Base | Representation |
---|---|
bin | 1011110100110100010101… |
… | …1111111101011110001011 |
3 | 1201000222111222002101112121 |
4 | 2331031011133331132023 |
5 | 3201011130143324042 |
6 | 43353015005352111 |
7 | 2511236321406214 |
oct | 275150537753613 |
9 | 51028458071477 |
10 | 13002032011147 |
11 | 4163149120853 |
12 | 155ba6a268637 |
13 | 73411795cc8a |
14 | 32d431c5440b |
15 | 17832cc6e367 |
hex | bd3457fd78b |
13002032011147 has 2 divisors, whose sum is σ = 13002032011148. Its totient is φ = 13002032011146.
The previous prime is 13002032011141. The next prime is 13002032011247. The reversal of 13002032011147 is 74111023020031.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 13002032011147 - 27 = 13002032011019 is a prime.
It is not a weakly prime, because it can be changed into another prime (13002032011141) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6501016005573 + 6501016005574.
It is an arithmetic number, because the mean of its divisors is an integer number (6501016005574).
Almost surely, 213002032011147 is an apocalyptic number.
13002032011147 is a deficient number, since it is larger than the sum of its proper divisors (1).
13002032011147 is an equidigital number, since it uses as much as digits as its factorization.
13002032011147 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1008, while the sum is 25.
Adding to 13002032011147 its reverse (74111023020031), we get a palindrome (87113055031178).
The spelling of 13002032011147 in words is "thirteen trillion, two billion, thirty-two million, eleven thousand, one hundred forty-seven".
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