Base | Representation |
---|---|
bin | 10011101100111011111… |
… | …101111001100110111011 |
3 | 11210102201001210220220011 |
4 | 103230323331321212323 |
5 | 134140312204440411 |
6 | 2513552125042351 |
7 | 166550263515631 |
oct | 23547375714673 |
9 | 4712631726804 |
10 | 1353920780731 |
11 | 482215a908a2 |
12 | 19a4953b63b7 |
13 | 9a89c23c95c |
14 | 4975c8cb351 |
15 | 25342ae2c21 |
hex | 13b3bf799bb |
1353920780731 has 2 divisors, whose sum is σ = 1353920780732. Its totient is φ = 1353920780730.
The previous prime is 1353920780713. The next prime is 1353920780783. The reversal of 1353920780731 is 1370870293531.
It is a happy number.
Together with previous prime (1353920780713) it forms an Ormiston pair, because they use the same digits, order apart.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 1353920780731 - 25 = 1353920780699 is a prime.
It is a super-2 number, since 2×13539207807312 (a number of 25 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (1353920780131) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 676960390365 + 676960390366.
It is an arithmetic number, because the mean of its divisors is an integer number (676960390366).
Almost surely, 21353920780731 is an apocalyptic number.
1353920780731 is a deficient number, since it is larger than the sum of its proper divisors (1).
1353920780731 is an equidigital number, since it uses as much as digits as its factorization.
1353920780731 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 952560, while the sum is 49.
The spelling of 1353920780731 in words is "one trillion, three hundred fifty-three billion, nine hundred twenty million, seven hundred eighty thousand, seven hundred thirty-one".
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