223011220110202313 |
5 | 20343021232141130201 |
6 | 223255405052034103 |
7 | 11444206151620464 |
oct | 1126530550244267 |
9 | 171588761262513 |
10 | 41140512442551 |
11 | 12121658819845 |
12 | 4745382462933 |
13 | 19c56c2645922 |
14 | a232d622466b |
15 | 4b525c1796d6 |
hex | 256ac5a148b7 |
41140512442551 has 4 divisors (see below), whose sum is σ = 54854016590072. Its totient is φ = 27427008295032.
The previous prime is 41140512442517. The next prime is 41140512442609. The reversal of 41140512442551 is 15524421504114.
It is a semiprime because it is the product of two primes.
It is not a de Polignac number, because 41140512442551 - 231 = 41138364958903 is a prime.
It is a Duffinian number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (41140512042551) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 6856752073756 + ... + 6856752073761.
It is an arithmetic number, because the mean of its divisors is an integer number (13713504147518).
Almost surely, 241140512442551 is an apocalyptic number.
41140512442551 is a deficient number, since it is larger than the sum of its proper divisors (13713504147521).
41140512442551 is a wasteful number, since it uses less digits than its factorization.
41140512442551 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 13713504147520.
The product of its (nonzero) digits is 128000, while the sum is 39.
Adding to 41140512442551 its reverse (15524421504114), we get a palindrome (56664933946665).
The spelling of 41140512442551 in words is "forty-one trillion, one hundred forty billion, five hundred twelve million, four hundred forty-two thousand, five hundred fifty-one".
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