Base | Representation |
---|---|
bin | 10011110011110001000… |
… | …100100100001000010001 |
3 | 11211010122010212202100211 |
4 | 103303301010210020101 |
5 | 134300322031434103 |
6 | 2521203533044121 |
7 | 200230101353632 |
oct | 23636104441021 |
9 | 4733563782324 |
10 | 1361254171153 |
11 | 485339545638 |
12 | 19b9a1326641 |
13 | 9b49a634926 |
14 | 49c5682d889 |
15 | 2562181436d |
hex | 13cf1124211 |
1361254171153 has 2 divisors, whose sum is σ = 1361254171154. Its totient is φ = 1361254171152.
The previous prime is 1361254171129. The next prime is 1361254171159. The reversal of 1361254171153 is 3511714521631.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 1129645871104 + 231608300049 = 1062848^2 + 481257^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1361254171153 is a prime.
It is not a weakly prime, because it can be changed into another prime (1361254171159) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 680627085576 + 680627085577.
It is an arithmetic number, because the mean of its divisors is an integer number (680627085577).
Almost surely, 21361254171153 is an apocalyptic number.
It is an amenable number.
1361254171153 is a deficient number, since it is larger than the sum of its proper divisors (1).
1361254171153 is an equidigital number, since it uses as much as digits as its factorization.
1361254171153 is an evil number, because the sum of its binary digits is even.
The product of its digits is 75600, while the sum is 40.
Adding to 1361254171153 its reverse (3511714521631), we get a palindrome (4872968692784).
The spelling of 1361254171153 in words is "one trillion, three hundred sixty-one billion, two hundred fifty-four million, one hundred seventy-one thousand, one hundred fifty-three".
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