Base | Representation |
---|---|
bin | 10110110010011111000000… |
… | …01001111000111000101111 |
3 | 20120102202210011210122120122 |
4 | 23121033200021320320233 |
5 | 23032023204344143434 |
6 | 254325354404435155 |
7 | 13361361266601446 |
oct | 1331174011707057 |
9 | 216382704718518 |
10 | 50113144131119 |
11 | 14a70957462473 |
12 | 575431a6a8abb |
13 | 21c6859429ba4 |
14 | c536bc1d1d5d |
15 | 5bd857cbbc2e |
hex | 2d93e0278e2f |
50113144131119 has 2 divisors, whose sum is σ = 50113144131120. Its totient is φ = 50113144131118.
The previous prime is 50113144131077. The next prime is 50113144131161. The reversal of 50113144131119 is 91113144131105.
It is a balanced prime because it is at equal distance from previous prime (50113144131077) and next prime (50113144131161).
It is a cyclic number.
It is not a de Polignac number, because 50113144131119 - 216 = 50113144065583 is a prime.
It is a super-2 number, since 2×501131441311192 (a number of 28 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (50113144181119) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 25056572065559 + 25056572065560.
It is an arithmetic number, because the mean of its divisors is an integer number (25056572065560).
Almost surely, 250113144131119 is an apocalyptic number.
50113144131119 is a deficient number, since it is larger than the sum of its proper divisors (1).
50113144131119 is an equidigital number, since it uses as much as digits as its factorization.
50113144131119 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 6480, while the sum is 35.
The spelling of 50113144131119 in words is "fifty trillion, one hundred thirteen billion, one hundred forty-four million, one hundred thirty-one thousand, one hundred nineteen".
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