Base | Representation |
---|---|
bin | 110111110010001011111010… |
… | …011010000010110100111001 |
3 | 1012011200101211122010210210012 |
4 | 313302023322122002310321 |
5 | 224124133012142412213 |
6 | 2225444125311031305 |
7 | 102451216461030023 |
oct | 6762137232026471 |
9 | 1164611748123705 |
10 | 245341323013433 |
11 | 7119a741895849 |
12 | 2362496080b535 |
13 | a6b878393392a |
14 | 4482824b87013 |
15 | 1d56d568ac4a8 |
hex | df22fa682d39 |
245341323013433 has 2 divisors, whose sum is σ = 245341323013434. Its totient is φ = 245341323013432.
The previous prime is 245341323013337. The next prime is 245341323013441. The reversal of 245341323013433 is 334310323143542.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 160464846870784 + 84876476142649 = 12667472^2 + 9212843^2 .
It is a cyclic number.
It is not a de Polignac number, because 245341323013433 - 28 = 245341323013177 is a prime.
It is not a weakly prime, because it can be changed into another prime (245341323013733) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 122670661506716 + 122670661506717.
It is an arithmetic number, because the mean of its divisors is an integer number (122670661506717).
Almost surely, 2245341323013433 is an apocalyptic number.
It is an amenable number.
245341323013433 is a deficient number, since it is larger than the sum of its proper divisors (1).
245341323013433 is an equidigital number, since it uses as much as digits as its factorization.
245341323013433 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 933120, while the sum is 41.
Adding to 245341323013433 its reverse (334310323143542), we get a palindrome (579651646156975).
The spelling of 245341323013433 in words is "two hundred forty-five trillion, three hundred forty-one billion, three hundred twenty-three million, thirteen thousand, four hundred thirty-three".
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