Base | Representation |
---|---|
bin | 101111100111101011… |
… | …000000001011101101 |
3 | 11212222102211210100212 |
4 | 233213223000023231 |
5 | 1314204122400333 |
6 | 35253420232205 |
7 | 3460041104033 |
oct | 574753001355 |
9 | 155872753325 |
10 | 51131450093 |
11 | 1a759409279 |
12 | 9aab99b665 |
13 | 4a8b2b2719 |
14 | 2690ad8553 |
15 | 14e3d8e448 |
hex | be7ac02ed |
51131450093 has 2 divisors, whose sum is σ = 51131450094. Its totient is φ = 51131450092.
The previous prime is 51131449991. The next prime is 51131450177. The reversal of 51131450093 is 39005413115.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 46477754569 + 4653695524 = 215587^2 + 68218^2 .
It is a cyclic number.
It is not a de Polignac number, because 51131450093 - 222 = 51127255789 is a prime.
It is a super-2 number, since 2×511314500932 (a number of 22 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (51131450893) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 25565725046 + 25565725047.
It is an arithmetic number, because the mean of its divisors is an integer number (25565725047).
Almost surely, 251131450093 is an apocalyptic number.
It is an amenable number.
51131450093 is a deficient number, since it is larger than the sum of its proper divisors (1).
51131450093 is an equidigital number, since it uses as much as digits as its factorization.
51131450093 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 8100, while the sum is 32.
The spelling of 51131450093 in words is "fifty-one billion, one hundred thirty-one million, four hundred fifty thousand, ninety-three".
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