Base | Representation |
---|---|
bin | 100110101010100100111… |
… | …0101010000101101111111 |
3 | 200211000122200121020100012 |
4 | 1031111021311100231333 |
5 | 1144031300442302111 |
6 | 15145134145253435 |
7 | 1055634452165534 |
oct | 115251165205577 |
9 | 20730580536305 |
10 | 5314113244031 |
11 | 17697804299a9 |
12 | 719ab193427b |
13 | 2c7170b01837 |
14 | 1452c103cd8b |
15 | 93373e3248b |
hex | 4d549d50b7f |
5314113244031 has 2 divisors, whose sum is σ = 5314113244032. Its totient is φ = 5314113244030.
The previous prime is 5314113243991. The next prime is 5314113244049. The reversal of 5314113244031 is 1304423114135.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-5314113244031 is a prime.
It is a super-3 number, since 3×53141132440313 (a number of 39 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (5314113244051) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2657056622015 + 2657056622016.
It is an arithmetic number, because the mean of its divisors is an integer number (2657056622016).
Almost surely, 25314113244031 is an apocalyptic number.
5314113244031 is a deficient number, since it is larger than the sum of its proper divisors (1).
5314113244031 is an equidigital number, since it uses as much as digits as its factorization.
5314113244031 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 17280, while the sum is 32.
Adding to 5314113244031 its reverse (1304423114135), we get a palindrome (6618536358166).
The spelling of 5314113244031 in words is "five trillion, three hundred fourteen billion, one hundred thirteen million, two hundred forty-four thousand, thirty-one".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.076 sec. • engine limits •