Base | Representation |
---|---|
bin | 100100000001000010011… |
… | …0100001010111011101011 |
3 | 122112012220010212021112222 |
4 | 1020002010310022323223 |
5 | 1122100130313000041 |
6 | 14310002422424255 |
7 | 1020425241550064 |
oct | 110020464127353 |
9 | 18465803767488 |
10 | 4950030593771 |
11 | 163932925001a |
12 | 67b42329208b |
13 | 29ba2a776883 |
14 | 13182313b56b |
15 | 88b658c174b |
hex | 48084d0aeeb |
4950030593771 has 2 divisors, whose sum is σ = 4950030593772. Its totient is φ = 4950030593770.
The previous prime is 4950030593737. The next prime is 4950030593839. The reversal of 4950030593771 is 1773950300594.
It is a weak prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-4950030593771 is a prime.
It is not a weakly prime, because it can be changed into another prime (4950030593731) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2475015296885 + 2475015296886.
It is an arithmetic number, because the mean of its divisors is an integer number (2475015296886).
Almost surely, 24950030593771 is an apocalyptic number.
4950030593771 is a deficient number, since it is larger than the sum of its proper divisors (1).
4950030593771 is an equidigital number, since it uses as much as digits as its factorization.
4950030593771 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 3572100, while the sum is 53.
The spelling of 4950030593771 in words is "four trillion, nine hundred fifty billion, thirty million, five hundred ninety-three thousand, seven hundred seventy-one".
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