Base | Representation |
---|---|
bin | 100101001101000010100… |
… | …0111110110110001000111 |
3 | 200002211011210122100120120 |
4 | 1022122011013312301013 |
5 | 1132233411043330203 |
6 | 14512554143233023 |
7 | 1035263652140205 |
oct | 112320507666107 |
9 | 20084153570516 |
10 | 5113244511303 |
11 | 16a1573251a75 |
12 | 6a6b932a8773 |
13 | 2b123b815a36 |
14 | 1396a7820075 |
15 | 8d0195aab53 |
hex | 4a6851f6c47 |
5113244511303 has 4 divisors (see below), whose sum is σ = 6817659348408. Its totient is φ = 3408829674200.
The previous prime is 5113244511301. The next prime is 5113244511329. The reversal of 5113244511303 is 3031154423115.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is not a de Polignac number, because 5113244511303 - 21 = 5113244511301 is a prime.
It is a super-2 number, since 2×51132445113032 (a number of 26 digits) contains 22 as substring.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (5113244511301) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 852207418548 + ... + 852207418553.
It is an arithmetic number, because the mean of its divisors is an integer number (1704414837102).
Almost surely, 25113244511303 is an apocalyptic number.
5113244511303 is a deficient number, since it is larger than the sum of its proper divisors (1704414837105).
5113244511303 is a wasteful number, since it uses less digits than its factorization.
5113244511303 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 1704414837104.
The product of its (nonzero) digits is 21600, while the sum is 33.
Adding to 5113244511303 its reverse (3031154423115), we get a palindrome (8144398934418).
The spelling of 5113244511303 in words is "five trillion, one hundred thirteen billion, two hundred forty-four million, five hundred eleven thousand, three hundred three".
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