Base | Representation |
---|---|
bin | 1100110101011000010110… |
… | …1011110001100010101011 |
3 | 1211222000111111221020112012 |
4 | 3031112011223301202223 |
5 | 3322144224331142303 |
6 | 50002333420110135 |
7 | 2654333643303422 |
oct | 315260553614253 |
9 | 54860444836465 |
10 | 14111210412203 |
11 | 4550592443363 |
12 | 16baa1b99534b |
13 | 7b48b214c698 |
14 | 36adb42d22b9 |
15 | 1970e93db5d8 |
hex | cd585af18ab |
14111210412203 has 2 divisors, whose sum is σ = 14111210412204. Its totient is φ = 14111210412202.
The previous prime is 14111210412199. The next prime is 14111210412209. The reversal of 14111210412203 is 30221401211141.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 14111210412203 - 22 = 14111210412199 is a prime.
It is a super-2 number, since 2×141112104122032 (a number of 27 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (14111210412209) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7055605206101 + 7055605206102.
It is an arithmetic number, because the mean of its divisors is an integer number (7055605206102).
Almost surely, 214111210412203 is an apocalyptic number.
14111210412203 is a deficient number, since it is larger than the sum of its proper divisors (1).
14111210412203 is an equidigital number, since it uses as much as digits as its factorization.
14111210412203 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 384, while the sum is 23.
Adding to 14111210412203 its reverse (30221401211141), we get a palindrome (44332611623344).
The spelling of 14111210412203 in words is "fourteen trillion, one hundred eleven billion, two hundred ten million, four hundred twelve thousand, two hundred three".
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