Base | Representation |
---|---|
bin | 110010010010010001… |
… | …111100010101111111 |
3 | 12011100220211110021211 |
4 | 302102101330111333 |
5 | 1341034402102303 |
6 | 40445430105251 |
7 | 3621005411635 |
oct | 622221742577 |
9 | 164326743254 |
10 | 53993784703 |
11 | 209980a1671 |
12 | a56a4ab227 |
13 | 51262c6441 |
14 | 2882d07355 |
15 | 16102dd66d |
hex | c9247c57f |
53993784703 has 2 divisors, whose sum is σ = 53993784704. Its totient is φ = 53993784702.
The previous prime is 53993784701. The next prime is 53993784737. The reversal of 53993784703 is 30748739935.
It is a happy number.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 53993784703 - 21 = 53993784701 is a prime.
It is a super-4 number, since 4×539937847034 (a number of 44 digits) contains 4444 as substring.
Together with 53993784701, it forms a pair of twin primes.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (53993784701) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 26996892351 + 26996892352.
It is an arithmetic number, because the mean of its divisors is an integer number (26996892352).
Almost surely, 253993784703 is an apocalyptic number.
53993784703 is a deficient number, since it is larger than the sum of its proper divisors (1).
53993784703 is an equidigital number, since it uses as much as digits as its factorization.
53993784703 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 17146080, while the sum is 58.
The spelling of 53993784703 in words is "fifty-three billion, nine hundred ninety-three million, seven hundred eighty-four thousand, seven hundred three".
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