Base | Representation |
---|---|
bin | 10011000100001000101… |
… | …101110000111110110010 |
3 | 11122020121220102200002202 |
4 | 103010020231300332302 |
5 | 132431101434140424 |
6 | 2441505022343202 |
7 | 163436530541465 |
oct | 23041055607662 |
9 | 4566556380082 |
10 | 1310111240114 |
11 | 465684705586 |
12 | 191aa972ab02 |
13 | 96709b62561 |
14 | 475a43c72dc |
15 | 2412b9582ae |
hex | 13108b70fb2 |
1310111240114 has 4 divisors (see below), whose sum is σ = 1965166860174. Its totient is φ = 655055620056.
The previous prime is 1310111240101. The next prime is 1310111240131. The reversal of 1310111240114 is 4110421110131.
It is a semiprime because it is the product of two primes.
It can be written as a sum of positive squares in only one way, i.e., 1081947387889 + 228163852225 = 1040167^2 + 477665^2 .
It is a super-2 number, since 2×13101112401142 (a number of 25 digits) contains 22 as substring.
It is an unprimeable number.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 327527810027 + ... + 327527810030.
Almost surely, 21310111240114 is an apocalyptic number.
1310111240114 is a deficient number, since it is larger than the sum of its proper divisors (655055620060).
1310111240114 is an equidigital number, since it uses as much as digits as its factorization.
1310111240114 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 655055620059.
The product of its (nonzero) digits is 96, while the sum is 20.
Adding to 1310111240114 its reverse (4110421110131), we get a palindrome (5420532350245).
The spelling of 1310111240114 in words is "one trillion, three hundred ten billion, one hundred eleven million, two hundred forty thousand, one hundred fourteen".
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