Base | Representation |
---|---|
bin | 10110011100010100000… |
… | …100111111101101111011 |
3 | 12110102202200001011100021 |
4 | 112130110010333231323 |
5 | 200231441402110042 |
6 | 3140253555351311 |
7 | 216264620444653 |
oct | 26342404775573 |
9 | 5412680034307 |
10 | 1542230113147 |
11 | 545068769449 |
12 | 20aa898b2537 |
13 | b257c511b39 |
14 | 549040b3363 |
15 | 2a1b4a4a867 |
hex | 1671413fb7b |
1542230113147 has 2 divisors, whose sum is σ = 1542230113148. Its totient is φ = 1542230113146.
The previous prime is 1542230113121. The next prime is 1542230113153. The reversal of 1542230113147 is 7413110322451.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1542230113147 is a prime.
It is not a weakly prime, because it can be changed into another prime (1542230113247) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 771115056573 + 771115056574.
It is an arithmetic number, because the mean of its divisors is an integer number (771115056574).
Almost surely, 21542230113147 is an apocalyptic number.
1542230113147 is a deficient number, since it is larger than the sum of its proper divisors (1).
1542230113147 is an equidigital number, since it uses as much as digits as its factorization.
1542230113147 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 20160, while the sum is 34.
Adding to 1542230113147 its reverse (7413110322451), we get a palindrome (8955340435598).
The spelling of 1542230113147 in words is "one trillion, five hundred forty-two billion, two hundred thirty million, one hundred thirteen thousand, one hundred forty-seven".
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