Base | Representation |
---|---|
bin | 1011110101000001011100… |
… | …1101010101011111001110 |
3 | 1201001022121011110020002001 |
4 | 2331100113031111133032 |
5 | 3201040331342003032 |
6 | 43354400051422514 |
7 | 2511422442632626 |
oct | 275202715253716 |
9 | 51038534406061 |
10 | 13005550344142 |
11 | 416469412a474 |
12 | 15606905b4a3a |
13 | 734558848015 |
14 | 32d687225886 |
15 | 178486a9d3e7 |
hex | bd4173557ce |
13005550344142 has 4 divisors (see below), whose sum is σ = 19508325516216. Its totient is φ = 6502775172070.
The previous prime is 13005550344131. The next prime is 13005550344173. The reversal of 13005550344142 is 24144305550031.
It is a semiprime because it is the product of two primes.
It is a super-3 number, since 3×130055503441423 (a number of 40 digits) contains 333 as substring.
It is a congruent number.
It is an unprimeable number.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 3251387586034 + ... + 3251387586037.
It is an arithmetic number, because the mean of its divisors is an integer number (4877081379054).
Almost surely, 213005550344142 is an apocalyptic number.
13005550344142 is a deficient number, since it is larger than the sum of its proper divisors (6502775172074).
13005550344142 is an equidigital number, since it uses as much as digits as its factorization.
13005550344142 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 6502775172073.
The product of its (nonzero) digits is 144000, while the sum is 37.
Adding to 13005550344142 its reverse (24144305550031), we get a palindrome (37149855894173).
The spelling of 13005550344142 in words is "thirteen trillion, five billion, five hundred fifty million, three hundred forty-four thousand, one hundred forty-two".
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