Base | Representation |
---|---|
bin | 10010000010111100010… |
… | …100101101000001011011 |
3 | 11101112221101222120210002 |
4 | 102002330110231001123 |
5 | 130304221143001042 |
6 | 2345410531443215 |
7 | 155411041066265 |
oct | 22027424550133 |
9 | 4345841876702 |
10 | 1240110125147 |
11 | 438a22921853 |
12 | 18041245650b |
13 | 8cc3241757a |
14 | 440435a0b35 |
15 | 223d11c7132 |
hex | 120bc52d05b |
1240110125147 has 2 divisors, whose sum is σ = 1240110125148. Its totient is φ = 1240110125146.
The previous prime is 1240110125131. The next prime is 1240110125207. The reversal of 1240110125147 is 7415210110421.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 1240110125147 - 24 = 1240110125131 is a prime.
It is not a weakly prime, because it can be changed into another prime (1240110125447) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 620055062573 + 620055062574.
It is an arithmetic number, because the mean of its divisors is an integer number (620055062574).
Almost surely, 21240110125147 is an apocalyptic number.
1240110125147 is a deficient number, since it is larger than the sum of its proper divisors (1).
1240110125147 is an equidigital number, since it uses as much as digits as its factorization.
1240110125147 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 2240, while the sum is 29.
Adding to 1240110125147 its reverse (7415210110421), we get a palindrome (8655320235568).
The spelling of 1240110125147 in words is "one trillion, two hundred forty billion, one hundred ten million, one hundred twenty-five thousand, one hundred forty-seven".
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