Base | Representation |
---|---|
bin | 1011111101101011011000… |
… | …1000010110000110100001 |
3 | 1201120112101222121212010121 |
4 | 2333122312020112012201 |
5 | 3211004343322020340 |
6 | 43550551005000241 |
7 | 2525235315623524 |
oct | 277326610260641 |
9 | 51515358555117 |
10 | 13154245501345 |
11 | 4211758722943 |
12 | 158546a253081 |
13 | 745594985140 |
14 | 336951594bbb |
15 | 17c28ad3294a |
hex | bf6b62161a1 |
13154245501345 has 64 divisors (see below), whose sum is σ = 17620256747520. Its totient is φ = 9364343040000.
The previous prime is 13154245501321. The next prime is 13154245501357. The reversal of 13154245501345 is 54310554245131.
13154245501345 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is not a de Polignac number, because 13154245501345 - 221 = 13154243404193 is a prime.
It is a super-3 number, since 3×131542455013453 (a number of 40 digits) contains 333 as substring.
It is an unprimeable number.
It is a polite number, since it can be written in 63 ways as a sum of consecutive naturals, for example, 115241061 + ... + 115355149.
It is an arithmetic number, because the mean of its divisors is an integer number (275316511680).
Almost surely, 213154245501345 is an apocalyptic number.
It is an amenable number.
13154245501345 is a deficient number, since it is larger than the sum of its proper divisors (4466011246175).
13154245501345 is a wasteful number, since it uses less digits than its factorization.
13154245501345 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 114586.
The product of its (nonzero) digits is 720000, while the sum is 43.
Adding to 13154245501345 its reverse (54310554245131), we get a palindrome (67464799746476).
The spelling of 13154245501345 in words is "thirteen trillion, one hundred fifty-four billion, two hundred forty-five million, five hundred one thousand, three hundred forty-five".
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