Base | Representation |
---|---|
bin | 10011100000000100111… |
… | …010001010101001001111 |
3 | 11202010001201102221221122 |
4 | 103200010322022221033 |
5 | 133424022202400132 |
6 | 2503350010151155 |
7 | 165551141232344 |
oct | 23400472125117 |
9 | 4663051387848 |
10 | 1340112153167 |
11 | 47737a4125a7 |
12 | 197880a224bb |
13 | 994ab487108 |
14 | 48c0c9cc7cb |
15 | 24cd56b2912 |
hex | 13804e8aa4f |
1340112153167 has 2 divisors, whose sum is σ = 1340112153168. Its totient is φ = 1340112153166.
The previous prime is 1340112153143. The next prime is 1340112153223. The reversal of 1340112153167 is 7613512110431.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 1340112153167 - 216 = 1340112087631 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (1340112153137) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 670056076583 + 670056076584.
It is an arithmetic number, because the mean of its divisors is an integer number (670056076584).
Almost surely, 21340112153167 is an apocalyptic number.
1340112153167 is a deficient number, since it is larger than the sum of its proper divisors (1).
1340112153167 is an equidigital number, since it uses as much as digits as its factorization.
1340112153167 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 15120, while the sum is 35.
Adding to 1340112153167 its reverse (7613512110431), we get a palindrome (8953624263598).
The spelling of 1340112153167 in words is "one trillion, three hundred forty billion, one hundred twelve million, one hundred fifty-three thousand, one hundred sixty-seven".
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