Base | Representation |
---|---|
bin | 100011000111101010000… |
… | …011010111100000110111 |
3 | 22112201102020021201010222 |
4 | 203013222003113200313 |
5 | 304020122232414333 |
6 | 5044411402404555 |
7 | 336235246314461 |
oct | 43075203274067 |
9 | 8481366251128 |
10 | 2413403404343 |
11 | 850578339258 |
12 | 32b897b0075b |
13 | 146776a083a9 |
14 | 84b48bcb131 |
15 | 42ba1460298 |
hex | 231ea0d7837 |
2413403404343 has 2 divisors, whose sum is σ = 2413403404344. Its totient is φ = 2413403404342.
The previous prime is 2413403404337. The next prime is 2413403404427. The reversal of 2413403404343 is 3434043043142.
2413403404343 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 2413403404343 - 234 = 2396223535159 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (2413403404843) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1206701702171 + 1206701702172.
It is an arithmetic number, because the mean of its divisors is an integer number (1206701702172).
Almost surely, 22413403404343 is an apocalyptic number.
2413403404343 is a deficient number, since it is larger than the sum of its proper divisors (1).
2413403404343 is an equidigital number, since it uses as much as digits as its factorization.
2413403404343 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 165888, while the sum is 35.
Adding to 2413403404343 its reverse (3434043043142), we get a palindrome (5847446447485).
The spelling of 2413403404343 in words is "two trillion, four hundred thirteen billion, four hundred three million, four hundred four thousand, three hundred forty-three".
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