Base | Representation |
---|---|
bin | 110001011001000001… |
… | …010001011000110101 |
3 | 12001212222011021122121 |
4 | 301121001101120311 |
5 | 1332102434110031 |
6 | 40210231355541 |
7 | 3555132011131 |
oct | 613101213065 |
9 | 161788137577 |
10 | 53033113141 |
11 | 2054489a928 |
12 | a340822bb1 |
13 | 5002279228 |
14 | 27d14b63c1 |
15 | 15a5cca011 |
hex | c59051635 |
53033113141 has 4 divisors (see below), whose sum is σ = 53044127532. Its totient is φ = 53022098752.
The previous prime is 53033113099. The next prime is 53033113157. The reversal of 53033113141 is 14131133035.
It is a semiprime because it is the product of two primes.
It can be written as a sum of positive squares in 2 ways, for example, as 24978538116 + 28054575025 = 158046^2 + 167495^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-53033113141 is a prime.
It is a super-2 number, since 2×530331131412 (a number of 22 digits) contains 22 as substring.
It is a Duffinian number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (53033117141) 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, 5499970 + ... + 5509603.
It is an arithmetic number, because the mean of its divisors is an integer number (13261031883).
Almost surely, 253033113141 is an apocalyptic number.
It is an amenable number.
53033113141 is a deficient number, since it is larger than the sum of its proper divisors (11014391).
53033113141 is a wasteful number, since it uses less digits than its factorization.
53033113141 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 11014390.
The product of its (nonzero) digits is 1620, while the sum is 25.
Adding to 53033113141 its reverse (14131133035), we get a palindrome (67164246176).
The spelling of 53033113141 in words is "fifty-three billion, thirty-three million, one hundred thirteen thousand, one hundred forty-one".
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