Base | Representation |
---|---|
bin | 10011101100011101011… |
… | …100111111001101111001 |
3 | 11210101101101111110101201 |
4 | 103230131130333031321 |
5 | 134133240130231343 |
6 | 2513425240214201 |
7 | 166531506204031 |
oct | 23543534771571 |
9 | 4711341443351 |
10 | 1353408836473 |
11 | 481a83006571 |
12 | 19a371a6a361 |
13 | 9a81a165065 |
14 | 497109068c1 |
15 | 25312bba94d |
hex | 13b1d73f379 |
1353408836473 has 2 divisors, whose sum is σ = 1353408836474. Its totient is φ = 1353408836472.
The previous prime is 1353408836471. The next prime is 1353408836489. The reversal of 1353408836473 is 3746388043531.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 1336366056169 + 17042780304 = 1156013^2 + 130548^2 .
It is a cyclic number.
It is not a de Polignac number, because 1353408836473 - 21 = 1353408836471 is a prime.
Together with 1353408836471, it forms a pair of twin primes.
It is not a weakly prime, because it can be changed into another prime (1353408836471) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 676704418236 + 676704418237.
It is an arithmetic number, because the mean of its divisors is an integer number (676704418237).
Almost surely, 21353408836473 is an apocalyptic number.
It is an amenable number.
1353408836473 is a deficient number, since it is larger than the sum of its proper divisors (1).
1353408836473 is an equidigital number, since it uses as much as digits as its factorization.
1353408836473 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 17418240, while the sum is 55.
The spelling of 1353408836473 in words is "one trillion, three hundred fifty-three billion, four hundred eight million, eight hundred thirty-six thousand, four hundred seventy-three".
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