Base | Representation |
---|---|
bin | 1100001101001101001100… |
… | …1101101010000100111111 |
3 | 1202112001000001112221110111 |
4 | 3003103103031222010333 |
5 | 3224342222434434003 |
6 | 44313310534250451 |
7 | 2553431264452045 |
oct | 303232315520477 |
9 | 52461001487414 |
10 | 13421021405503 |
11 | 4304906775198 |
12 | 1609100ab6a27 |
13 | 76479b49462a |
14 | 34581bdd0395 |
15 | 1841a1809e6d |
hex | c34d336a13f |
13421021405503 has 2 divisors, whose sum is σ = 13421021405504. Its totient is φ = 13421021405502.
The previous prime is 13421021405479. The next prime is 13421021405573. The reversal of 13421021405503 is 30550412012431.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 13421021405503 - 25 = 13421021405471 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (13421021405573) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6710510702751 + 6710510702752.
It is an arithmetic number, because the mean of its divisors is an integer number (6710510702752).
Almost surely, 213421021405503 is an apocalyptic number.
13421021405503 is a deficient number, since it is larger than the sum of its proper divisors (1).
13421021405503 is an equidigital number, since it uses as much as digits as its factorization.
13421021405503 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 14400, while the sum is 31.
Adding to 13421021405503 its reverse (30550412012431), we get a palindrome (43971433417934).
The spelling of 13421021405503 in words is "thirteen trillion, four hundred twenty-one billion, twenty-one million, four hundred five thousand, five hundred three".
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