Base | Representation |
---|---|
bin | 10010111101111001111… |
… | …001001001010011101001 |
3 | 11121121100120220020211201 |
4 | 102331321321021103221 |
5 | 132323401022123103 |
6 | 2434441040324201 |
7 | 163111652111563 |
oct | 22757171112351 |
9 | 4547316806751 |
10 | 1303420114153 |
11 | 462860746613 |
12 | 190740913661 |
13 | 95bb1843621 |
14 | 4712b8b0333 |
15 | 23d8930251d |
hex | 12f79e494e9 |
1303420114153 has 2 divisors, whose sum is σ = 1303420114154. Its totient is φ = 1303420114152.
The previous prime is 1303420114099. The next prime is 1303420114159. The reversal of 1303420114153 is 3514110243031.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 1283423891689 + 19996222464 = 1132883^2 + 141408^2 .
It is a cyclic number.
It is not a de Polignac number, because 1303420114153 - 229 = 1302883243241 is a prime.
It is not a weakly prime, because it can be changed into another prime (1303420114159) 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, 651710057076 + 651710057077.
It is an arithmetic number, because the mean of its divisors is an integer number (651710057077).
Almost surely, 21303420114153 is an apocalyptic number.
It is an amenable number.
1303420114153 is a deficient number, since it is larger than the sum of its proper divisors (1).
1303420114153 is an equidigital number, since it uses as much as digits as its factorization.
1303420114153 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 4320, while the sum is 28.
Adding to 1303420114153 its reverse (3514110243031), we get a palindrome (4817530357184).
The spelling of 1303420114153 in words is "one trillion, three hundred three billion, four hundred twenty million, one hundred fourteen thousand, one hundred fifty-three".
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