Base | Representation |
---|---|
bin | 10011010110101011111… |
… | …110001011111010000111 |
3 | 11201011001002021121011220 |
4 | 103112223332023322013 |
5 | 133242400201200403 |
6 | 2455001332531423 |
7 | 165043246213545 |
oct | 23265376137207 |
9 | 4634032247156 |
10 | 1330030100103 |
11 | 473076354941 |
12 | 195928480b73 |
13 | 9856279ac43 |
14 | 485339cad95 |
15 | 248e54e7953 |
hex | 135abf8be87 |
1330030100103 has 4 divisors (see below), whose sum is σ = 1773373466808. Its totient is φ = 886686733400.
The previous prime is 1330030100093. The next prime is 1330030100107. The reversal of 1330030100103 is 3010010300331.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is not a de Polignac number, because 1330030100103 - 217 = 1330029969031 is a prime.
It is a super-2 number, since 2×13300301001032 (a number of 25 digits) contains 22 as substring.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (1330030100107) 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, 221671683348 + ... + 221671683353.
It is an arithmetic number, because the mean of its divisors is an integer number (443343366702).
Almost surely, 21330030100103 is an apocalyptic number.
1330030100103 is a deficient number, since it is larger than the sum of its proper divisors (443343366705).
1330030100103 is an equidigital number, since it uses as much as digits as its factorization.
1330030100103 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 443343366704.
The product of its (nonzero) digits is 81, while the sum is 15.
Adding to 1330030100103 its reverse (3010010300331), we get a palindrome (4340040400434).
The spelling of 1330030100103 in words is "one trillion, three hundred thirty billion, thirty million, one hundred thousand, one hundred three".
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