Base | Representation |
---|---|
bin | 10011000101111111110… |
… | …111010001111010011001 |
3 | 11122102210012212112002222 |
4 | 103011333313101322121 |
5 | 132444200204114001 |
6 | 2442435231433425 |
7 | 163540212606104 |
oct | 23057767217231 |
9 | 4572705775088 |
10 | 1312110223001 |
11 | 466510017a56 |
12 | 192367085875 |
13 | 96968041976 |
14 | 47713a9973b |
15 | 241e71b961b |
hex | 1317fdd1e99 |
1312110223001 has 2 divisors, whose sum is σ = 1312110223002. Its totient is φ = 1312110223000.
The previous prime is 1312110222977. The next prime is 1312110223003. The reversal of 1312110223001 is 1003220112131.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 777491880025 + 534618342976 = 881755^2 + 731176^2 .
It is a cyclic number.
It is not a de Polignac number, because 1312110223001 - 234 = 1294930353817 is a prime.
Together with 1312110223003, it forms a pair of twin primes.
It is a Chen prime.
It is not a weakly prime, because it can be changed into another prime (1312110223003) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 656055111500 + 656055111501.
It is an arithmetic number, because the mean of its divisors is an integer number (656055111501).
Almost surely, 21312110223001 is an apocalyptic number.
It is an amenable number.
1312110223001 is a deficient number, since it is larger than the sum of its proper divisors (1).
1312110223001 is an equidigital number, since it uses as much as digits as its factorization.
1312110223001 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 72, while the sum is 17.
Adding to 1312110223001 its reverse (1003220112131), we get a palindrome (2315330335132).
The spelling of 1312110223001 in words is "one trillion, three hundred twelve billion, one hundred ten million, two hundred twenty-three thousand, one".
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