Base | Representation |
---|---|
bin | 100011111100100111010… |
… | …1001110010100101100001 |
3 | 122111022101011112010001120 |
4 | 1013321032221302211201 |
5 | 1121421202140441403 |
6 | 14301352110122453 |
7 | 1016640662555442 |
oct | 107711651624541 |
9 | 18438334463046 |
10 | 4940531968353 |
11 | 16352a4551173 |
12 | 6796121b0429 |
13 | 29ab75903067 |
14 | 1311a180b2c9 |
15 | 887aba4de53 |
hex | 47e4ea72961 |
4940531968353 has 4 divisors (see below), whose sum is σ = 6587375957808. Its totient is φ = 3293687978900.
The previous prime is 4940531968301. The next prime is 4940531968387. The reversal of 4940531968353 is 3538691350494.
It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.
It is a cyclic number.
It is not a de Polignac number, because 4940531968353 - 221 = 4940529871201 is a prime.
It is a super-2 number, since 2×49405319683532 (a number of 26 digits) contains 22 as substring.
It is not an unprimeable number, because it can be changed into a prime (4940531968153) 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, 823421994723 + ... + 823421994728.
It is an arithmetic number, because the mean of its divisors is an integer number (1646843989452).
Almost surely, 24940531968353 is an apocalyptic number.
It is an amenable number.
4940531968353 is a deficient number, since it is larger than the sum of its proper divisors (1646843989455).
4940531968353 is a wasteful number, since it uses less digits than its factorization.
4940531968353 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 1646843989454.
The product of its (nonzero) digits is 41990400, while the sum is 60.
The spelling of 4940531968353 in words is "four trillion, nine hundred forty billion, five hundred thirty-one million, nine hundred sixty-eight thousand, three hundred fifty-three".
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