Base | Representation |
---|---|
bin | 111111000100101110011… |
… | …100111001111101101111 |
3 | 120100100211212210102012200 |
4 | 333010232130321331233 |
5 | 1032003323220043210 |
6 | 13115110235032543 |
7 | 625102335265335 |
oct | 77045634717557 |
9 | 16310755712180 |
10 | 4334401331055 |
11 | 1421233610084 |
12 | 5a005229b753 |
13 | 255969aca223 |
14 | 10db0126db55 |
15 | 77b338910c0 |
hex | 3f12e739f6f |
4334401331055 has 72 divisors (see below), whose sum is σ = 8267174588400. Its totient is φ = 2094795115776.
The previous prime is 4334401331023. The next prime is 4334401331059. The reversal of 4334401331055 is 5501331044334.
4334401331055 is a `hidden beast` number, since 4 + 33 + 440 + 133 + 1 + 0 + 55 = 666.
It is not a de Polignac number, because 4334401331055 - 25 = 4334401331023 is a prime.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (4334401331059) by changing a digit.
It is a polite number, since it can be written in 71 ways as a sum of consecutive naturals, for example, 4339270 + ... + 5243859.
Almost surely, 24334401331055 is an apocalyptic number.
4334401331055 is a gapful number since it is divisible by the number (45) formed by its first and last digit.
4334401331055 is a deficient number, since it is larger than the sum of its proper divisors (3932773257345).
4334401331055 is a wasteful number, since it uses less digits than its factorization.
4334401331055 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 9583205 (or 9583179 counting only the distinct ones).
The product of its (nonzero) digits is 129600, while the sum is 36.
Adding to 4334401331055 its reverse (5501331044334), we get a palindrome (9835732375389).
The spelling of 4334401331055 in words is "four trillion, three hundred thirty-four billion, four hundred one million, three hundred thirty-one thousand, fifty-five".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.065 sec. • engine limits •