Base | Representation |
---|---|
bin | 11000000110101000010100… |
… | …10101011101111111001111 |
3 | 20221200011112022120100111102 |
4 | 30003110022111131333033 |
5 | 23421410414342044301 |
6 | 304421511413413315 |
7 | 14110303400026415 |
oct | 1403241225357717 |
9 | 227604468510442 |
10 | 53004364799951 |
11 | 15986030121654 |
12 | 5b4072663a23b |
13 | 23763a0610036 |
14 | d135d40952b5 |
15 | 61db72687b6b |
hex | 30350a55dfcf |
53004364799951 has 2 divisors, whose sum is σ = 53004364799952. Its totient is φ = 53004364799950.
The previous prime is 53004364799849. The next prime is 53004364800097. The reversal of 53004364799951 is 15999746340035.
It is a weak prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-53004364799951 is a prime.
It is a super-4 number, since 4×530043647999514 (a number of 56 digits) contains 4444 as substring. Note that it is a super-d number also for d = 2.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (53004364999951) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 26502182399975 + 26502182399976.
It is an arithmetic number, because the mean of its divisors is an integer number (26502182399976).
Almost surely, 253004364799951 is an apocalyptic number.
53004364799951 is a deficient number, since it is larger than the sum of its proper divisors (1).
53004364799951 is an equidigital number, since it uses as much as digits as its factorization.
53004364799951 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 110224800, while the sum is 65.
The spelling of 53004364799951 in words is "fifty-three trillion, four billion, three hundred sixty-four million, seven hundred ninety-nine thousand, nine hundred fifty-one".
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