Base | Representation |
---|---|
bin | 11000101100101100011010… |
… | …10111100011110011100001 |
3 | 21010022012100220202000201011 |
4 | 30112112031113203303201 |
5 | 24104322423201040340 |
6 | 311302410120540521 |
7 | 14303633525134141 |
oct | 1426261527436341 |
9 | 233265326660634 |
10 | 54312233221345 |
11 | 1633a763811240 |
12 | 61120a9021741 |
13 | 243c8108cb408 |
14 | d5aa24656921 |
15 | 642bbc2670ea |
hex | 31658d5e3ce1 |
54312233221345 has 16 divisors (see below), whose sum is σ = 71999646347520. Its totient is φ = 38999808432000.
The previous prime is 54312233221319. The next prime is 54312233221369.
54312233221345 is nontrivially palindromic in base 10.
It is not a de Polignac number, because 54312233221345 - 25 = 54312233221313 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 54312233221298 and 54312233221307.
It is an unprimeable number.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 6249964956 + ... + 6249973645.
It is an arithmetic number, because the mean of its divisors is an integer number (4499977896720).
Almost surely, 254312233221345 is an apocalyptic number.
54312233221345 is a gapful number since it is divisible by the number (55) formed by its first and last digit.
It is an amenable number.
54312233221345 is a deficient number, since it is larger than the sum of its proper divisors (17687413126175).
54312233221345 is a wasteful number, since it uses less digits than its factorization.
54312233221345 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 12499938696.
The product of its digits is 518400, while the sum is 40.
It can be divided in two parts, 5431223 and 3221345, that added together give a palindrome (8652568).
The spelling of 54312233221345 in words is "fifty-four trillion, three hundred twelve billion, two hundred thirty-three million, two hundred twenty-one thousand, three hundred forty-five".
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