Base | Representation |
---|---|
bin | 10011000110000111100… |
… | …111111000001011100011 |
3 | 11122110010021211100201202 |
4 | 103012013213320023203 |
5 | 132444432020400141 |
6 | 2442500201551415 |
7 | 163543353250115 |
oct | 23060747701343 |
9 | 4573107740652 |
10 | 1312240403171 |
11 | 466577552275 |
12 | 1923a27a556b |
13 | 96988cc1459 |
14 | 477270a53b5 |
15 | 2420383149b |
hex | 131879f82e3 |
1312240403171 has 2 divisors, whose sum is σ = 1312240403172. Its totient is φ = 1312240403170.
The previous prime is 1312240403147. The next prime is 1312240403237. The reversal of 1312240403171 is 1713040422131.
It is a weak prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1312240403171 is a prime.
It is not a weakly prime, because it can be changed into another prime (1312240403101) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 656120201585 + 656120201586.
It is an arithmetic number, because the mean of its divisors is an integer number (656120201586).
Almost surely, 21312240403171 is an apocalyptic number.
1312240403171 is a deficient number, since it is larger than the sum of its proper divisors (1).
1312240403171 is an equidigital number, since it uses as much as digits as its factorization.
1312240403171 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 4032, while the sum is 29.
The spelling of 1312240403171 in words is "one trillion, three hundred twelve billion, two hundred forty million, four hundred three thousand, one hundred seventy-one".
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