Base | Representation |
---|---|
bin | 1100000000111011100010… |
… | …0101101110010110110011 |
3 | 1201202212122001212011022111 |
4 | 3000032320211232112303 |
5 | 3212413312030112042 |
6 | 44032351305550151 |
7 | 2532254056430365 |
oct | 300167045562663 |
9 | 51685561764274 |
10 | 13210121332147 |
11 | 4233421117603 |
12 | 1594262aa5357 |
13 | 74a92ab6710c |
14 | 339532410d35 |
15 | 17d95b484517 |
hex | c03b896e5b3 |
13210121332147 has 2 divisors, whose sum is σ = 13210121332148. Its totient is φ = 13210121332146.
The previous prime is 13210121332081. The next prime is 13210121332153. The reversal of 13210121332147 is 74123312101231.
It is a happy number.
13210121332147 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 13210121332147 - 211 = 13210121330099 is a prime.
It is a super-2 number, since 2×132101213321472 (a number of 27 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (13210121336147) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6605060666073 + 6605060666074.
It is an arithmetic number, because the mean of its divisors is an integer number (6605060666074).
Almost surely, 213210121332147 is an apocalyptic number.
13210121332147 is a deficient number, since it is larger than the sum of its proper divisors (1).
13210121332147 is an equidigital number, since it uses as much as digits as its factorization.
13210121332147 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 6048, while the sum is 31.
Adding to 13210121332147 its reverse (74123312101231), we get a palindrome (87333433433378).
The spelling of 13210121332147 in words is "thirteen trillion, two hundred ten billion, one hundred twenty-one million, three hundred thirty-two thousand, one hundred forty-seven".
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