Base | Representation |
---|---|
bin | 101101110100100000… |
… | …100110100100111011 |
3 | 11200222210020121222101 |
4 | 231310200212210323 |
5 | 1301230013001211 |
6 | 34334000404231 |
7 | 3361130344333 |
oct | 556440464473 |
9 | 150883217871 |
10 | 49199343931 |
11 | 19957824321 |
12 | 9650916677 |
13 | 4840c1a33a |
14 | 254a2740c3 |
15 | 142e43d5c1 |
hex | b7482693b |
49199343931 has 2 divisors, whose sum is σ = 49199343932. Its totient is φ = 49199343930.
The previous prime is 49199343923. The next prime is 49199344001. The reversal of 49199343931 is 13934399194.
49199343931 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 49199343931 - 23 = 49199343923 is a prime.
It is a super-2 number, since 2×491993439312 (a number of 22 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (49199343331) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 24599671965 + 24599671966.
It is an arithmetic number, because the mean of its divisors is an integer number (24599671966).
Almost surely, 249199343931 is an apocalyptic number.
49199343931 is a deficient number, since it is larger than the sum of its proper divisors (1).
49199343931 is an equidigital number, since it uses as much as digits as its factorization.
49199343931 is an evil number, because the sum of its binary digits is even.
The product of its digits is 2834352, while the sum is 55.
The spelling of 49199343931 in words is "forty-nine billion, one hundred ninety-nine million, three hundred forty-three thousand, nine hundred thirty-one".
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