Base | Representation |
---|---|
bin | 1100111011010001010100… |
… | …1100101100010111010010 |
3 | 1212022200200202120102202121 |
4 | 3032310111030230113102 |
5 | 3330324000010301232 |
6 | 50121030524204454 |
7 | 2664545403260605 |
oct | 316642514542722 |
9 | 55280622512677 |
10 | 14212402431442 |
11 | 458a49a724126 |
12 | 171656099972a |
13 | 7c12ba8bab47 |
14 | 371c537a633c |
15 | 199a6d0d6d97 |
hex | ced1532c5d2 |
14212402431442 has 4 divisors (see below), whose sum is σ = 21318603647166. Its totient is φ = 7106201215720.
The previous prime is 14212402431349. The next prime is 14212402431509. The reversal of 14212402431442 is 24413420421241.
14212402431442 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a semiprime because it is the product of two primes.
It can be written as a sum of positive squares in only one way, i.e., 12261265588881 + 1951136842561 = 3501609^2 + 1396831^2 .
It is a junction number, because it is equal to n+sod(n) for n = 14212402431398 and 14212402431407.
It is an unprimeable number.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 3553100607859 + ... + 3553100607862.
Almost surely, 214212402431442 is an apocalyptic number.
14212402431442 is a deficient number, since it is larger than the sum of its proper divisors (7106201215724).
14212402431442 is an equidigital number, since it uses as much as digits as its factorization.
14212402431442 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 7106201215723.
The product of its (nonzero) digits is 49152, while the sum is 34.
Adding to 14212402431442 its reverse (24413420421241), we get a palindrome (38625822852683).
The spelling of 14212402431442 in words is "fourteen trillion, two hundred twelve billion, four hundred two million, four hundred thirty-one thousand, four hundred forty-two".
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