Base | Representation |
---|---|
bin | 110001010111000110… |
… | …110001100010111111 |
3 | 12001210201201211021122 |
4 | 301113012301202333 |
5 | 1332021224014101 |
6 | 40203123221155 |
7 | 3554262035150 |
oct | 612706614277 |
9 | 161721654248 |
10 | 53001001151 |
11 | 2052875769a |
12 | a331b177bb |
13 | 4cc8714b57 |
14 | 27cb117927 |
15 | 15a308551b |
hex | c571b18bf |
53001001151 has 4 divisors (see below), whose sum is σ = 60572572752. Its totient is φ = 45429429552.
The previous prime is 53001001139. The next prime is 53001001187. The reversal of 53001001151 is 15110010035.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is not a de Polignac number, because 53001001151 - 234 = 35821131967 is a prime.
It is a super-2 number, since 2×530010011512 (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 (53001003151) 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, 3785785790 + ... + 3785785803.
It is an arithmetic number, because the mean of its divisors is an integer number (15143143188).
Almost surely, 253001001151 is an apocalyptic number.
53001001151 is a deficient number, since it is larger than the sum of its proper divisors (7571571601).
53001001151 is an equidigital number, since it uses as much as digits as its factorization.
53001001151 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 7571571600.
The product of its (nonzero) digits is 75, while the sum is 17.
Adding to 53001001151 its reverse (15110010035), we get a palindrome (68111011186).
The spelling of 53001001151 in words is "fifty-three billion, one million, one thousand, one hundred fifty-one".
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