Base | Representation |
---|---|
bin | 10011101100001110000… |
… | …010110111011000100011 |
3 | 11210100201101001222201211 |
4 | 103230032002313120203 |
5 | 134132222441033103 |
6 | 2513343443410551 |
7 | 166522220014024 |
oct | 23541602673043 |
9 | 4710641058654 |
10 | 1353150330403 |
11 | 4819600a3093 |
12 | 19a2bb383a57 |
13 | 9a7a8734aa1 |
14 | 496c8454c4b |
15 | 252ea15636d |
hex | 13b0e0b7623 |
1353150330403 has 16 divisors (see below), whose sum is σ = 1495268099136. Its totient is φ = 1217996144640.
The previous prime is 1353150330401. The next prime is 1353150330509. The reversal of 1353150330403 is 3040330513531.
It is a cyclic number.
It is not a de Polignac number, because 1353150330403 - 21 = 1353150330401 is a prime.
It is a super-2 number, since 2×13531503304032 (a number of 25 digits) contains 22 as substring.
It is not an unprimeable number, because it can be changed into a prime (1353150330401) 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, 2344675 + ... + 2864227.
It is an arithmetic number, because the mean of its divisors is an integer number (93454256196).
Almost surely, 21353150330403 is an apocalyptic number.
1353150330403 is a deficient number, since it is larger than the sum of its proper divisors (142117768733).
1353150330403 is a wasteful number, since it uses less digits than its factorization.
1353150330403 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 526254.
The product of its (nonzero) digits is 24300, while the sum is 31.
Adding to 1353150330403 its reverse (3040330513531), we get a palindrome (4393480843934).
The spelling of 1353150330403 in words is "one trillion, three hundred fifty-three billion, one hundred fifty million, three hundred thirty thousand, four hundred three".
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