Base | Representation |
---|---|
bin | 10011101101000010101… |
… | …010111001111000110011 |
3 | 11210102222220102112021102 |
4 | 103231002222321320303 |
5 | 134141030002213101 |
6 | 2514011223315015 |
7 | 166553130440216 |
oct | 23550252717063 |
9 | 4712886375242 |
10 | 1354033241651 |
11 | 482273513402 |
12 | 19a506baba6b |
13 | 9a8b962718c |
14 | 4976d80377d |
15 | 2534c90986b |
hex | 13b42ab9e33 |
1354033241651 has 2 divisors, whose sum is σ = 1354033241652. Its totient is φ = 1354033241650.
The previous prime is 1354033241647. The next prime is 1354033241653. The reversal of 1354033241651 is 1561423304531.
It is a strong prime.
It is an emirp because it is prime and its reverse (1561423304531) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 1354033241651 - 22 = 1354033241647 is a prime.
It is a super-2 number, since 2×13540332416512 (a number of 25 digits) contains 22 as substring.
It is a Sophie Germain prime.
Together with 1354033241653, 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 (1354033241653) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 677016620825 + 677016620826.
It is an arithmetic number, because the mean of its divisors is an integer number (677016620826).
Almost surely, 21354033241651 is an apocalyptic number.
1354033241651 is a deficient number, since it is larger than the sum of its proper divisors (1).
1354033241651 is an equidigital number, since it uses as much as digits as its factorization.
1354033241651 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 129600, while the sum is 38.
The spelling of 1354033241651 in words is "one trillion, three hundred fifty-four billion, thirty-three million, two hundred forty-one thousand, six hundred fifty-one".
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