Base | Representation |
---|---|
bin | 111110101110000001010… |
… | …101001100000010111001 |
3 | 120021000220211210202010010 |
4 | 332232001111030002321 |
5 | 1031103411113404423 |
6 | 13055555145412133 |
7 | 623250235432545 |
oct | 76560125140271 |
9 | 16230824722103 |
10 | 4310022013113 |
11 | 1411963078733 |
12 | 59738979a049 |
13 | 2535830512cb |
14 | 10c86b4d4a25 |
15 | 771a8424693 |
hex | 3eb8154c0b9 |
4310022013113 has 16 divisors (see below), whose sum is σ = 5771144171520. Its totient is φ = 2861140377984.
The previous prime is 4310022013091. The next prime is 4310022013123. The reversal of 4310022013113 is 3113102200134.
It is not a de Polignac number, because 4310022013113 - 210 = 4310022012089 is a prime.
It is a super-3 number, since 3×43100220131133 (a number of 39 digits) contains 333 as substring.
It is a self number, because there is not a number n which added to its sum of digits gives 4310022013113.
It is not an unprimeable number, because it can be changed into a prime (4310022013123) by changing a digit.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 1006393 + ... + 3103686.
It is an arithmetic number, because the mean of its divisors is an integer number (360696510720).
Almost surely, 24310022013113 is an apocalyptic number.
It is an amenable number.
4310022013113 is a deficient number, since it is larger than the sum of its proper divisors (1461122158407).
4310022013113 is a wasteful number, since it uses less digits than its factorization.
4310022013113 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 4111568.
The product of its (nonzero) digits is 432, while the sum is 21.
Adding to 4310022013113 its reverse (3113102200134), we get a palindrome (7423124213247).
The spelling of 4310022013113 in words is "four trillion, three hundred ten billion, twenty-two million, thirteen thousand, one hundred thirteen".
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