Base | Representation |
---|---|
bin | 1111100000001… |
… | …00110000010101 |
3 | 100001200212022202 |
4 | 13300010300111 |
5 | 231242333101 |
6 | 20523134245 |
7 | 3136226552 |
oct | 760046025 |
9 | 301625282 |
10 | 130042901 |
11 | 6745117a |
12 | 37674385 |
13 | 20c32181 |
14 | 133b1a29 |
15 | b63b36b |
hex | 7c04c15 |
130042901 has 4 divisors (see below), whose sum is σ = 133557612. Its totient is φ = 126528192.
The previous prime is 130042841. The next prime is 130042907. The reversal of 130042901 is 109240031.
130042901 is nontrivially palindromic in base 15.
It is a semiprime because it is the product of two primes.
It can be written as a sum of positive squares in 2 ways, for example, as 56100100 + 73942801 = 7490^2 + 8599^2 .
It is a cyclic number.
It is not a de Polignac number, because 130042901 - 26 = 130042837 is a prime.
It is a super-2 number, since 2×1300429012 = 33822312200991602, which contains 22 as substring.
It is a Duffinian number.
It is a Curzon number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (130042907) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (11) of ones.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 1757300 + ... + 1757373.
It is an arithmetic number, because the mean of its divisors is an integer number (33389403).
Almost surely, 2130042901 is an apocalyptic number.
It is an amenable number.
130042901 is a deficient number, since it is larger than the sum of its proper divisors (3514711).
130042901 is an equidigital number, since it uses as much as digits as its factorization.
130042901 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 3514710.
The product of its (nonzero) digits is 216, while the sum is 20.
The square root of 130042901 is about 11403.6354291077. The cubic root of 130042901 is about 506.6354208474.
Adding to 130042901 its reverse (109240031), we get a palindrome (239282932).
The spelling of 130042901 in words is "one hundred thirty million, forty-two thousand, nine hundred one".
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