Base | Representation |
---|---|
bin | 1010000000010011000011… |
… | …1101000010000001011101 |
3 | 1102221121112010020110121122 |
4 | 2200010300331002001131 |
5 | 2420211434013103433 |
6 | 35221234223540325 |
7 | 2213512005631154 |
oct | 240046075020135 |
9 | 42847463213548 |
10 | 11000232550493 |
11 | 35611995545aa |
12 | 1297b0375a6a5 |
13 | 61a4186671b4 |
14 | 2a05b23cc59b |
15 | 14121bce3198 |
hex | a0130f4205d |
11000232550493 has 2 divisors, whose sum is σ = 11000232550494. Its totient is φ = 11000232550492.
The previous prime is 11000232550487. The next prime is 11000232550561. The reversal of 11000232550493 is 39405523200011.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 8405111310649 + 2595121239844 = 2899157^2 + 1610938^2 .
It is a cyclic number.
It is not a de Polignac number, because 11000232550493 - 224 = 11000215773277 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (11000232555493) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5500116275246 + 5500116275247.
It is an arithmetic number, because the mean of its divisors is an integer number (5500116275247).
Almost surely, 211000232550493 is an apocalyptic number.
It is an amenable number.
11000232550493 is a deficient number, since it is larger than the sum of its proper divisors (1).
11000232550493 is an equidigital number, since it uses as much as digits as its factorization.
11000232550493 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 32400, while the sum is 35.
The spelling of 11000232550493 in words is "eleven trillion, two hundred thirty-two million, five hundred fifty thousand, four hundred ninety-three".
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